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✓ 40 results · all verified · 16 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 24 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Extraspecial $p$ -Groups and Central Products

1 · Prerequisites

2 · Summary

Published notions of centre, commutator subgroup, quotient group, and semidirect product supply the ambient group theory, while the prerequisite Frattini page supplies Elementary abelian $p$ -groups, $Φ (P) = P^{'} P^{p}$ for a finite $p$ -group, The Frattini quotient is the largest elementary abelian quotient of a finite $p$ -group, and Burnside Basis Theorem. These are the inputs that turn the three standard descriptions of an extraspecial group into equivalent ones and that let the central quotient carry a genuine $F_{p}$ -linear structure.

The page defines special and extraspecial $p$ -groups, external and internal central products, the commutator pairing, the square map for the two-group case, and the plus/minus types. It then proves the class-two commutator formulas, the central-product universal property and recognition theorem, the decomposition of every extraspecial group into order- $p^{3}$ factors, the order and maximal-abelian consequences of that decomposition, the classification of the nonabelian groups of order $p^{3}$ , the odd- and two-primary classification theorems, the exponent dichotomy, and the two automorphism results that act on the Frattini quotient and on the commutator pairing.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Commutator identities in a group whose derived subgroup is central

Statement

Let $G$ be a group (Group and abelian group) whose derived subgroup $[G, G]$ is contained in the centre $Z (G)$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , The center $Z (G)$ of a group). If $[G, G] \leq Z (G)$ then $[x y, w] = [x, w] [y, w]$ , $[x, y w] = [x, y] [x, w]$ , and $[x^{n}, y] = [x, y]^{n} = [x, y^{n}]$ for every integer $n$ and all $x, y, w \in G$ , integer powers being those of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ .

Facts & Assumptions

Given: A group $G$ with $[G, G] \leq Z (G)$ , and elements $x, y, w \in G$ .

[F1]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ , and $[G, G] : = ⟨ {[g, h] : g, h \in G} ⟩$ is the subgroup generated by all commutators (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[L1]

For all $m, n \in Z$ one has $g^{m + n} = g^{m} g^{n}$ and $g^{- m} = (g^{m})^{- 1}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L2]

$(g h)^{- 1} = h^{- 1} g^{- 1}$ (In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential).

Proof

technique · direct

1.1F1F2given

Every commutator $[g, h]$ lies in $[G, G]$ , hence in $Z (G)$ , so it commutes with every element of $G$ and may be moved to any position in a product without changing that product.

2.1F1L2step 1.1algebra

Expanding, $[x y, w] = (x y) w (x y)^{- 1} w^{- 1} = x y w y^{- 1} x^{- 1} w^{- 1} = x (y w y^{- 1} w^{- 1}) w x^{- 1} w^{- 1} = x [y, w] w x^{- 1} w^{- 1} = [y, w] x w x^{- 1} w^{- 1} = [y, w] [x, w] = [x, w] [y, w]$ , the last two equalities moving the central factor $[y, w]$ past $x$ and then past $[x, w]$ .

2.2F1L2step 1.1algebra

Expanding in the second variable, $[x, y w] = x (y w) x^{- 1} (y w)^{- 1} = x y w x^{- 1} w^{- 1} y^{- 1} = (x y x^{- 1}) (x w x^{- 1} w^{- 1}) y^{- 1} = (x y x^{- 1}) [x, w] y^{- 1} = [x, w] x y x^{- 1} y^{- 1} = [x, y] [x, w]$ , where the central factor $[x, w]$ is moved past $y^{- 1}$ and then past $[x, y]$ .

3.1F1L1step 2.1algebra

For $n \in N$ the identity $[x^{n}, y] = [x, y]^{n}$ follows by induction: at $n = 0$ both sides are the identity, since $x^{0}$ is the identity and $[e, y] = e y e^{- 1} y^{- 1} = e$ , and $[x^{n + 1}, y] = [x^{n} x, y] = [x^{n}, y] [x, y] = [x, y]^{n} [x, y] = [x, y]^{n + 1}$ .

4.1L1step 2.1step 2.2step 3.1algebra∎

For a negative integer $n$ put $k = - n$ ; then $e = [x^{0}, y] = [x^{k} x^{- k}, y] = [x^{k}, y] [x^{- k}, y]$ by step 2.1, so $[x^{n}, y] = [x^{- k}, y] = ([x, y]^{k})^{- 1} = [x, y]^{- k} = [x, y]^{n}$ , and the identity $[x, y^{n}] = [x, y]^{n}$ is obtained the same way from step 2.2.

Remarks

The hypothesis $[G, G] \leq Z (G)$ is exactly the vanishing of the third term of the lower central series: $γ_{3} (G) = [G, [G, G]]$ is trivial precisely when every commutator commutes with every element of $G$ (Subgroup commutators and the lower central series).

The first two identities are additivity in each variable separately, and they fail without the hypothesis: in general $[x y, w] = x [y, w] x^{- 1} \cdot [x, w]$ , and the conjugating factor is what the hypothesis removes.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

In a group with central derived subgroup, $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$

Statement

Let $G$ be a group with $[G, G] \leq Z (G)$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , The center $Z (G)$ of a group) and let $x, y \in G$ . If $[G, G] \leq Z (G)$ then $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ for every $n \in N$ , where $(\binom{n}{2})$ is the binomial coefficient of The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ and the powers are those of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ .

The commutator on the right is $[y, x]$ , not $[x, y]$ : in the convention $[g, h] = g h g^{- 1} h^{- 1}$ fixed by Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ one has $y x = [y, x] x y$ , and it is that factor which accumulates.

Facts & Assumptions

Given: A group $G$ with $[G, G] \leq Z (G)$ , elements $x, y \in G$ , and $n \in N$ .

[F1]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

$(\binom{n}{k}) : = ∣ [n]^{k} ∣$ , the number of $k$ -element subsets of $n$ ; in particular $(\binom{n}{1}) = n$ , and $(\binom{n}{k}) = 0$ whenever $k > n$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L1]

If $[G, G] \leq Z (G)$ then $[x y, w] = [x, w] [y, w]$ , $[x, y w] = [x, y] [x, w]$ , and $[x^{n}, y] = [x, y]^{n} = [x, y^{n}]$ for every integer $n$ (Commutator identities in a group whose derived subgroup is central).

[L2]

$(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , with no restriction relating $k$ to $n$ (Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ ).

[L3]

For all $m, n \in Z$ one has $g^{m + n} = g^{m} g^{n}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

Proof

technique · induction

1.1F3base

At $n = 0$ the exponent $(\binom{0}{2})$ is zero because $2 > 0$ , and the asserted identity reads $e = e$ ; this is the base case.

1.2F3L2algebra

For every $n \in N$ , Pascal's rule at $k = 1$ gives $(\binom{n + 1}{2}) = (\binom{n}{1}) + (\binom{n}{2}) = n + (\binom{n}{2})$ .

1.3F1algebra

For all $g, h \in G$ one has $[g, h] h g = (g h g^{- 1} h^{- 1}) h g = g h$ , hence $g h = [g, h] h g$ ; applied to the pair $y, x$ this reads $y x = [y, x] x y$ .

1.4ih

Assume the identity at a given $n \in N$ , that is $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ .

2.1L1step 1.3algebra

Applying step 1.3 to the pair $y^{n}, x$ gives $y^{n} x = [y^{n}, x] x y^{n}$ , and $[y^{n}, x] = [y, x]^{n}$ , so $y^{n} x = [y, x]^{n} x y^{n}$ .

2.2step 1.4algebra

Multiplying the assumption on the right by $x y$ gives $(x y)^{n + 1} = (x y)^{n} (x y) = [y, x]^{(\binom{n}{2})} x^{n} y^{n} x y$ .

3.1F2L3step 2.1step 2.2algebra

Substituting step 2.1 into step 2.2 and moving the central factor $[y, x]^{n}$ to the front gives $(x y)^{n + 1} = [y, x]^{(\binom{n}{2})} x^{n} [y, x]^{n} x y^{n} y = [y, x]^{(\binom{n}{2}) + n} x^{n + 1} y^{n + 1}$ .

4.1step 1.2step 3.1discharge-induction∎

By step 1.2 the exponent $(\binom{n}{2}) + n$ equals $(\binom{n + 1}{2})$ , so the identity holds at $n + 1$ and therefore at every natural number.

Remarks

The formula is the reason the $p$ -th power map behaves differently at $p = 2$ : the coefficient $(\binom{2}{2})$ equals $1$ , so the commutator factor survives, whereas for odd $p$ the coefficient $(\binom{p}{2})$ is a multiple of $p$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

For an odd prime $p$ , the $p$ -th power map is a homomorphism on a finite group whose derived subgroup is central of exponent dividing $p$

Statement

Let $p$ be an odd prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and let $G$ be a finite group with $[G, G] \leq Z (G)$ whose derived subgroup has exponent dividing $p$ (The exponent of a finite group, Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , The center $Z (G)$ of a group). Then

$(x y)^{p} = x^{p} y^{p} for all x, y \in G,$

so $x \mapsto x^{p}$ is a group homomorphism from $G$ to $G$ .

Facts & Assumptions

Given: An odd prime $p$ and a finite group $G$ with $[G, G] \leq Z (G)$ and $\exp ([G, G])$ dividing $p$ ; elements $x, y \in G$ .

[F1]

For a finite group $H$ , $\exp (H) = \min {n \in N : n > 0 and g^{n} = e for every g \in H}$ (The exponent of a finite group).

[F2]

$(\binom{n}{k}) : = ∣ [n]^{k} ∣$ , the number of $k$ -element subsets of $n$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[F3]

$2! = 2$ and $n^{\underline{2}} = n (n - 1)$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L1]

If $[G, G] \leq Z (G)$ then $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ for every $n \in N$ (In a group with central derived subgroup, $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ ).

[L2]

For $k \leq n$ one has $(\binom{n}{k}) \cdot k! = n^{\underline{k}}$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L3]

For all $m, n \in Z$ one has $(g^{m})^{n} = g^{m n}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

Proof

technique · direct

1.1L1

Taking $n = p$ in the product formula gives $(x y)^{p} = [y, x]^{(\binom{p}{2})} x^{p} y^{p}$ .

1.2F2F3L2algebra

Since $p$ is odd and $p \geq 3$ , the closed formula at $n = p$ and $k = 2$ gives $(\binom{p}{2}) \cdot 2 = p (p - 1)$ , and writing $p - 1 = 2 m$ with $m \in N$ turns this into $(\binom{p}{2}) \cdot 2 = 2 p m$ , so $(\binom{p}{2}) = p m$ .

1.3F1L3algebra

The element $c = [y, x]$ lies in $[G, G]$ , and $\exp ([G, G])$ divides $p$ , say $p = \exp ([G, G]) \cdot t$ ; hence $c^{p} = (c^{\exp ([G, G])})^{t} = e$ .

2.1step 1.1step 1.2step 1.3algebra∎

Combining, $[y, x]^{(\binom{p}{2})} = (c^{p})^{m} = e$ , so step 1.1 reads $(x y)^{p} = x^{p} y^{p}$ ; as this holds for all $x, y \in G$ , the map $x \mapsto x^{p}$ is a homomorphism.

Remarks

Only the oddness of $p$ is used, in step 1.2; primality enters through the hypothesis on the derived subgroup rather than through the arithmetic. At $p = 2$ the conclusion fails at the first step: $(\binom{2}{2}) = 1$ , so $(x y)^{2} = [y, x] x^{2} y^{2}$ and the commutator factor survives.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Special and extraspecial $p$ -groups

Definition

Let $p$ be a prime and let $P$ be a finite $p$ -group (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ). Write $P^{'} = [P, P]$ for its derived subgroup (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ), $Z (P)$ for its centre (The center $Z (G)$ of a group) and $Φ (P)$ for its Frattini subgroup (The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group).

A finite $p$ -group $P$ is special when $Z (P) = P^{'} = Φ (P)$ is elementary abelian, and extraspecial when in addition $P$ is nonabelian and this common subgroup has order $p$ . Elementary abelian $p$ -groups are those of Elementary abelian $p$ -groups, and the trivial group is one of them.

Remarks

The nonabelian clause is stated rather than left implicit. It is not redundant for every formulation in the literature: the informal description " $P / Z (P)$ is elementary abelian and $∣ Z (P) ∣ = p$ " is satisfied by the cyclic group of order $p$ , whose centre is the whole group, so a definition phrased that way must exclude the abelian case by hand. With the clause $Z (P) = P^{'}$ in force the exclusion is automatic, since an abelian group has trivial derived subgroup.

Two source conventions for extraspecial groups are in circulation and agree under the order- $p$ centre hypothesis. Craven asks that $Z (P) = P^{'} = Φ (P)$ be elementary abelian and then that this common subgroup have order $p$ ; van Beek asks that $P^{'} = Z (P)$ have order $p$ and that $P / Z (P)$ have exponent $p$ . Their equivalence for nonabelian $P$ is the content of Three equivalent descriptions of an extraspecial $p$ -group.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Three equivalent descriptions of an extraspecial $p$ -group

Statement

For a finite $p$ -group $P$ the following are equivalent: $P$ is extraspecial; $P$ is nonabelian, $∣ Z (P) ∣ = p$ and $P / Z (P)$ is elementary abelian; $P$ is nonabelian and $Z (P) = P^{'} = Φ (P)$ has order $p$ .

Here $P^{'} = [P, P]$ , $Φ (P)$ is the Frattini subgroup, and quotients are those of The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ .

Facts & Assumptions

Given: A prime $p$ and a finite $p$ -group $P$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[F1]

A finite $p$ -group $P$ is special when $Z (P) = P^{'} = Φ (P)$ is elementary abelian, and extraspecial when in addition $P$ is nonabelian and this common subgroup has order $p$ (Special and extraspecial $p$ -groups).

[L1]

For a finite $p$ -group $P$ , the quotient $P / Φ (P)$ is elementary abelian, and for $N ⊴ P$ the quotient $P / N$ is elementary abelian if and only if $Φ (P) \leq N$ (The Frattini quotient is the largest elementary abelian quotient of a finite $p$ -group).

[L2]

For every finite $p$ -group $P$ , $Φ (P) = P^{'} P^{p}$ , where $P^{p} = ⟨ g^{p} : g \in P ⟩$ ( $Φ (P) = P^{'} P^{p}$ for a finite $p$ -group, The $p$ th-power subgroup $G^{p}$ ).

[L3]

For $N ⊴ G$ , the quotient $G / N$ is abelian if and only if $[G, G] \subseteq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

[L4]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ ; the trivial group is permitted (Elementary abelian $p$ -groups).

[L5]

For every group $G$ , the center $Z (G)$ is a normal subgroup of $G$ (The center of a group is a normal subgroup).

[L6]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L7]

In a finite group whose order is prime, every $g \neq e$ has order $∣ G ∣$ and generates $G$ (A finite group of prime order is cyclic and every nonidentity element generates it).

Proof

technique · direct

1.1F1L1L4L5

Suppose $P$ is extraspecial. Then $P$ is nonabelian, $Φ (P) = Z (P)$ has order $p$ , and $P / Φ (P)$ is elementary abelian; since $Φ (P) = Z (P)$ this says $P / Z (P)$ is elementary abelian, the quotient being formed along a normal subgroup. So the second description holds.

1.2L1L2L3L5L6

Suppose $P$ is nonabelian with $∣ Z (P) ∣ = p$ and $P / Z (P)$ elementary abelian. Applying the elementary abelian criterion to the normal subgroup $Z (P)$ gives $Φ (P) \leq Z (P)$ , and an elementary abelian quotient is abelian, so $P^{'} \leq Z (P)$ . As $P$ is nonabelian, $P^{'} \neq 1$ ; a subgroup of the group $Z (P)$ of order $p$ has order $1$ or $p$ , so $P^{'} = Z (P)$ . Then $Φ (P) = P^{'} P^{p}$ contains $P^{'} = Z (P)$ and is contained in $Z (P)$ , so $Φ (P) = P^{'} = Z (P)$ has order $p$ , which is the third description.

1.3F1L4L7

Suppose $P$ is nonabelian with $Z (P) = P^{'} = Φ (P)$ of order $p$ . A group of prime order is cyclic, hence abelian, and each of its nonidentity elements has order $p$ , so this common subgroup is elementary abelian; it is a finite $p$ -group because its order is $p$ . Thus $P$ meets the definition and is extraspecial.

2.1step 1.1step 1.2step 1.3∎

The three implications close a cycle, so the three descriptions are equivalent.

Remarks

The nonabelian hypothesis does real work exactly once, in step 1.2, where it supplies $P^{'} \neq 1$ . Dropping it leaves the cyclic group of order $p$ satisfying the second description with $Z (P) = P$ of order $p$ and trivial quotient, while its derived subgroup is trivial and the third description fails.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$

Statement

Let $P$ be an extraspecial $p$ -group. Then $P$ is nilpotent of nilpotency class exactly two, its derived subgroup satisfies $P^{'} = Z (P)$ and has order $p$ , and every nonidentity commutator of $P$ has order $p$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ (Special and extraspecial $p$ -groups).

[L1]

[L2]

For subgroups $A, B \leq G$ the subgroup commutator is $[A, B] = ⟨ [a, b] : a \in A, b \in B ⟩$ , and the lower central series is $γ_{1} (G) = G$ , $γ_{r + 1} (G) = [G, γ_{r} (G)]$ (Subgroup commutators and the lower central series).

[L3]

$G$ is nilpotent exactly when its lower central series reaches $1$ , and the least $c$ with $γ_{c + 1} (G) = 1$ is its nilpotency class (Nilpotence via central series, the upper central series, and the lower central series, Nilpotent groups and nilpotency class).

[L4]

In a finite group whose order is prime, every $g \neq e$ has order $∣ G ∣$ (A finite group of prime order is cyclic and every nonidentity element generates it).

[F1]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

Proof

technique · direct

1.1L1

By the third description in the characterisation, $P^{'} = Z (P)$ has order $p$ and $P$ is nonabelian.

1.2L2

The second term of the lower central series is $γ_{2} (P) = [P, γ_{1} (P)] = [P, P] = P^{'}$ , and the third is $γ_{3} (P) = [P, P^{'}]$ .

2.1F1step 1.1step 1.2

Every element of $P^{'} = Z (P)$ commutes with every element of $P$ , so each generator $[g, z]$ of $[P, P^{'}]$ is the identity and $γ_{3} (P) = 1$ .

3.1L3step 1.1step 1.2step 2.1

Hence $P$ is nilpotent of class at most two; the class is not zero or one, since class at most one would give $γ_{2} (P) = P^{'} = 1$ and $P$ is nonabelian, so the class is exactly two.

4.1L4step 1.1∎

Every commutator lies in $P^{'}$ , a group of order $p$ , so a nonidentity commutator has order $p$ .

Remarks

Both conclusions are hypotheses of the class-two commutator calculus: the derived subgroup is central, which is what class two says, and it has exponent $p$ , which is what the order- $p$ conclusion says.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The centre of an extraspecial $p$ -group has no complement

Statement

Let $P$ be an extraspecial $p$ -group. Then $Z (P)$ has no complement in $P$ : there is no subgroup $H \leq P$ with $P = Z (P) H$ and $Z (P) \cap H = 1$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ (Special and extraspecial $p$ -groups).

[F1]

For subgroups $N, H$ of $G$ with $N ⊴ G$ , $G = N H$ and $N \cap H = {1}$ , the group $G$ is the internal semidirect product of $N$ by $H$ , and $H$ is called a complement to $N$ in $G$ (An internal semidirect product and a complement to a normal subgroup).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[L1]

[L2]

For every homomorphism $f : G \to H$ , the rule $g \ker f \mapsto f (g)$ is an isomorphism from $G / \ker f$ onto $im f$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

[L3]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

Proof

technique · contradiction

1.1F1L1assume-contra

Suppose $H \leq P$ is a complement to $Z (P)$ , so that $P = Z (P) H$ and $Z (P) \cap H = 1$ ; the centre is normal, so the quotient $P / Z (P)$ is defined.

2.1L2step 1.1

Let $π : P \to P / Z (P)$ be the quotient map and restrict it to $H$ . Its kernel is $H \cap Z (P) = 1$ , and its image is all of $P / Z (P)$ because every $g \in P$ is $z h$ with $z \in Z (P)$ and $h \in H$ , whence $π (g) = π (h)$ ; so $H ≅ P / Z (P)$ .

3.1L1L3step 2.1

The quotient $P / Z (P)$ is elementary abelian, hence abelian, so $H$ is abelian.

4.1F2step 1.1step 3.1discharge-contradiction∎

Every element of $P$ is $z h$ with $z$ central and $h \in H$ , and $(z_{1} h_{1}) (z_{2} h_{2}) = z_{1} z_{2} h_{1} h_{2} = z_{2} z_{1} h_{2} h_{1} = (z_{2} h_{2}) (z_{1} h_{1})$ , so $P$ is abelian; this contradicts the nonabelianness of an extraspecial group.

Remarks

The argument uses no bound on the order of $P$ , so the conclusion holds for every extraspecial group and not only for those of order $p^{3}$ . What fails is not that a complement is hard to find but that its existence would make the group abelian, which the definition forbids.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Every conjugacy class of an extraspecial $p$ -group outside the centre has exactly $p$ elements

Statement

Every conjugacy class of an extraspecial $p$ -group whose representative is not central has exactly $p$ elements. That is, for an extraspecial $p$ -group $P$ and $x \in P ∖ Z (P)$ ,

$∣ {Cl}_{P} (x) ∣ = p .$

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ (Special and extraspecial $p$ -groups) and an element $x \in P$ with $x \notin Z (P)$ .

[F1]

${Cl}_{G} (x) : = {g x g^{- 1} : g \in G}$ and $C_{G} (x) : = {g \in G : g x = x g} = {g \in G : g x g^{- 1} = x}$ (The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F4]

For $H \leq G$ and $g \in G$ , the right coset is $H g : = {h g : h \in H}$ (Left and right cosets $g H$ and $H g$ of a subgroup).

[L1]

An extraspecial $p$ -group has derived subgroup $P^{'} = Z (P)$ of order $p$ (An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$ ).

[L2]

$∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ for a finite group $G$ ( $G / C_{G} (x) \to {Cl}_{G} (x)$ is a bijection, so $∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ whenever these cardinalities are finite).

[L3]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L4]

$[G : H] : = ∣ G / H ∣$ , the number of left cosets of $H$ in $G$ (The coset set $G / H$ and the index $[G : H]$ of a subgroup).

[L5]

A finite $p$ -group is a finite group whose order has the form $∣ P ∣ = p^{n}$ for some $n \in N$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

Proof

technique · direct

1.1F3algebra

For every $g \in P$ one has $[g, x] x = (g x g^{- 1} x^{- 1}) x = g x g^{- 1}$ , so each conjugate of $x$ has the form $[g, x] x$ .

1.2L1L5

The derived subgroup of $P$ is $Z (P)$ and has order $p$ , and $P$ is a finite $p$ -group, so $∣ P ∣ = p^{n}$ for some $n$ .

1.3F1L2L3L4

The size of the class of $x$ is the index $[P : C_{P} (x)]$ , and by Lagrange that index divides $∣ P ∣$ .

2.1F4step 1.1step 1.2algebra

Each $[g, x]$ lies in $P^{'} = Z (P)$ , so every conjugate of $x$ lies in the right coset $Z (P) x$ ; the map $z \mapsto z x$ is a bijection from $Z (P)$ onto that coset, so the coset has $p$ elements and the class of $x$ has at most $p$ .

2.2F1F2step 1.3

Since $x \notin Z (P)$ , some $g \in P$ fails to commute with $x$ , so $C_{P} (x) \neq P$ , its index is greater than one, and the class of $x$ has more than one element.

3.1step 1.2step 1.3step 2.1step 2.2∎

The class size divides $p^{n}$ , so it is a power of $p$ ; it lies strictly between $1$ and $p$ inclusive, and the only such power of $p$ is $p$ itself.

Remarks

The hypothesis $x \notin Z (P)$ is used only at step 2.2, and it is used to rule out the class of size one. A central $x$ runs through the same computation and comes out with the class ${x}$ , which is consistent with step 2.1 and shows that the two cases exhaust the group.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A noncentral element of an extraspecial $p$ -group has centraliser of index $p$

Statement

Let $P$ be an extraspecial $p$ -group and let $x \in P ∖ Z (P)$ . Then

$[P : C_{P} (x)] = p, equivalently ∣ C_{P} (x) ∣ = \frac{∣ P ∣}{p} .$

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ and an element $x \in P$ with $x \notin Z (P)$ .

[F1]

$C_{G} (x) : = {g \in G : g x = x g} = {g \in G : g x g^{- 1} = x}$ (The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

[F2]

$[G : H] : = ∣ G / H ∣$ , the number of left cosets of $H$ in $G$ (The coset set $G / H$ and the index $[G : H]$ of a subgroup).

[L1]

Every conjugacy class of an extraspecial $p$ -group whose representative is not central has exactly $p$ elements (Every conjugacy class of an extraspecial $p$ -group outside the centre has exactly $p$ elements).

[L2]

[L3]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1L1

The class of $x$ has exactly $p$ elements, because $x$ is not central.

1.2F1F2L2

The size of that class is the index of the centraliser of $x$ in $P$ .

2.1L3step 1.1step 1.2∎

Hence $[P : C_{P} (x)] = p$ , and Lagrange turns this into $∣ P ∣ = p ∣ C_{P} (x) ∣$ .

Remarks

Every centraliser named here is proper, since $x$ is noncentral, and maximal in the order sense: index $p$ is the smallest index a proper subgroup of a finite $p$ -group can have.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups

Definition

Let $G$ and $H$ be groups, let $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ be subgroups of their centres (The center $Z (G)$ of a group, Subgroup), and let $α : Z_{1} \to Z_{2}$ be an isomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ). Inside the external direct product $G \times H$ (The external direct product $G \times H$ with componentwise multiplication, $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections) put

$N : = {(z, α (z)^{- 1}) : z \in Z_{1}} .$

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ , formed as in The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ :

$G \circ_{α} H : = (G \times H) / N .$

That $N$ is a normal subgroup, so that the quotient is defined (Normal subgroup: invariance under conjugation), is proved in The identified subgroup used to form a central product is central, hence normal ↗.

Write $π : G \times H \to G \circ_{α} H$ for the quotient map and $\overset{ˉ}{g} : = π (g, e)$ , $\overset{ˉ}{h} : = π (e, h)$ for the canonical images of $g \in G$ and $h \in H$ .

Remarks

The identification is along $α$ and reverses the second coordinate: killing $(z, α (z)^{- 1})$ is exactly what makes $\overset{ˉ}{z} = \overline{α (z)}$ hold in the quotient, so the two identified central subgroups become one. Killing $(z, α (z))$ instead would identify $z$ with $α (z)^{- 1}$ , which is the central product along $α^{- 1}$ composed with inversion rather than along $α$ .

Craven writes $G * H$ for a central product and van Beek writes $G \circ H$ ; the notation $\circ_{α}$ is used here because the isomorphism is part of the data and different choices of $α$ can give non-isomorphic quotients.

Taking $Z_{1} = Z_{2} = 1$ gives $N = 1$ and recovers the direct product, so the construction is a genuine generalisation and not a separate object.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The identified subgroup used to form a central product is central, hence normal

Statement

Let $G$ and $H$ be groups with central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ . The subgroup $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ of $G \times H$ is central, hence normal, so the quotient $(G \times H) / N$ of The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups is defined.

Facts & Assumptions

Given: Groups $G, H$ , central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ , and an isomorphism $α : Z_{1} \to Z_{2}$ .

[F1]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F2]

A subset $H \subseteq G$ is a subgroup when $e \in H$ , $H$ is closed under the operation, and $H$ is closed under inverses (Subgroup).

[F3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F4]

A subgroup $N \leq G$ is normal in $G$ when $g N g^{- 1} = N$ for every $g \in G$ , where $g N g^{- 1} : = {g n g^{- 1} : n \in N}$ (Normal subgroup: invariance under conjugation).

[L1]

A group homomorphism $f$ satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed).

[L2]

The external direct product $G \times H : = {(g, h) : g \in G, h \in H}$ carries the componentwise operation $(g, h) (g^{'}, h^{'}) = (g g^{'}, h h^{'})$ (The external direct product $G \times H$ with componentwise multiplication).

[L3]

The componentwise operation makes $G \times H$ a group with identity $(e_{G}, e_{H})$ and $(g, h)^{- 1} = (g^{- 1}, h^{- 1})$ ( $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

Proof

technique · direct

1.1F3L1algebra

The isomorphism $α$ is in particular a homomorphism, so $α (e) = e$ and $α (z^{- 1}) = α (z)^{- 1}$ ; moreover $Z_{2} \leq Z (H)$ , so any two elements of $Z_{2}$ commute and $α (z_{1} z_{2})^{- 1} = α (z_{1})^{- 1} α (z_{2})^{- 1}$ .

2.1F1F2L2L3step 1.1

Hence $(e, e) = (e, α (e)^{- 1})$ lies in $N$ ; the product $(z_{1}, α (z_{1})^{- 1}) (z_{2}, α (z_{2})^{- 1}) = (z_{1} z_{2}, α (z_{1})^{- 1} α (z_{2})^{- 1}) = (z_{1} z_{2}, α (z_{1} z_{2})^{- 1})$ lies in $N$ ; and $(z, α (z)^{- 1})^{- 1} = (z^{- 1}, α (z)) = (z^{- 1}, α (z^{- 1})^{- 1})$ lies in $N$ . So $N$ is a subgroup of $G \times H$ .

3.1F1F3L2step 2.1

Every element of $N$ has first coordinate in $Z_{1} \leq Z (G)$ and second coordinate in $Z_{2} \leq Z (H)$ , and the operation is componentwise, so $(z, α (z)^{- 1}) (g, h) = (z g, α (z)^{- 1} h) = (g z, h α (z)^{- 1}) = (g, h) (z, α (z)^{- 1})$ for every $(g, h)$ ; thus $N \leq Z (G \times H)$ .

4.1F4step 3.1∎

For $n \in N$ and $x \in G \times H$ centrality gives $x n x^{- 1} = n x x^{- 1} = n$ , so $x N x^{- 1} = N$ and $N$ is normal; the quotient $(G \times H) / N$ is therefore defined.

Remarks

Centrality is used twice. It makes the inverse-coordinate rule $z \mapsto (z, α (z)^{- 1})$ multiplicative, so that $N$ is a subgroup, and it then makes that subgroup central and hence normal. For merely isomorphic subgroups the displayed antidiagonal need be neither a subgroup nor a normal subset, so the quotient construction does not apply.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The two canonical maps into a central product are injective homomorphisms whose images commute, generate it, and meet in the identified centre

Statement

Let $G$ and $H$ be groups with central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , and let $π : G \times H \to G \circ_{α} H$ be the quotient map. The canonical maps $G \to G \circ_{α} H$ and $H \to G \circ_{α} H$ are injective homomorphisms whose images commute elementwise, generate $G \circ_{α} H$ , and meet in the image of $Z_{1}$ . They are $g \mapsto \overset{ˉ}{g} = π (g, e)$ and $h \mapsto \overset{ˉ}{h} = π (e, h)$ , and the intersection of their images is ${\overset{ˉ}{z} : z \in Z_{1}} = {\overline{α (z)} : z \in Z_{1}}$ .

Facts & Assumptions

Given: Groups $G, H$ , central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ , an isomorphism $α : Z_{1} \to Z_{2}$ , and the quotient map $π : G \times H \to G \circ_{α} H$ .

[F1]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F2]

The quotient group $G / N$ has the left cosets $g N$ as elements with product $(g N) (h N) : = g h N$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[F3]

For a group homomorphism $f : G \to H$ , $\ker f : = {g \in G : f (g) = e_{H}}$ and $im f : = {f (g) : g \in G}$ (The kernel and image of a group homomorphism).

[L1]

The subgroup $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ of $G \times H$ is central, hence normal (The identified subgroup used to form a central product is central, hence normal).

[L2]

[L3]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ , namely $⋂ {K : K \leq G and S \subseteq K}$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L4]

An isomorphism is a bijective group homomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Proof

technique · direct

1.1F1F2L1L2

The coordinate maps $g \mapsto (g, e)$ and $h \mapsto (e, h)$ are homomorphisms into $G \times H$ , because the operation there is componentwise, and $π$ is a homomorphism onto the quotient; so both canonical maps are homomorphisms.

2.1F1F3L4step 1.1

The kernel of $g \mapsto \overset{ˉ}{g}$ is ${g \in G : (g, e) \in N}$ ; an equality $(g, e) = (z, α (z)^{- 1})$ gives $g = z$ and $α (z) = e$ , so $z = e$ because $α$ is injective, and the kernel is trivial. Likewise $(e, h) = (z, α (z)^{- 1})$ gives $z = e$ and then $h = α (e)^{- 1} = e$ . Both canonical maps are therefore injective.

2.2L2step 1.1

In $G \times H$ one has $(g, e) (e, h) = (g, h) = (e, h) (g, e)$ , so $\overset{ˉ}{g} \overset{ˉ}{h} = \overset{ˉ}{h} \overset{ˉ}{g}$ for all $g \in G$ and $h \in H$ : the two images commute elementwise.

2.3F2L2L3step 1.1

Every element of $G \circ_{α} H$ is $π (g, h) = π ((g, e) (e, h)) = \overset{ˉ}{g} \overset{ˉ}{h}$ , so the two images together generate $G \circ_{α} H$ .

3.1F1step 2.1step 2.2∎

If $\overset{ˉ}{g} = \overset{ˉ}{h}$ then $(g, e) (e, h)^{- 1} = (g, h^{- 1})$ lies in $N$ , so $g = z \in Z_{1}$ and $h^{- 1} = α (z)^{- 1}$ , that is $h = α (g)$ ; conversely $\overset{ˉ}{z} = \overline{α (z)}$ for every $z \in Z_{1}$ , since $(z, α (z)^{- 1}) \in N$ . Hence the two images meet exactly in ${\overset{ˉ}{z} : z \in Z_{1}}$ .

Remarks

Injectivity is what makes the central product an honest amalgam: each factor embeds, and the only collapsing is the prescribed identification of $Z_{1}$ with $Z_{2}$ . If $α$ were merely a surjective homomorphism, $N$ would meet the first coordinate copy of $G$ in $\ker α \times 1$ and that copy would not embed.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Order, centre and derived subgroup of a central product

Statement

Let $G$ and $H$ be groups with central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , and write $\overset{ˉ}{g}, \overset{ˉ}{h}$ for the canonical images in $P = G \circ_{α} H$ . Then:

if $G$ and $H$ are finite, $∣ P ∣ = \frac{∣ G ∣ ∣ H ∣}{∣ Z_{1} ∣}$ ;
$Z (P) = {\overset{ˉ}{g} \overset{ˉ}{h} : g \in Z (G), h \in Z (H)}$ , the image of $Z (G) \times Z (H)$ ;
$[P, P] = {\overline{c} \overline{d} : c \in [G, G], d \in [H, H]}$ , the image of $[G, G] \times [H, H]$ .

Facts & Assumptions

Given: Groups $G, H$ , central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ , an isomorphism $α : Z_{1} \to Z_{2}$ , the quotient map $π : G \times H \to P$ , and the canonical images $\overset{ˉ}{g} = π (g, e)$ , $\overset{ˉ}{h} = π (e, h)$ .

[F1]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ , and $[G, G]$ is the subgroup generated by all commutators (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[L1]

The canonical maps $G \to G \circ_{α} H$ and $H \to G \circ_{α} H$ are injective homomorphisms whose images commute elementwise, generate $G \circ_{α} H$ , and meet in the image of $Z_{1}$ (The two canonical maps into a central product are injective homomorphisms whose images commute, generate it, and meet in the identified centre).

[L2]

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ (For finite groups $G$ and $H$ , $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ ).

[L3]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L4]

$⟨ S ⟩$ is the smallest subgroup containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L5]

The subgroup $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ of $G \times H$ is central, hence normal (The identified subgroup used to form a central product is central, hence normal).

Proof

technique · direct

1.1F1L2L3L5algebra

The map $z \mapsto (z, α (z)^{- 1})$ is a bijection from $Z_{1}$ onto $N$ , so $∣ N ∣ = ∣ Z_{1} ∣$ ; with $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ and Lagrange applied to $N \leq G \times H$ , the number of cosets is $∣ G ∣ ∣ H ∣ / ∣ Z_{1} ∣$ , which is the order of $P$ .

1.2L1

The images of $G$ and of $H$ in $P$ commute elementwise, generate $P$ , and each canonical map is injective.

2.1F2F3step 1.2

Let $\overset{ˉ}{g} \overset{ˉ}{h}$ be central in $P$ and let $g^{'} \in G$ . Then $\overset{ˉ}{h}$ commutes with $\overline{g^{'}}$ , so $\overset{ˉ}{g}$ commutes with $\overline{g^{'}}$ , hence $\overline{[g, g^{'}]} = \overline{g} \overline{g^{'}} {\overline{g}}^{- 1} {\overline{g^{'}}}^{- 1}$ is trivial; injectivity of the canonical map gives $[g, g^{'}] = e$ , so $g \in Z (G)$ , and symmetrically $h \in Z (H)$ .

2.2F2step 1.2

Conversely, if $g \in Z (G)$ and $h \in Z (H)$ then $\overset{ˉ}{g} \overset{ˉ}{h}$ commutes with every $\overline{g^{'}}$ and with every $\overline{h^{'}}$ , and those elements generate $P$ , so $\overset{ˉ}{g} \overset{ˉ}{h}$ is central.

2.3F3step 1.2

Because the two images commute elementwise, $[{\overset{ˉ}{g}}_{1} {\overset{ˉ}{h}}_{1}, {\overset{ˉ}{g}}_{2} {\overset{ˉ}{h}}_{2}] = [{\overset{ˉ}{g}}_{1}, {\overset{ˉ}{g}}_{2}] [{\overset{ˉ}{h}}_{1}, {\overset{ˉ}{h}}_{2}] = \overline{[g_{1}, g_{2}]} \overline{[h_{1}, h_{2}]}$ for all $g_{i} \in G$ and $h_{i} \in H$ .

3.1F3L4step 1.1step 2.1step 2.2step 2.3∎

Every element of $P$ is $\overset{ˉ}{g} \overset{ˉ}{h}$ by step 1.2, so steps 2.1 and 2.2 identify $Z (P)$ as the image of $Z (G) \times Z (H)$ ; and by step 2.3 the commutators of $P$ are exactly the products $\overline{[g_{1}, g_{2}]} \overline{[h_{1}, h_{2}]}$ , whose generated subgroup is the image of $[G, G] \times [H, H]$ .

Remarks

Clause 1 needs both factors finite; clauses 2 and 3 do not, since they use only that the two images commute and generate. The order formula divides by $∣ Z_{1} ∣$ and not by $∣ Z_{1} ∣^{2}$ : one copy of the identified subgroup survives inside the product, and it is the image described in the intersection clause of the canonical-maps proposition.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Homomorphisms out of a central product

Statement

Let $G$ and $H$ be groups with central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , and let $K$ be a group. If $φ : G \to K$ and $ψ : H \to K$ are homomorphisms with commuting images and $φ ∣_{Z_{1}} = ψ \circ α$ , then there is a unique homomorphism $G \circ_{α} H \to K$ restricting to $φ$ and $ψ$ along the canonical maps. Explicitly it sends $\overset{ˉ}{g} \overset{ˉ}{h}$ to $φ (g) ψ (h)$ .

Facts & Assumptions

Given: Groups $G, H, K$ , central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ , an isomorphism $α : Z_{1} \to Z_{2}$ , and homomorphisms $φ : G \to K$ , $ψ : H \to K$ with $φ (g) ψ (h) = ψ (h) φ (g)$ for all $g, h$ and $φ (z) = ψ (α (z))$ for all $z \in Z_{1}$ .

[F1]

[F2]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F3]

For a group homomorphism $f : G \to H$ , $\ker f : = {g \in G : f (g) = e_{H}}$ (The kernel and image of a group homomorphism).

[F4]

The quotient group $G / N$ has the left cosets $g N$ as elements, with product $(g N) (h N) : = g h N$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L1]

[L2]

For $H \leq G$ the rule $(a H) (b H) : = a b H$ on left cosets is independent of the representatives $a$ and $b$ if and only if $H ⊴ G$ (Coset multiplication $(g H) (h H) = g h H$ is well defined if and only if $H$ is normal).

[L3]

The subgroup $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ of $G \times H$ is central, hence normal (The identified subgroup used to form a central product is central, hence normal).

[L4]

Proof

technique · direct

1.1F1givenalgebra

The assignment $θ (g, h) : = φ (g) ψ (h)$ satisfies $θ ((g, h) (g^{'}, h^{'})) = φ (g) φ (g^{'}) ψ (h) ψ (h^{'}) = φ (g) ψ (h) φ (g^{'}) ψ (h^{'}) = θ (g, h) θ (g^{'}, h^{'})$ , the middle equality being the commuting-images hypothesis; so $θ : G \times H \to K$ is a homomorphism.

1.2F2F3L1given

For $z \in Z_{1}$ , $θ (z, α (z)^{- 1}) = φ (z) ψ (α (z)^{- 1}) = φ (z) ψ (α (z))^{- 1} = φ (z) φ (z)^{- 1} = e$ , so $N \leq \ker θ$ .

2.1F4L2L3step 1.1step 1.2

If $x N = y N$ in $G \times H$ then $x^{- 1} y \in N$ , so $θ (x)^{- 1} θ (y) = θ (x^{- 1} y) = e$ and $θ (x) = θ (y)$ ; hence $\overset{ˉ}{θ} (x N) : = θ (x)$ is a well-defined function on $G \circ_{α} H$ , and it is a homomorphism because $N$ is normal and $(x N) (y N) = x y N$ .

3.1L1L4step 2.1∎

On the canonical images, $\overset{ˉ}{θ} (\overset{ˉ}{g}) = θ (g, e) = φ (g)$ and $\overset{ˉ}{θ} (\overset{ˉ}{h}) = θ (e, h) = ψ (h)$ ; and any homomorphism agreeing with $φ$ and $ψ$ on the two images agrees with $\overset{ˉ}{θ}$ on a generating set of $G \circ_{α} H$ , hence everywhere.

Remarks

Both hypotheses are needed and neither is implied by the other. Without commuting images the assignment of step 1.1 is not a homomorphism on the direct product; without the agreement on $Z_{1}$ the subgroup $N$ need not lie in the kernel, so nothing descends to the quotient.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Internal central products of a finite family of subgroups

Definition

Let $G$ be a group and let $G_{1}, \dots, G_{r}$ be subgroups of $G$ , where $r \in N$ (Subgroup). Subgroups $G_{1}, \dots, G_{r}$ of $G$ form an internal central product when they generate $G$ and $[G_{i}, G_{j}] = 1$ for $i \neq j$ , where $⟨ \cdot ⟩$ is the generated subgroup of The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups and $[G_{i}, G_{j}]$ is the subgroup commutator of Subgroup commutators and the lower central series. The empty family is an internal central product of the trivial group.

Since the factors commute pairwise, $G = G_{1} G_{2} \dots G_{r}$ as a set of products, and for $i \neq j$ every element of $G_{i} \cap G_{j}$ commutes with both $G_{i}$ and $G_{j}$ , so $G_{i} \cap G_{j} \leq Z (G_{i}) \cap Z (G_{j})$ (The center $Z (G)$ of a group).

Remarks

The condition differs from that of an internal direct product (Internal direct products of finitely many normal subgroups) in exactly one place: there the factors are required to intersect trivially, here they are allowed to share a central subgroup. An internal direct product of normal subgroups is in particular an internal central product, since distinct factors of a direct product commute elementwise.

No hypothesis is placed on the intersections beyond what the commuting condition already forces. That is deliberate: the intersections are what the recognition theorem computes, rather than data prescribed in advance.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Internal central products are the images of external ones

Statement

Let $G$ be a group and $G_{1}, \dots, G_{r} \leq G$ . Subgroups $G_{1}, \dots, G_{r}$ form an internal central product of $G$ if and only if the multiplication map $G_{1} \times \dots \times G_{r} \to G$ , $(g_{1}, \dots, g_{r}) \mapsto g_{1} \dots g_{r}$ , is a surjective homomorphism; each factor then meets its kernel trivially.

For two factors this identifies the internal notion with the external one: if $G_{1}, G_{2}$ form an internal central product of $G$ and $D = G_{1} \cap G_{2}$ , then $D \leq Z (G_{1})$ and $D \leq Z (G_{2})$ , and

$G ≅ G_{1} \circ_{{id}_{D}} G_{2},$

the external central product of The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups taken along the identity isomorphism of $D$ .

Facts & Assumptions

Given: A group $G$ and subgroups $G_{1}, \dots, G_{r} \leq G$ ; in the second half, $r = 2$ and $D = G_{1} \cap G_{2}$ .

[F1]

Subgroups $G_{1}, \dots, G_{r}$ of $G$ form an internal central product when they generate $G$ and $[G_{i}, G_{j}] = 1$ for $i \neq j$ (Internal central products of a finite family of subgroups).

[F2]

The external direct product $G \times H : = {(g, h) : g \in G, h \in H}$ carries the componentwise operation (The external direct product $G \times H$ with componentwise multiplication).

[F3]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F4]

For a group homomorphism $f : G \to H$ , $\ker f : = {g \in G : f (g) = e_{H}}$ and $im f : = {f (g) : g \in G}$ (The kernel and image of a group homomorphism).

[L1]

For every homomorphism $f : G \to H$ , the rule $g \ker f \mapsto f (g)$ is an isomorphism from $G / \ker f$ onto $im f$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

[L2]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

Proof

technique · direct

1.1F1F2F4L2algebra

Suppose the subgroups generate $G$ and commute pairwise. Writing $μ (g_{1}, \dots, g_{r}) = g_{1} \dots g_{r}$ , the commuting hypothesis lets the factors of $μ (x) μ (y)$ be sorted by index, so $μ (x y) = μ (x) μ (y)$ and $μ$ is a homomorphism; its image is a subgroup containing every $G_{i}$ , hence equals $⟨ G_{1}, \dots, G_{r} ⟩ = G$ , so $μ$ is surjective.

1.2F1F2L2algebra

For the converse, suppose $μ$ is a surjective homomorphism. Surjectivity gives $G = G_{1} \dots G_{r} \subseteq ⟨ G_{1}, \dots, G_{r} ⟩$ , so the subgroups generate. For $i \neq j$ , $a \in G_{i}$ and $b \in G_{j}$ , the tuples $x$ with $a$ in place $i$ and $y$ with $b$ in place $j$ commute in the direct product, so $a b = μ (x) μ (y) = μ (x y) = μ (y x) = μ (y) μ (x) = b a$ ; hence $[G_{i}, G_{j}] = 1$ .

1.3F4

In either case a tuple with a single nonidentity entry $g_{i}$ has $μ$ -value $g_{i}$ , so it lies in $\ker μ$ only if $g_{i} = e$ : each factor meets the kernel trivially.

2.1F1L3step 1.1

Now let $r = 2$ and let $G_{1}, G_{2}$ form an internal central product with $D = G_{1} \cap G_{2}$ . An element $d \in D$ lies in $G_{2}$ , so it commutes with every element of $G_{1}$ , giving $D \leq Z (G_{1})$ ; symmetrically $D \leq Z (G_{2})$ .

2.2F4step 1.1

The kernel of $μ : G_{1} \times G_{2} \to G$ is ${(g_{1}, g_{2}) : g_{1} g_{2} = e} = {(d, d^{- 1}) : d \in D}$ , since $g_{1} = g_{2}^{- 1}$ lies in both subgroups.

3.1F3L1step 1.1step 2.1step 2.2∎

That kernel is exactly the subgroup $N$ used to build $G_{1} \circ_{{id}_{D}} G_{2}$ , so the first isomorphism theorem gives $G ≅ (G_{1} \times G_{2}) / N = G_{1} \circ_{{id}_{D}} G_{2}$ .

Remarks

The commuting condition is imposed only for $i \neq j$ . A single factor is not required to be abelian, which is what allows a nonabelian group to be an internal central product of one factor, namely itself.

The identity isomorphism of $D$ is forced here rather than chosen: the kernel of the multiplication map is ${(d, d^{- 1})}$ , and that is the identified subgroup of the external product along ${id}_{D}$ and along no other map.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A central product of extraspecial $p$ -groups identified along their centres is extraspecial

Statement

Let $E_{1}$ and $E_{2}$ be extraspecial $p$ -groups and let $α : Z (E_{1}) \to Z (E_{2})$ be an isomorphism. Then $P = E_{1} \circ_{α} E_{2}$ is extraspecial, of order $∣ E_{1} ∣ ∣ E_{2} ∣ / p$ , and its centre and derived subgroup are the common image of $Z (E_{1})$ and $Z (E_{2})$ .

Facts & Assumptions

Given: Extraspecial $p$ -groups $E_{1}, E_{2}$ , an isomorphism $α : Z (E_{1}) \to Z (E_{2})$ , and $P = E_{1} \circ_{α} E_{2}$ with canonical images $\overset{ˉ}{g}$ for $g \in E_{1}$ and $\overset{ˉ}{h}$ for $h \in E_{2}$ .

[F1]

[F2]

A finite $p$ -group is a finite group whose order has the form $∣ P ∣ = p^{n}$ for some $n \in N$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L1]

[L2]

For a central product of finite groups, $∣ G \circ_{α} H ∣ = ∣ G ∣ ∣ H ∣ / ∣ Z_{1} ∣$ ; without a finiteness hypothesis, its centre is the image of $Z (G) \times Z (H)$ and its derived subgroup is the image of $[G, G] \times [H, H]$ (Order, centre and derived subgroup of a central product).

[L3]

[L4]

For $N ⊴ G$ , the quotient $G / N$ is abelian if and only if $[G, G] \subseteq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

[L5]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

Proof

technique · direct

1.1L1

Each $E_{i}$ is nonabelian, has $Z (E_{i}) = [E_{i}, E_{i}]$ of order $p$ , and has elementary abelian central quotient $E_{i} / Z (E_{i})$ .

1.2L3

The canonical maps embed $E_{1}$ and $E_{2}$ in $P$ ; their images commute elementwise and generate $P$ .

2.1F2L2step 1.1algebra

The central-product formulas give $∣ P ∣ = ∣ E_{1} ∣ ∣ E_{2} ∣ / p$ , which is a power of $p$ ; the centre of $P$ is the image of $Z (E_{1}) \times Z (E_{2})$ and the derived subgroup of $P$ is the image of $[E_{1}, E_{1}] \times [E_{2}, E_{2}]$ . Those two subgroups of $E_{1} \times E_{2}$ coincide by step 1.1, so $Z (P) = [P, P]$ ; the subgroup $Z (E_{1}) \times Z (E_{2})$ has order $p^{2}$ and contains the identified subgroup of order $p$ , so its image has order $p$ .

2.2step 1.1step 1.2

The group $P$ is nonabelian, because the canonical map embeds the nonabelian group $E_{1}$ into it.

3.1L4L5step 1.1step 1.2step 2.1

Since $[P, P] = Z (P)$ , the quotient $P / Z (P)$ is abelian; and for $g \in E_{1}$ , $h \in E_{2}$ the commuting images give $(\overset{ˉ}{g} \overset{ˉ}{h})^{p} = {\overset{ˉ}{g}}^{p} {\overset{ˉ}{h}}^{p} = \overline{g^{p}} \overline{h^{p}}$ , where $g^{p} \in Z (E_{1})$ and $h^{p} \in Z (E_{2})$ because the central quotients are elementary abelian, so $(\overset{ˉ}{g} \overset{ˉ}{h})^{p}$ lies in $Z (P)$ . Every element of $P$ has this form, so every element of the finite abelian $p$ -group $P / Z (P)$ has order dividing $p$ and $P / Z (P)$ is elementary abelian.

4.1F1L1step 2.1step 2.2step 3.1∎

So $P$ is a nonabelian finite $p$ -group with $∣ Z (P) ∣ = p$ and elementary abelian central quotient, which is the second description in the characterisation; hence $P$ is extraspecial.

Remarks

The identification must be along the full centres: if a proper subgroup of $Z (E_{1})$ were identified it would be trivial, the product would be the direct product, and its centre would have order $p^{2}$ . That is the case recorded on the companion page as a special group which is not extraspecial.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre

Definition

Let $P$ be an extraspecial $p$ -group (Special and extraspecial $p$ -groups) and fix a generator $z$ of its centre, so that $Z (P) = ⟨ z ⟩$ has order $p$ . Write

$V : = P / Z (P),$

which is elementary abelian (Three equivalent descriptions of an extraspecial $p$ -group, Elementary abelian $p$ -groups, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ), and give it its canonical $F_{p}$ -vector-space structure (An elementary abelian $p$ -group has a canonical $F_{p}$ -vector-space structure), where $F_{p} = Z / p$ is the field of For every prime $p$ , the two operations on $Z / p$ make it a field and For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, the group operation of $V$ is vector addition and the scalar action is $\overset{ˉ}{a} \cdot \overset{ˉ}{x} = {\overset{ˉ}{x}}^{a}$ . Elements of $V$ are written multiplicatively, $\overset{ˉ}{x}$ denoting the coset $x Z (P)$ ; scalars are written additively.

The commutator pairing of $P$ relative to $z$ is the map

$b_{z} : V \times V \to F_{p}, [x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})} .$

Why the exponent exists and is unique. Every commutator of $P$ lies in $[P, P] = Z (P) = ⟨ z ⟩$ (An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$ , Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , The center $Z (G)$ of a group), so $[x, y] = z^{k}$ for some integer $k$ . Since $z$ has order $p$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, A finite group of prime order is cyclic and every nonidentity element generates it), $z^{k}$ depends only on the class of $k$ in $Z / p$ , and $z^{k} = z^{k^{'}}$ forces $k \equiv k^{'}$ ; so the class $\overset{ˉ}{k} \in F_{p}$ is determined by $[x, y]$ , and that class is $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ .

That the value depends only on the cosets $\overset{ˉ}{x}$ and $\overset{ˉ}{y}$ , and not on the representatives $x$ and $y$ , is proved in The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating ↗.

Remarks

The pairing depends on the choice of $z$ , and only on it: replacing $z$ by $z^{c}$ with $c \neq 0$ replaces $b_{z}$ by $c^{- 1} b_{z}$ . So the radical, the orthogonality relation and every statement about a subspace being self-orthogonal are independent of the choice, while the individual values are not. The companion page records the false statement that no choice is needed.

Nothing here is imported from a theory of bilinear forms. The target $F_{p}$ is the field $Z / p$ , the vector-space structure on $V$ is the canonical scalar action of an elementary abelian $p$ -group, and every property of $b_{z}$ used below is proved from the commutator identities of a group whose derived subgroup is central.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating

Statement

Let $P$ be an extraspecial $p$ -group with $Z (P) = ⟨ z ⟩$ and $V = P / Z (P)$ . The commutator pairing $b_{z}$ is well defined on $V \times V$ : the value of $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ does not depend on the representatives $x$ and $y$ . It is $F_{p}$ -bilinear,

$b_{z} (\overset{ˉ}{x} \overset{ˉ}{y}, \overset{ˉ}{w}) = b_{z} (\overset{ˉ}{x}, \overset{ˉ}{w}) + b_{z} (\overset{ˉ}{y}, \overset{ˉ}{w}), b_{z} (\overset{ˉ}{w}, \overset{ˉ}{x} \overset{ˉ}{y}) = b_{z} (\overset{ˉ}{w}, \overset{ˉ}{x}) + b_{z} (\overset{ˉ}{w}, \overset{ˉ}{y}), b_{z} ({\overset{ˉ}{x}}^{a}, \overset{ˉ}{y}) = a b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) = b_{z} (\overset{ˉ}{x}, {\overset{ˉ}{y}}^{a}),$

and alternating: $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{x}) = 0$ for every $\overset{ˉ}{x} \in V$ . Consequently $b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x}) = - b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ with $Z (P) = ⟨ z ⟩$ of order $p$ , the quotient $V = P / Z (P)$ with its canonical $F_{p}$ -structure, and the pairing $b_{z}$ defined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ .

[F1]

The commutator pairing of $P$ relative to $z$ is the map $b_{z} : V \times V \to F_{p}$ determined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ , where $V = P / Z (P)$ carries the canonical scalar action $\overset{ˉ}{a} \cdot \overset{ˉ}{x} = {\overset{ˉ}{x}}^{a}$ (The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre).

[F2]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F3]

The rule $\overset{ˉ}{a} \cdot x = x^{a}$ gives every elementary abelian $p$ -group its canonical $F_{p}$ -vector-space structure, with the group operation as vector addition and the identity as zero (An elementary abelian $p$ -group has a canonical $F_{p}$ -vector-space structure).

[L1]

An extraspecial $p$ -group is nilpotent of class exactly two, so its derived subgroup is central, and every nonidentity commutator has order $p$ (An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$ ).

[L2]

[L3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[L4]

For every prime $p$ , the operations of addition and multiplication on $Z / p$ make it a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

Proof

technique · direct

1.1L1L2

The derived subgroup of $P$ is central, so the two expansion identities and the power identity are available for all elements of $P$ .

1.2F1F2

Alternation: $[x, x] = x x x^{- 1} x^{- 1} = e = z^{0}$ , so $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{x}) = 0$ .

2.1F1F2L3step 1.1

If $\overset{ˉ}{x} = \overline{x^{'}}$ then $x^{'} = x u$ with $u \in Z (P)$ , and $[x u, y] = [x, y] [u, y] = [x, y]$ because $u$ is central makes $[u, y] = u y u^{- 1} y^{- 1} = e$ ; the second variable is the same computation with the other expansion identity. So $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ depends only on the two cosets.

2.2F1F3step 1.1

Additivity in the first variable: $z^{b_{z} (\overset{ˉ}{x} \overset{ˉ}{y}, \overset{ˉ}{w})} = [x y, w] = [x, w] [y, w] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{w})} z^{b_{z} (\overset{ˉ}{y}, \overset{ˉ}{w})} = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{w}) + b_{z} (\overset{ˉ}{y}, \overset{ˉ}{w})}$ , and exponents of $z$ are determined modulo $p$ ; the second variable is symmetric.

2.3F1F3L4step 1.1

Compatibility with scalars: for an integer $a$ , $z^{b_{z} ({\overset{ˉ}{x}}^{a}, \overset{ˉ}{y})} = [x^{a}, y] = [x, y]^{a} = z^{a b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ , and both sides depend only on $a$ modulo $p$ because $z$ has order $p$ ; the scalar action on $V$ is $\overset{ˉ}{a} \cdot \overset{ˉ}{x} = {\overset{ˉ}{x}}^{a}$ , so this is exactly $b_{z} (a \cdot \overset{ˉ}{x}, \overset{ˉ}{y}) = a b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ , and likewise in the second variable.

3.1step 2.2step 1.2∎

Expanding $0 = b_{z} (\overset{ˉ}{x} \overset{ˉ}{y}, \overset{ˉ}{x} \overset{ˉ}{y})$ by steps 2.2 and 1.2 gives $0 = b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) + b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x})$ , so $b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x}) = - b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ .

Remarks

Alternation is the primitive property and skew symmetry is derived from it, not the other way round. At $p = 2$ the two are not interchangeable: there $- 1 = 1$ , so skew symmetry says only that the pairing is symmetric, and it is the vanishing of $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{x})$ that carries content.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The commutator pairing of an extraspecial $p$ -group has trivial radical

Statement

Let $P$ be an extraspecial $p$ -group with $Z (P) = ⟨ z ⟩$ and $V = P / Z (P)$ , and let $b_{z}$ be its commutator pairing. The radical of $b_{z}$ is trivial: if $\overset{ˉ}{x} \in V$ satisfies $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) = 0$ for every $\overset{ˉ}{y} \in V$ , then $\overset{ˉ}{x}$ is the identity of $V$ . Conversely the identity of $V$ pairs to zero with every element.

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ with $Z (P) = ⟨ z ⟩$ , the quotient $V = P / Z (P)$ , and the commutator pairing $b_{z}$ .

[F1]

The commutator pairing of $P$ relative to $z$ is the map $b_{z} : V \times V \to F_{p}$ determined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ , where $V = P / Z (P)$ (The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

The quotient group $G / N$ has the left cosets $g N$ as elements (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L1]

The commutator pairing is well defined on $V \times V$ , is $F_{p}$ -bilinear, and is alternating (The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating).

[L2]

Proof

technique · direct

1.1F1L1

Let $x \in P$ represent $\overset{ˉ}{x}$ and suppose $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) = 0$ for every $\overset{ˉ}{y} \in V$ . Since every element of $P$ represents some coset, this says $[x, y] = z^{0} = e$ for every $y \in P$ , that is $x y = y x$ for every $y \in P$ .

2.1F2F3L2step 1.1

Hence $x \in Z (P)$ , so $\overset{ˉ}{x} = x Z (P)$ is the identity coset of $V$ .

3.1F1F2F3step 2.1∎

Conversely, if $\overset{ˉ}{x}$ is the identity of $V$ then $x \in Z (P)$ , so $[x, y] = e$ and $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) = 0$ for every $\overset{ˉ}{y}$ .

Remarks

Triviality of the radical, rather than merely its smallness, is exactly the statement that the centre is the whole kernel of the quotient map. It is what lets a single element of $V$ be detected by pairing it against the others, and it is used in that form by every counting argument below.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Two elements of an extraspecial $p$ -group with nontrivial commutator generate an extraspecial subgroup of order $p^{3}$

Statement

Let $P$ be an extraspecial $p$ -group with $Z (P) = ⟨ z ⟩$ , and let $x, y \in P$ satisfy $[x, y] \neq e$ . Then $Q = ⟨ x, y ⟩$ contains $Z (P)$ , has order $p^{3}$ , is nonabelian, and is extraspecial with $Z (Q) = Z (P)$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ with $Z (P) = ⟨ z ⟩$ of order $p$ , the quotient $V = P / Z (P)$ with its commutator pairing $b_{z}$ , and elements $x, y \in P$ with $[x, y] \neq e$ .

[F1]

The commutator pairing of $P$ relative to $z$ is the map $b_{z} : V \times V \to F_{p}$ determined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ (The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

A subset $S$ of an elementary abelian $p$ -group is independent when $\prod_{s \in S} s^{a_{s}} = e$ with finite support forces every $a_{s} = 0$ , and it spans when every element is such a product ( $F_{p}$ -spanning sets, independence, and bases in an elementary abelian $p$ -group).

[F4]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[L1]

[L2]

The commutator pairing is well defined, $F_{p}$ -bilinear and alternating, and satisfies $b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x}) = - b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ (The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating).

[L3]

In a finite group whose order is prime, every $g \neq e$ has order $∣ G ∣$ and generates $G$ (A finite group of prime order is cyclic and every nonidentity element generates it).

[L4]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L5]

If the quotient group $G / Z (G)$ is cyclic, then $G$ is abelian (If $G / Z (G)$ is cyclic, then $G$ is abelian).

[L6]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L7]

The rule $\overset{ˉ}{a} \cdot x = x^{a}$ gives every elementary abelian $p$ -group its canonical $F_{p}$ -vector-space structure (An elementary abelian $p$ -group has a canonical $F_{p}$ -vector-space structure).

[L8]

If $P$ is a finite $p$ -group and $H \leq P$ , then $∣ H ∣ = p^{k}$ for some $k \in N$ (Every subgroup of a finite $p$ -group has order a power of $p$ ).

Proof

technique · direct

1.1F2L1L3L6

The element $[x, y]$ lies in $[P, P] = Z (P)$ and is not the identity, so it generates the group $Z (P)$ of order $p$ ; since $[x, y] \in Q$ , this gives $Z (P) \leq Q$ .

1.2F1

Write $c = b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ ; then $z^{c} = [x, y] \neq e$ , so $c \neq 0$ in $F_{p}$ .

2.1F3L2L7step 1.2

The pair $\overset{ˉ}{x}, \overset{ˉ}{y}$ is independent in $V$ : if ${\overset{ˉ}{x}}^{a} {\overset{ˉ}{y}}^{d}$ is the identity of $V$ , pairing with $\overset{ˉ}{y}$ gives $a c = 0$ and hence $a = 0$ , and pairing with $\overset{ˉ}{x}$ gives $- d c = 0$ and hence $d = 0$ .

3.1F3L4L6L7step 1.1step 2.1

Since $Z (P) \leq Q$ , the image of $Q$ in $V$ is $Q / Z (P)$ , and it equals $⟨ \overset{ˉ}{x}, \overset{ˉ}{y} ⟩ = {{\overset{ˉ}{x}}^{a} {\overset{ˉ}{y}}^{d} : a, d \in F_{p}}$ ; independence makes the $p^{2}$ displayed products pairwise distinct, so $∣ Q / Z (P) ∣ = p^{2}$ and Lagrange gives $∣ Q ∣ = p \cdot p^{2} = p^{3}$ .

4.1F2L3L4L5L8step 1.1step 3.1

$Q$ is nonabelian because $[x, y] \neq e$ , and $Z (P) \leq Z (Q)$ because an element central in $P$ is central in the subgroup $Q$ containing it. The order of $Z (Q)$ is a power of $p$ dividing $p^{3}$ and is not $p^{3}$ ; were it $p^{2}$ , the quotient $Q / Z (Q)$ would have order $p$ and hence be cyclic, forcing $Q$ abelian. So $∣ Z (Q) ∣ = p$ and $Z (Q) = Z (P)$ .

5.1F4L1step 3.1step 4.1∎

The quotient $Q / Z (Q) = Q / Z (P)$ is a subgroup of the elementary abelian group $V$ , hence is itself a finite abelian $p$ -group all of whose nonidentity elements have order $p$ ; so $Q$ is a nonabelian finite $p$ -group with centre of order $p$ and elementary abelian central quotient, and the characterisation makes it extraspecial.

Remarks

The hypothesis is on the pair, not on either element separately: $x$ and $y$ are automatically noncentral, since a central element commutes with everything, but two noncentral elements can commute and then generate an abelian subgroup.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$

Statement

Let $P$ be an extraspecial $p$ -group and $Z (P) = ⟨ z ⟩$ .

Splitting. If $x, y \in P$ satisfy $[x, y] \neq e$ and $F = ⟨ x, y ⟩$ , then $P = F C_{P} (F)$ and $F \cap C_{P} (F) = Z (P)$ , so $F$ and $C_{P} (F)$ form an internal central product of $P$ (Internal central products of a finite family of subgroups, The centralizer $C_{G} (H)$ of a subgroup). Here $F$ is extraspecial of order $p^{3}$ with $Z (F) = Z (P)$ ; and $C_{P} (F)$ has order $∣ P ∣ / p^{2}$ with $Z (C_{P} (F)) = Z (P)$ , and is extraspecial when $∣ P ∣ > p^{3}$ .

Decomposition. There are $n \geq 1$ subgroups $P_{1}, \dots, P_{n}$ of $P$ , each nonabelian of order $p^{3}$ with $Z (P_{i}) = Z (P)$ , which form an internal central product of $P$ ; call such a family admissible. Moreover $∣ P ∣ = p^{1 + 2 n}$ .

Peeling. For an admissible family, $P_{i} \cap P_{j} = Z (P)$ whenever $i \neq j$ ; and when $n \geq 2$ , for each index $i$ the subgroup $C_{i} = ⟨ P_{j} : j \neq i ⟩$ satisfies $[P_{i}, C_{i}] = 1$ , $P = P_{i} C_{i}$ , $P_{i} \cap C_{i} = Z (P)$ and $Z (C_{i}) = Z (P)$ , is extraspecial of order $p^{1 + 2 (n - 1)}$ , and has $(P_{j})_{j \neq i}$ as an admissible family for itself.

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ with $Z (P) = ⟨ z ⟩$ of order $p$ , the quotient $V = P / Z (P)$ and its commutator pairing $b_{z}$ .

[F1]

Subgroups $G_{1}, \dots, G_{r}$ of $G$ form an internal central product when they generate $G$ and $[G_{i}, G_{j}] = 1$ for $i \neq j$ (Internal central products of a finite family of subgroups).

[F2]

[F3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F4]

$C_{G} (H) : = {g \in G : g h = h g for every h \in H}$ , and $C_{G} (H) \cap H = Z (H)$ (The centralizer $C_{G} (H)$ of a subgroup).

[L1]

[L2]

The commutator pairing is well defined, $F_{p}$ -bilinear and alternating, with $b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x}) = - b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ (The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating).

[L3]

The radical of the commutator pairing is trivial (The commutator pairing of an extraspecial $p$ -group has trivial radical).

[L4]

If $[x, y] \neq e$ in an extraspecial $p$ -group $P$ , then $⟨ x, y ⟩$ contains $Z (P)$ , has order $p^{3}$ , is nonabelian, and is extraspecial with centre $Z (P)$ (Two elements of an extraspecial $p$ -group with nontrivial commutator generate an extraspecial subgroup of order $p^{3}$ ).

[L5]

Subgroups $G_{1}, \dots, G_{r}$ form an internal central product of $G$ if and only if the multiplication map $G_{1} \times \dots \times G_{r} \to G$ is a surjective homomorphism; for two factors this identifies the internal product with the external central product along the identity on their intersection (Internal central products are the images of external ones).

[L6]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L7]

For every group $G$ and $x \in G$ , the centralizer $C_{G} (x)$ is a subgroup of $G$ ( $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ).

[L8]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[L9]

For every homomorphism $f : G \to H$ , the rule $g \ker f \mapsto f (g)$ is an isomorphism from $G / \ker f$ onto $im f$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

[L10]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L11]

For every prime $p$ , $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

Proof

technique · induction

1.1F1F3F4L4L6base

If $∣ P ∣ = p^{3}$ the one-member family $P_{1} = P$ is admissible, and $p^{3} = p^{1 + 2 \cdot 1}$ ; moreover any $F = ⟨ x, y ⟩$ with $[x, y] \neq e$ then has order $p^{3}$ and equals $P$ , so $C_{P} (F) = Z (P)$ has order $p = ∣ P ∣ / p^{2}$ and the splitting clause holds.

1.2ih

Assume all three clauses of the theorem hold for every extraspecial $p$ -group of order smaller than $∣ P ∣$ .

1.3F1F3F4L1L5L6L8L9

Suppose $A, B \leq P$ satisfy $[A, B] = 1$ , $P = A B$ , $A \cap B = Z (P)$ and $∣ A ∣ = p^{3}$ . An element of $Z (B)$ commutes with $B$ and, lying in $B$ , with $A$ , hence with $A B = P$ , so $Z (B) \leq Z (P)$ ; and $Z (P) = A \cap B$ lies in $B$ and is central in $P$ , so $Z (B) = Z (P)$ . The multiplication map $μ : A \times B \to P$ is a surjective homomorphism. If $μ (a, b) = e$ , then $a = b^{- 1} \in A \cap B$ , while every $(z, z^{- 1})$ with $z \in A \cap B$ lies in the kernel. Thus $\ker μ$ is the antidiagonal of $A \cap B$ and has order $p$ , so $∣ P ∣ = p^{3} ∣ B ∣ / p$ and $∣ B ∣ = ∣ P ∣ / p^{2}$ . If $B$ were abelian it would equal $Z (B) = Z (P)$ and $∣ P ∣$ would be $p^{3}$ ; so for $∣ P ∣ > p^{3}$ the group $B$ is nonabelian, and $B / Z (B) = B / Z (P)$ is a subgroup of the elementary abelian group $V$ , hence elementary abelian, so $B$ is extraspecial.

1.4F2L4L10L11

If $[x, y] \neq e$ then $F = ⟨ x, y ⟩$ is extraspecial of order $p^{3}$ with $Z (F) = Z (P)$ , and $β = b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ is a nonzero element of the field $F_{p}$ , hence invertible.

1.5F1F3F4L7L10

Let $(P_{1}, \dots, P_{n})$ be an admissible family and fix an index $i$ , writing $C_{i} = ⟨ P_{j} : j \neq i ⟩$ . Each generator of $C_{i}$ commutes with every element of $P_{i}$ , and the centraliser of an element is a subgroup, so $[P_{i}, C_{i}] = 1$ and $C_{i} \leq C_{P} (P_{i})$ ; then $P_{i} C_{i}$ is a subgroup containing every member of the family, so $P = P_{i} C_{i}$ ; and $P_{i} \cap C_{i} \leq P_{i} \cap C_{P} (P_{i}) = Z (P_{i}) = Z (P)$ , while $Z (P) = Z (P_{j}) \leq P_{j} \leq C_{i}$ gives the reverse inclusion. Taking $C_{i}$ to be a single $P_{j}$ shows $P_{i} \cap P_{j} = Z (P)$ .

2.1F2L2L11step 1.4

For $g \in P$ put $s = β^{- 1} b_{z} (\overset{ˉ}{g}, \overset{ˉ}{y})$ and $r = - β^{- 1} b_{z} (\overset{ˉ}{g}, \overset{ˉ}{x})$ and $u = x^{s} y^{r}$ . Bilinearity and alternation give $b_{z} (\overset{ˉ}{u}, \overset{ˉ}{x}) = r b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x}) = - r β = b_{z} (\overset{ˉ}{g}, \overset{ˉ}{x})$ and $b_{z} (\overset{ˉ}{u}, \overset{ˉ}{y}) = s b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) = s β = b_{z} (\overset{ˉ}{g}, \overset{ˉ}{y})$ , so $b_{z} (\overline{g u^{- 1}}, \overset{ˉ}{x}) = 0$ and $b_{z} (\overline{g u^{- 1}}, \overset{ˉ}{y}) = 0$ .

2.2F1F3ihstep 1.2step 1.3step 1.5

Applying step 1.3 with $A = P_{i}$ and $B = C_{i}$ gives $Z (C_{i}) = Z (P)$ and $∣ C_{i} ∣ = ∣ P ∣ / p^{2}$ ; when $n \geq 2$ the subgroup $C_{i}$ contains the nonabelian $P_{j}$ , so $∣ P ∣ > p^{3}$ and $C_{i}$ is extraspecial. The members $P_{j}$ with $j \neq i$ generate it, commute pairwise and have centre $Z (C_{i})$ , so they form an admissible family for it; since $C_{i}$ has smaller order, the induction hypothesis applied to that $(n - 1)$ -member family gives $∣ C_{i} ∣ = p^{1 + 2 (n - 1)}$ .

3.1F1F4L7L10step 1.3step 1.4step 2.1

Hence $g u^{- 1}$ commutes with $x$ and with $y$ ; the centraliser of $g u^{- 1}$ is a subgroup containing both, hence contains $F$ , so $g u^{- 1} \in C_{P} (F)$ and $g = (g u^{- 1}) u \in C_{P} (F) F$ . Therefore $P = F C_{P} (F)$ , and $F \cap C_{P} (F) = Z (F) = Z (P)$ , so $F$ and $C_{P} (F)$ form an internal central product of $P$ ; step 1.3 with $A = F$ and $B = C_{P} (F)$ supplies the order of $C_{P} (F)$ , its centre, and that it is extraspecial when $∣ P ∣ > p^{3}$ .

4.1F1F2F3L1L3L10ihstep 1.1step 1.2step 3.1discharge-induction∎

If $∣ P ∣ > p^{3}$ , choose $x \notin Z (P)$ , which exists because $P$ is nonabelian, and then $y$ with $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) \neq 0$ , which exists because the radical is trivial; put $F = ⟨ x, y ⟩$ and $C = C_{P} (F)$ . By step 3.1 the group $C$ is extraspecial of order $∣ P ∣ / p^{2}$ , so the induction hypothesis gives an admissible family $P_{2}, \dots, P_{n}$ for $C$ with $∣ C ∣ = p^{1 + 2 (n - 1)}$ . Then $F, P_{2}, \dots, P_{n}$ generate $P$ , commute pairwise, are nonabelian of order $p^{3}$ and have centre $Z (C) = Z (P)$ , so they form an admissible family for $P$ , and $∣ P ∣ = p^{2} ∣ C ∣ = p^{1 + 2 n}$ ; with step 1.1 this completes the induction.

Remarks

The factors are not canonical: the subgroup $F$ depends on the choice of $x$ and of a partner $y$ , and the companion page records two decompositions of one group with different factors. What the order formula does fix is the number of factors.

The splitting and peeling clauses are stated for an arbitrary noncommuting pair and an arbitrary admissible family because the classification arguments need to remove a factor of a prescribed isomorphism type, not the one this proof happens to construct.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$

Statement

Let $P$ be an extraspecial $p$ -group. Then $∣ P ∣ = p^{1 + 2 n}$ for some integer $n \geq 1$ , and $∣ P / Z (P) ∣ = p^{2 n}$ . In particular no extraspecial group has order $p^{2 m}$ , and none has order $p$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ .

[F1]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[L1]

Every extraspecial $p$ -group is an internal central product of $n \geq 1$ nonabelian subgroups of order $p^{3}$ with pairwise intersections $Z (P)$ , and $∣ P ∣ = p^{1 + 2 n}$ (Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$ ).

[L2]

[L3]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L4]

$[G : H] : = ∣ G / H ∣$ , the number of left cosets of $H$ in $G$ (The coset set $G / H$ and the index $[G : H]$ of a subgroup).

Proof

technique · direct

1.1L1

The decomposition theorem writes $P$ as an internal central product of $n \geq 1$ nonabelian subgroups of order $p^{3}$ and gives $∣ P ∣ = p^{1 + 2 n}$ .

1.2F1L2

The centre of $P$ has order $p$ .

2.1L3L4step 1.1step 1.2

Lagrange applied to $Z (P) \leq P$ gives $∣ P / Z (P) ∣ = ∣ P ∣ / p = p^{2 n}$ .

3.1step 1.1step 2.1∎

Since $1 + 2 n$ is odd, no extraspecial group has order an even power of $p$ ; and $n \geq 1$ excludes the order $p$ , which corresponds to $n = 0$ .

Remarks

The exponent $n$ is determined by the order and therefore by the group, so it can be used as an invariant even though the decomposition producing it is not unique. That $n \geq 1$ is what the nonabelian clause of the definition buys: an abelian group with a centre of order $p$ would be the cyclic group of order $p$ , of order $p^{1 + 2 \cdot 0}$ .

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

An extraspecial $p$ -group of order $p^{1 + 2 n}$ has generator rank $2 n$

Statement

Let $P$ be an extraspecial $p$ -group of order $p^{1 + 2 n}$ . Then $Φ (P) = Z (P)$ has order $p$ , the Frattini quotient $P / Φ (P)$ has order $p^{2 n}$ , the generator rank is $d (P) = 2 n$ , and every minimal generating set of $P$ has exactly $2 n$ elements.

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ of order $p^{1 + 2 n}$ .

[F1]

For a finite $p$ -group $P$ , the generator rank $d (P)$ is the common size of a basis of $P / Φ (P)$ (The generator rank $d (P)$ of a finite $p$ -group).

[F2]

A subset $S$ of an elementary abelian $p$ -group spans when every element is a product $\prod_{s \in S} s^{a_{s}}$ with coefficients in $F_{p}$ , and is independent when such a product is the identity only for zero coefficients; a basis is an independent spanning subset ( $F_{p}$ -spanning sets, independence, and bases in an elementary abelian $p$ -group).

[L1]

[L2]

An extraspecial $p$ -group has $∣ P ∣ = p^{1 + 2 n}$ with $n \geq 1$ and $∣ P / Z (P) ∣ = p^{2 n}$ (An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$ ).

[L3]

Every finite elementary abelian $p$ -group has a basis; every independent subset extends to a basis, every spanning subset contains a basis, and all bases have the same finite size (Finite elementary abelian $p$ -groups have bases, basis extension, and a well-defined dimension).

[L4]

A subset $X$ of a finite $p$ -group $P$ is a minimal generating set if and only if the quotient map restricts to a bijection from $X$ onto a basis of $P / Φ (P)$ (Burnside Basis Theorem).

[L5]

Proof

technique · direct

1.1L1

By the third description in the characterisation, $Φ (P) = Z (P)$ has order $p$ , so the Frattini quotient is the central quotient.

1.2L1L2

The central quotient has order $p^{2 n}$ and is elementary abelian.

2.1F2L3L5step 1.1step 1.2

It therefore has a basis, and all of its bases have the same size, say $k$ ; independence and spanning make the map sending a coefficient family $(a_{1}, \dots, a_{k})$ to the product of the corresponding powers a bijection from the $p^{k}$ coefficient families onto the group, so $p^{k} = p^{2 n}$ and $k = 2 n$ .

3.1F1L4step 2.1∎

Hence $d (P) = 2 n$ by the definition of the generator rank, and by the Burnside basis theorem a minimal generating set of $P$ is carried bijectively onto a basis of $P / Φ (P)$ , so it has $2 n$ elements.

Remarks

The two clauses say different things. The first is about the quotient and is a count of a basis; the second is about $P$ itself and needs the Burnside basis theorem, because a generating set of $P$ of size $2 n$ could a priori collapse in the quotient. It is the restricted-bijection clause of that theorem which rules that out.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A subgroup of the central quotient and its orthogonal complement have orders multiplying to the order of the quotient

Statement

Let $P$ be an extraspecial $p$ -group of order $p^{1 + 2 n}$ , let $V = P / Z (P)$ and let $b_{z}$ be its commutator pairing. For a subgroup $U \leq V$ put

$U^{⊥} : = {v \in V : b_{z} (v, u) = 0 for every u \in U} .$

Then $U^{⊥}$ is a subgroup of $V$ and

$∣ U ∣ ∣ U^{⊥} ∣ = ∣ V ∣ = p^{2 n} .$

This is a statement about this pairing on this quotient, proved by counting inside $V$ ; no theory of bilinear forms on a vector space is used.

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ of order $p^{1 + 2 n}$ , the quotient $V = P / Z (P)$ with its commutator pairing $b_{z}$ , and a subgroup $U \leq V$ .

[F1]

[F2]

[F3]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[L1]

The commutator pairing is well defined, $F_{p}$ -bilinear and alternating (The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating).

[L2]

The radical of the commutator pairing is trivial (The commutator pairing of an extraspecial $p$ -group has trivial radical).

[L3]

An extraspecial $p$ -group has $∣ P ∣ = p^{1 + 2 n}$ with $n \geq 1$ and $∣ P / Z (P) ∣ = p^{2 n}$ (An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$ ).

[L4]

[L5]

[L6]

For every homomorphism $f : G \to H$ , the rule $g \ker f \mapsto f (g)$ is an isomorphism from $G / \ker f$ onto $im f$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

[L7]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1F3L3L5

The group $V$ is elementary abelian of order $p^{2 n}$ , and $U$ is a subgroup of it, hence also a finite abelian $p$ -group whose nonidentity elements have order $p$ .

1.2F1L1

For each fixed $u \in V$ the map $v \mapsto b_{z} (v, u)$ is a homomorphism from $V$ to the additive group of $F_{p}$ , by additivity of $b_{z}$ in its first variable; $U^{⊥}$ is the intersection of the kernels of these maps as $u$ runs over $U$ , hence a subgroup of $V$ .

2.1F2L4L5step 1.1

Choose a basis $u_{1}, \dots, u_{k}$ of $U$ and extend it to a basis $u_{1}, \dots, u_{m}$ of $V$ . Unique representation in a basis makes the assignment of coefficient families to elements a bijection, so $∣ U ∣ = p^{k}$ and $p^{m} = ∣ V ∣ = p^{2 n}$ , whence $m = 2 n$ .

3.1F1F2L1L2step 2.1

Define $Ψ : V \to V$ by $Ψ (v) = \prod_{i = 1}^{m} u_{i}^{b_{z} (v, u_{i})}$ . It is a homomorphism, again by additivity in the first variable. If $Ψ (v)$ is the identity then independence of the basis forces $b_{z} (v, u_{i}) = 0$ for every $i$ , and then additivity in the second variable gives $b_{z} (v, w) = 0$ for every $w \in V$ , so $v$ is the identity by triviality of the radical. Thus $Ψ$ is injective, hence bijective because $V$ is finite.

4.1F2step 2.1step 3.1

Let $π : V \to U$ send $\prod_{i = 1}^{m} u_{i}^{a_{i}}$ to $\prod_{i = 1}^{k} u_{i}^{a_{i}}$ ; unique representation makes $π$ a well-defined homomorphism, and it is surjective onto $U$ because it fixes each $u_{i}$ with $i \leq k$ . The composite $Φ = π \circ Ψ$ is therefore a surjective homomorphism from $V$ onto $U$ , and $Φ (v)$ is the identity exactly when $b_{z} (v, u_{i}) = 0$ for every $i \leq k$ , which by additivity in the second variable is exactly $v \in U^{⊥}$ .

5.1L6L7step 2.1step 4.1∎

The first isomorphism theorem and Lagrange applied to $Φ$ give $∣ V ∣ = ∣ U^{⊥} ∣ ∣ U ∣$ , that is $∣ U ∣ ∣ U^{⊥} ∣ = p^{2 n}$ .

Remarks

The two ends of the range behave as the formula predicts and are worth naming. For the trivial subgroup the formula reads $∣ V ∣ = ∣ V ∣$ , since the perpendicular of the trivial subgroup is all of $V$ ; for $U = V$ it reads $∣ V^{⊥} ∣ = 1$ , which is triviality of the radical.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

In an extraspecial $p$ -group of order $p^{1 + 2 n}$ every maximal abelian subgroup has order $p^{1 + n}$

Statement

Let $P$ be an extraspecial $p$ -group of order $p^{1 + 2 n}$ . Call an abelian subgroup of $P$ maximal abelian when it is not properly contained in any abelian subgroup of $P$ . Then maximal abelian subgroups exist, every one of them contains $Z (P)$ , and every one of them has order $p^{1 + n}$ . Under the correspondence $A \mapsto A / Z (P)$ they are exactly the subgroups $A$ with $Z (P) \leq A \leq P$ whose image $U = A / Z (P)$ in $V = P / Z (P)$ satisfies $U = U^{⊥}$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ of order $p^{1 + 2 n}$ , the quotient $V = P / Z (P)$ with its commutator pairing $b_{z}$ , and for $U \leq V$ the subgroup $U^{⊥} = {v \in V : b_{z} (v, u) = 0 for all u \in U}$ .

[F1]

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L1]

[L2]

An extraspecial $p$ -group has $∣ P ∣ = p^{1 + 2 n}$ with $n \geq 1$ and $∣ P / Z (P) ∣ = p^{2 n}$ (An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$ ).

[L3]

For a subgroup $U$ of $V$ , $∣ U ∣ ∣ U^{⊥} ∣ = ∣ V ∣ = p^{2 n}$ (A subgroup of the central quotient and its orthogonal complement have orders multiplying to the order of the quotient).

[L4]

For $N ⊴ G$ the maps $H \mapsto H / N$ and $K \mapsto π^{- 1} (K)$ are inverse inclusion-preserving bijections between subgroups $H$ with $N \leq H \leq G$ and subgroups $K \leq G / N$ (Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

[L5]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L6]

Proof

technique · direct

1.1F2algebra

If $B \leq P$ is abelian then $B Z (P)$ is a subgroup, because $Z (P)$ is normal, and it is abelian, because $(b u) (b^{'} u^{'}) = b b^{'} u u^{'} = (b^{'} u^{'}) (b u)$ for $u, u^{'} \in Z (P)$ ; it contains $B$ . So a maximal abelian subgroup equals $B Z (P)$ and contains $Z (P)$ .

1.2F1L1

For $Z (P) \leq A \leq P$ with image $U = A / Z (P)$ : two elements $a, a^{'}$ of $A$ commute exactly when $[a, a^{'}] = e$ , that is exactly when $b_{z} (\overset{ˉ}{a}, {\overset{ˉ}{a}}^{'}) = 0$ ; so $A$ is abelian exactly when $U \leq U^{⊥}$ .

1.3F2L4L5L6

The correspondence $A \mapsto A / Z (P)$ is an inclusion-preserving bijection between the subgroups of $P$ containing $Z (P)$ and the subgroups of $V$ , and $∣ A ∣ = p ∣ A / Z (P) ∣$ by Lagrange.

1.4L2L3

If $U \leq U^{⊥}$ then $∣ U ∣^{2} \leq ∣ U ∣ ∣ U^{⊥} ∣ = p^{2 n}$ , so $∣ U ∣ \leq p^{n}$ .

1.5L2

The trivial subgroup satisfies $U \leq U^{⊥}$ , and $V$ is finite, so among the subgroups with $U \leq U^{⊥}$ there is one of largest order, and it is maximal with that property.

2.1step 1.1step 1.2step 1.3

Combining the previous three observations, $A \mapsto A / Z (P)$ carries the maximal abelian subgroups of $P$ bijectively onto the subgroups $U$ of $V$ that are maximal subject to $U \leq U^{⊥}$ .

2.2F1F3L1L3step 1.4

Let $U$ be maximal subject to $U \leq U^{⊥}$ and suppose $U \neq U^{⊥}$ . Pick $v \in U^{⊥}$ with $v \notin U$ and set $U^{'} = ⟨ U, v ⟩$ , whose elements are the products $u v^{a}$ . For such elements, $b_{z} (u v^{a}, u^{'} v^{a^{'}}) = b_{z} (u, u^{'}) + a^{'} b_{z} (u, v) + a b_{z} (v, u^{'}) + a a^{'} b_{z} (v, v) = 0$ , using $U \leq U^{⊥}$ , the choice of $v$ , skew symmetry and alternation. So $U^{'} \leq U^{' ⊥}$ and $U^{'}$ properly contains $U$ , contradicting maximality. Hence $U = U^{⊥}$ and $∣ U ∣^{2} = p^{2 n}$ , so $∣ U ∣ = p^{n}$ .

3.1step 1.5step 2.1step 2.2∎

Therefore maximal abelian subgroups exist, each contains $Z (P)$ , each corresponds to a subgroup $U$ with $U = U^{⊥}$ of order $p^{n}$ , and each has order $p \cdot p^{n} = p^{1 + n}$ .

Remarks

Maximality is under inclusion, not merely maximality of order, and the two agree here only because step 2.2 shows every maximal self-orthogonal subgroup has the same order. That is what makes the conclusion a statement about every maximal abelian subgroup rather than about a largest one.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

An extraspecial $p$ -group is the product of two maximal abelian subgroups meeting in its centre

Statement

Let $P$ be an extraspecial $p$ -group of order $p^{1 + 2 n}$ . Then there are maximal abelian subgroups $A$ and $B$ of $P$ , each of order $p^{1 + n}$ , with

$P = A B, A \cap B = Z (P), [A, B] = Z (P) .$

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ of order $p^{1 + 2 n}$ with $Z (P) = ⟨ z ⟩$ , the quotient $V = P / Z (P)$ and its commutator pairing $b_{z}$ .

[F1]

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[F4]

For subgroups $A, B \leq G$ , $[A, B] = ⟨ [a, b] : a \in A, b \in B ⟩$ (Subgroup commutators and the lower central series).

[L1]

Every extraspecial $p$ -group is an internal central product of $n \geq 1$ nonabelian subgroups $P_{1}, \dots, P_{n}$ of order $p^{3}$ with $Z (P_{i}) = Z (P)$ and $P_{i} \cap P_{j} = Z (P)$ for $i \neq j$ (Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$ , Internal central products of a finite family of subgroups).

[L2]

The commutator pairing is well defined, $F_{p}$ -bilinear and alternating (The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating).

[L3]

For a subgroup $U$ of $V$ , $∣ U ∣ ∣ U^{⊥} ∣ = ∣ V ∣ = p^{2 n}$ (A subgroup of the central quotient and its orthogonal complement have orders multiplying to the order of the quotient).

[L4]

Maximal abelian subgroups of $P$ contain $Z (P)$ , have order $p^{1 + n}$ , and under the correspondence $A \mapsto A / Z (P)$ are exactly the subgroups containing $Z (P)$ whose image $U$ satisfies $U = U^{⊥}$ (In an extraspecial $p$ -group of order $p^{1 + 2 n}$ every maximal abelian subgroup has order $p^{1 + n}$ ).

[L5]

In a finite group whose order is prime, every $g \neq e$ has order $∣ G ∣$ and generates $G$ (A finite group of prime order is cyclic and every nonidentity element generates it).

[L6]

[L7]

If $H \leq G$ and $N ⊴ G$ , then $H / (H \cap N) ≅ H N / N$ (Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$ ).

[L8]

For every group $G$ and $x \in G$ , the centralizer $C_{G} (x)$ is a subgroup of $G$ ( $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ).

[L9]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L10]

Proof

technique · direct

1.1F3L1

Fix an internal central product decomposition $P = ⟨ P_{1}, \dots, P_{n} ⟩$ with each $P_{i}$ nonabelian of order $p^{3}$ , $Z (P_{i}) = Z (P)$ , $[P_{i}, P_{j}] = 1$ for $i \neq j$ , and $P_{i} \cap P_{j} = Z (P)$ .

2.1F1F2L2L5L10step 1.1

Each $P_{i}$ is nonabelian, so it contains elements with nontrivial commutator; choose $x_{i}, w_{i} \in P_{i}$ with $[x_{i}, w_{i}] \neq e$ . That commutator lies in $Z (P)$ and is not the identity, so it generates $Z (P)$ and equals $z^{c_{i}}$ with $c_{i} \neq 0$ ; putting $y_{i} = w_{i}^{t_{i}}$ with $t_{i} c_{i} = 1$ in $F_{p}$ gives $[x_{i}, y_{i}] = z$ .

3.1F1L2step 1.1step 2.1

The pairing values are $b_{z} ({\overset{ˉ}{x}}_{i}, {\overset{ˉ}{y}}_{i}) = 1$ , while $b_{z} ({\overset{ˉ}{x}}_{i}, {\overset{ˉ}{y}}_{j}) = 0$ , $b_{z} ({\overset{ˉ}{x}}_{i}, {\overset{ˉ}{x}}_{j}) = 0$ and $b_{z} ({\overset{ˉ}{y}}_{i}, {\overset{ˉ}{y}}_{j}) = 0$ whenever the two elements lie in different factors or are equal, because distinct factors commute elementwise and the pairing is alternating.

4.1F2F3L8step 2.1step 3.1

Put $A = ⟨ Z (P), x_{1}, \dots, x_{n} ⟩$ and $B = ⟨ Z (P), y_{1}, \dots, y_{n} ⟩$ . Their generators commute pairwise, so the centraliser of each generator is a subgroup containing all of them and hence contains $A$ (respectively $B$ ); therefore each generator is central in $A$ (respectively $B$ ), and $A$ and $B$ are abelian.

5.1F3L2step 3.1step 4.1

The images $\overset{ˉ}{A} = ⟨ {\overset{ˉ}{x}}_{1}, \dots, {\overset{ˉ}{x}}_{n} ⟩$ and $\overset{ˉ}{B} = ⟨ {\overset{ˉ}{y}}_{1}, \dots, {\overset{ˉ}{y}}_{n} ⟩$ consist of the products $\prod_{i} {\overset{ˉ}{x}}_{i}^{a_{i}}$ and $\prod_{i} {\overset{ˉ}{y}}_{i}^{c_{i}}$ . If $\prod_{i} {\overset{ˉ}{x}}_{i}^{a_{i}}$ is the identity, pairing with ${\overset{ˉ}{y}}_{k}$ gives $a_{k} = 0$ , so the ${\overset{ˉ}{x}}_{i}$ are independent and $∣ \overset{ˉ}{A} ∣ = p^{n}$ ; the same argument with the roles exchanged gives $∣ \overset{ˉ}{B} ∣ = p^{n}$ . If $\prod_{i} {\overset{ˉ}{x}}_{i}^{a_{i}} = \prod_{j} {\overset{ˉ}{y}}_{j}^{c_{j}}$ , pairing with ${\overset{ˉ}{y}}_{k}$ gives $a_{k} = 0$ and pairing with ${\overset{ˉ}{x}}_{k}$ gives $c_{k} = 0$ , so $\overset{ˉ}{A} \cap \overset{ˉ}{B}$ is trivial.

6.1L3L4L9step 4.1step 5.1

Since $A$ is abelian its image satisfies $\overset{ˉ}{A} \leq {\overset{ˉ}{A}}^{⊥}$ , and the counting formula gives $∣ {\overset{ˉ}{A}}^{⊥} ∣ = p^{2 n} / p^{n} = p^{n} = ∣ \overset{ˉ}{A} ∣$ , so $\overset{ˉ}{A} = {\overset{ˉ}{A}}^{⊥}$ and $A$ is maximal abelian of order $p^{1 + n}$ ; likewise for $B$ .

7.1L6L7L9step 5.1step 6.1

In the abelian group $V$ the product $\overset{ˉ}{A} \overset{ˉ}{B}$ is a subgroup and the second isomorphism theorem gives $∣ \overset{ˉ}{A} \overset{ˉ}{B} ∣ = ∣ \overset{ˉ}{A} ∣ ∣ \overset{ˉ}{B} ∣ / ∣ \overset{ˉ}{A} \cap \overset{ˉ}{B} ∣ = p^{2 n} = ∣ V ∣$ , so $\overset{ˉ}{A} \overset{ˉ}{B} = V$ ; lifting along the correspondence, every $g \in P$ has $\overset{ˉ}{g} = \overset{ˉ}{a} \overset{ˉ}{b}$ , hence $g \in a b Z (P) \subseteq A B$ because $Z (P) \leq A$ , and $P = A B$ . The correspondence also gives $(A \cap B) / Z (P) = \overset{ˉ}{A} \cap \overset{ˉ}{B}$ trivial, so $A \cap B = Z (P)$ .

8.1F4L5L10step 2.1step 7.1∎

Finally $[A, B]$ is generated by commutators of elements of $P$ , so it lies in $[P, P] = Z (P)$ ; and it contains $[x_{1}, y_{1}] = z$ , which generates $Z (P)$ . Hence $[A, B] = Z (P)$ .

Remarks

The two subgroups are built from a chosen decomposition and are not canonical; another decomposition may yield the same pair or a different one. What the statement fixes is that a factorisation of this shape exists, with both factors as large as an abelian subgroup can be.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The Heisenberg group of order $p^{3}$ over $Z / p$

Definition

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and let $Z / p$ carry the addition and multiplication of For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, which make it a field (For every prime $p$ , the two operations on $Z / p$ make it a field). The Heisenberg group of order $p^{3}$ is the set

$H_{p} : = {(a, b, c) : a, b, c \in Z / p}$

with the multiplication

$(a, b, c) (a^{'}, b^{'}, c^{'}) : = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'}) .$

That this is a group law (Group and abelian group), that it is not commutative, and that $H_{p}$ has $p^{3}$ elements are proved in The Heisenberg multiplication is a group law, nonabelian, on a set of $p^{3}$ elements ↗.

Remarks

The same multiplication is written on the published example The finite Heisenberg group is the unique Sylow $p$ -subgroup of its coordinate upper-triangular group, where the triple $(a, b, c)$ records the three entries above the diagonal of a unipotent upper-triangular $3 \times 3$ matrix over $Z / p$ ; the two constructions produce the same group, and the notation is kept identical so that a reader meets one object rather than two.

The asymmetry of the third coordinate, $c + c^{'} + a b^{'}$ rather than $c + c^{'} + a^{'} b$ , is a choice of convention: the opposite choice gives the group with the roles of the first two coordinates exchanged, and the map $(a, b, c) \mapsto (b, a, c)$ carries one to the other. Nothing below depends on which is taken, provided one is taken throughout.

At $p = 2$ the construction does not produce a group of exponent two: the element $(1, 1, 0)$ squares to $(0, 0, 1)$ , so $H_{2}$ has an element of order four.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The Heisenberg multiplication is a group law, nonabelian, on a set of $p^{3}$ elements

Statement

Let $p$ be a prime. The multiplication $(a, b, c) (a^{'}, b^{'}, c^{'}) = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'})$ makes the set $H_{p}$ a group with identity $(0, 0, 0)$ and inverse $(a, b, c)^{- 1} = (- a, - b, - c + a b)$ ; the group is not abelian; and $∣ H_{p} ∣ = p^{3}$ . Moreover $(1, 0, 0)^{a} = (a, 0, 0)$ , $(0, 1, 0)^{b} = (0, b, 0)$ and $(0, 0, 1)^{c} = (0, 0, c)$ for all $a, b, c$ , so those three elements generate $H_{p}$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups) and each has order $p$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

Facts & Assumptions

Given: A prime $p$ and the set $H_{p}$ with the multiplication above.

[F1]

The Heisenberg group of order $p^{3}$ is the set $H_{p} = {(a, b, c) : a, b, c \in Z / p}$ with $(a, b, c) (a^{'}, b^{'}, c^{'}) : = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'})$ (The Heisenberg group of order $p^{3}$ over $Z / p$ ).

[F2]

A group is a set with an associative operation having a two-sided identity and two-sided inverses (Group and abelian group).

[L1]

For every $n \in N$ , $(Z / n, +, [0]_{n})$ is an abelian group, $(Z / n, \cdot, [1]_{n})$ is a commutative monoid, and multiplication distributes over addition on both sides (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L2]

If $A$ and $B$ are finite then $A \times B$ is finite and $∣ A \times B ∣ = ∣ A ∣ \cdot ∣ B ∣$ (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ).

[L3]

$∣ A ∣$ is the unique natural number $n$ with $A \approx n$ (The cardinality $∣ A ∣$ of a finite set).

[L4]

For every prime $p$ , $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

Proof

technique · direct

1.1F1L1algebra

Associativity: both $((a, b, c) (a^{'}, b^{'}, c^{'})) (a^{''}, b^{''}, c^{''})$ and $(a, b, c) ((a^{'}, b^{'}, c^{'}) (a^{''}, b^{''}, c^{''}))$ have first coordinate $a + a^{'} + a^{''}$ , second coordinate $b + b^{'} + b^{''}$ , and third coordinate $c + c^{'} + c^{''} + a b^{'} + a b^{''} + a^{'} b^{''}$ , the two computations differing only in the order in which the three products are formed.

1.2F1L1algebra

The triple $(0, 0, 0)$ is a two-sided identity: $(a, b, c) (0, 0, 0) = (a, b, c + 0 + a \cdot 0) = (a, b, c)$ and $(0, 0, 0) (a, b, c) = (a, b, 0 + c + 0 \cdot b) = (a, b, c)$ .

1.3F1L1algebra

The triple $(- a, - b, - c + a b)$ is a two-sided inverse of $(a, b, c)$ : the product in one order is $(0, 0, c + (- c + a b) + a (- b)) = (0, 0, 0)$ , and in the other it is $(0, 0, (- c + a b) + c + (- a) b) = (0, 0, 0)$ .

1.4F1L2L3algebra

The quotient set $Z / p$ has the $p$ distinct classes $[0]_{p}, \dots, [p - 1]_{p}$ , so it has $p$ elements. The underlying set of $H_{p}$ is the threefold product of those $p$ -element sets, and [L2] therefore gives $∣ H_{p} ∣ = p^{3}$ .

1.5F1L1algebra

The three displayed power formulas hold because $(a, 0, 0) (1, 0, 0) = (a + 1, 0, 0 + 0 + a \cdot 0) = (a + 1, 0, 0)$ , $(0, b, 0) (0, 1, 0) = (0, b + 1, 0 + 0 + 0 \cdot 1) = (0, b + 1, 0)$ and $(0, 0, c) (0, 0, 1) = (0, 0, c + 1)$ , so each power is obtained from the previous one by adding one in the relevant coordinate.

2.1F2step 1.1step 1.2step 1.3

By steps 1.1 to 1.3 the multiplication makes $H_{p}$ a group.

2.2F1L4step 1.4

It is not abelian: $(1, 0, 0) (0, 1, 0) = (1, 1, 1)$ while $(0, 1, 0) (1, 0, 0) = (1, 1, 0)$ , and these differ because $1 \neq 0$ in $Z / p$ for every prime $p$ .

3.1L1step 1.5step 2.1step 2.2∎

Each of $(1, 0, 0)$ , $(0, 1, 0)$ and $(0, 0, 1)$ has $p$ -th power $(0, 0, 0)$ by step 1.5, hence order $p$ since each is not the identity; and $(a, 0, 0) (0, b, 0) (0, 0, c - a b) = (a, b, a b) (0, 0, c - a b) = (a, b, c)$ , so the three elements generate $H_{p}$ .

Remarks

The verification of associativity is where the third coordinate earns its shape: the two bracketings produce the cross terms $a b^{'} + a b^{''}$ and $a b^{'} + a b^{''}$ respectively together with the common term $a^{'} b^{''}$ , and they agree because multiplication in $Z / p$ distributes over addition.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The Heisenberg group of order $p^{3}$ is extraspecial, and for odd $p$ it has exponent $p$

Statement

Let $p$ be a prime and let $H_{p}$ be the Heisenberg group of order $p^{3}$ . Then

$Z (H_{p}) = [H_{p}, H_{p}] = {(0, 0, c) : c \in Z / p},$

a subgroup of order $p$ , and $H_{p}$ is extraspecial. For odd $p$ the exponent of $H_{p}$ is $p$ . At $p = 2$ the exponent is $4$ , since $(1, 1, 0)^{2} = (0, 0, 1)$ .

Facts & Assumptions

Given: A prime $p$ and the Heisenberg group $H_{p}$ .

[F1]

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

[F4]

For a finite group $G$ , $\exp (G) = \min {n \in N : n > 0 and g^{n} = e for every g \in G}$ (The exponent of a finite group).

[L1]

The Heisenberg multiplication is a group law with identity $(0, 0, 0)$ and inverse $(a, b, c)^{- 1} = (- a, - b, - c + a b)$ ; the group is not abelian, $∣ H_{p} ∣ = p^{3}$ , and $(1, 0, 0)^{a} = (a, 0, 0)$ , $(0, 1, 0)^{b} = (0, b, 0)$ , $(0, 0, 1)^{c} = (0, 0, c)$ (The Heisenberg multiplication is a group law, nonabelian, on a set of $p^{3}$ elements).

[L2]

[L3]

For $N ⊴ G$ , the quotient $G / N$ is abelian if and only if $[G, G] \subseteq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

[L4]

If the quotient group $G / Z (G)$ is cyclic, then $G$ is abelian (If $G / Z (G)$ is cyclic, then $G$ is abelian).

[L5]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L6]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[L7]

For an odd prime $p$ and a finite group $G$ with $[G, G] \leq Z (G)$ of exponent dividing $p$ , $(x y)^{p} = x^{p} y^{p}$ for all $x, y$ (For an odd prime $p$ , the $p$ -th power map is a homomorphism on a finite group whose derived subgroup is central of exponent dividing $p$ ).

[L8]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L9]

A finite $p$ -group is a finite group whose order has the form $p^{n}$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

Proof

technique · direct

1.1F1F2L1

A triple $(a, b, c)$ commutes with $(a^{'}, b^{'}, c^{'})$ exactly when $a b^{'} = a^{'} b$ . Taking $(a^{'}, b^{'}, c^{'}) = (0, 1, 0)$ gives $a = 0$ , and taking $(a^{'}, b^{'}, c^{'}) = (1, 0, 0)$ gives $b = 0$ ; conversely every $(0, 0, c)$ commutes with everything. So $Z (H_{p}) = {(0, 0, c)}$ , a subgroup of order $p$ .

1.2F1F3L1

For $g = (a, b, c)$ and $h = (a^{'}, b^{'}, c^{'})$ one has $g h = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'})$ and $h g = (a + a^{'}, b + b^{'}, c + c^{'} + a^{'} b)$ , so $g h = h g \cdot (0, 0, a b^{'} - a^{'} b)$ and therefore $[g, h] = (g h) (h g)^{- 1} = (0, 0, a b^{'} - a^{'} b)$ .

1.3L1L9

$H_{p}$ is a finite $p$ -group of order $p^{3}$ and is not abelian.

2.1F3L8step 1.1step 1.2

Every commutator lies in ${(0, 0, t)}$ , and taking $a = b^{'} = 1$ with $a^{'} = b = 0$ gives the commutator $(0, 0, 1)$ , which generates that subgroup; hence $[H_{p}, H_{p}] = Z (H_{p})$ , of order $p$ .

3.1L3L4L5L6step 2.1step 1.3

By Lagrange the quotient $H_{p} / Z (H_{p})$ has order $p^{2}$ ; it is abelian because $[H_{p}, H_{p}] \leq Z (H_{p})$ , and it is not cyclic, since a cyclic central quotient would make $H_{p}$ abelian. An abelian group of order $p^{2}$ that is not cyclic has no element of order $p^{2}$ , so every one of its nonidentity elements has order $p$ and it is elementary abelian.

4.1L2step 1.1step 1.3step 3.1

By the second description in the characterisation, $H_{p}$ is extraspecial.

5.1F4L1L7L8step 2.1step 4.1

For odd $p$ , the derived subgroup is central of order $p$ , so the $p$ -th power map is a homomorphism; the three generators $(1, 0, 0)$ , $(0, 1, 0)$ and $(0, 0, 1)$ have $p$ -th power the identity, so the $p$ -th power map is trivial on a generating set and hence on $H_{p}$ . Since $H_{p}$ is nontrivial, its exponent is $p$ .

6.1F1F4L1L5step 1.3∎

At $p = 2$ the exponent is not $2$ : $(1, 1, 0)^{2} = (1 + 1, 1 + 1, 0 + 0 + 1 \cdot 1) = (0, 0, 1)$ , which is not the identity, while $(0, 0, 1)^{2} = (0, 0, 0)$ , so $(1, 1, 0)$ has order $4$ and the exponent is $4$ .

Remarks

The group is extraspecial at every prime, $p = 2$ included; only the exponent depends on the parity of $p$ . The step that fails at $p = 2$ is the $p$ -th power homomorphism, whose hypothesis is that $p$ be odd, and the failure is visible in the element $(1, 1, 0)$ of order four.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Raising to the power $1 + p$ is an automorphism of order $p$ of a cyclic group of order $p^{2}$

Statement

Let $p$ be a prime. In $Z / p^{2}$ the class of $1 + p$ is a unit of multiplicative order $p$ , and $(1 + p)^{k} = 1 + k p$ for every $k \in N$ . Consequently, if $A$ is a cyclic group of order $p^{2}$ , the map $x \mapsto x^{1 + p}$ is an automorphism of $A$ of order $p$ (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Facts & Assumptions

Given: A prime $p$ , the ring $Z / p^{2}$ , and a cyclic group $A$ of order $p^{2}$ .

[F1]

[F2]

$ord (g) : = \min {k \in N : k \geq 1 and g^{k} = e}$ when that set is nonempty (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[F3]

$p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L1]

$Aut (C_{n}) ≅ (Z / n)^{\times}$ , and if $C_{n} = ⟨ g ⟩$ the unit class $[a]$ corresponds to the automorphism $g \mapsto g^{a}$ ( $Aut (C_{n}) ≅ (Z / n Z)^{\times}$ ).

[L2]

A cyclic group with a generator of finite order $n$ is isomorphic to $(Z / n, +)$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

Proof

technique · induction

1.1F1base

At $k = 0$ the claim reads $(1 + p)^{0} = 1 = 1 + 0 \cdot p$ in $Z / p^{2}$ .

1.2ih

Assume $(1 + p)^{k} = 1 + k p$ in $Z / p^{2}$ for a given $k \in N$ .

1.3F1F3algebra

In $Z / p^{2}$ the class of $p^{2}$ is zero, so $k p = 0$ holds exactly when $p^{2}$ divides $k p$ , that is exactly when $p$ divides $k$ .

1.4L1L2

An automorphism of a cyclic group of order $p^{2}$ is $x \mapsto x^{a}$ for a unit class $[a]$ of $Z / p^{2}$ , and this correspondence is an isomorphism of groups, so it preserves orders.

2.1F1step 1.2algebra

Then $(1 + p)^{k + 1} = (1 + k p) (1 + p) = 1 + (k + 1) p + k p^{2} = 1 + (k + 1) p$ , the last equality because $p^{2} = 0$ in $Z / p^{2}$ .

3.1F2step 1.3step 1.4step 2.1discharge-induction∎

Hence $(1 + p)^{k} = 1 + k p$ for every $k \in N$ ; by step 1.3 this equals $1$ exactly when $p$ divides $k$ , so the least positive such $k$ is $p$ and the class of $1 + p$ is a unit of order $p$ . Under the correspondence of step 1.4 it is the automorphism $x \mapsto x^{1 + p}$ , which therefore has order $p$ .

Remarks

The computation holds at $p = 2$ as well: there $1 + p = 3$ and $3^{2} = 9 = 1$ in $Z / 4$ , so the automorphism $x \mapsto x^{3} = x^{- 1}$ of a cyclic group of order four is inversion and has order two.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The modular group of order $p^{3}$ as a semidirect product $C_{p^{2}} ⋊ C_{p}$

Definition

Let $p$ be a prime, let $A = ⟨ a ⟩$ be a cyclic group of order $p^{2}$ and let $B = ⟨ s ⟩$ be a cyclic group of order $p$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ). Define

$α : B ⟶ Aut (A), α_{s^{i}} (x) = x^{(1 + p)^{i}} .$

Why this is well defined and is an action by automorphisms. Each $x \mapsto x^{1 + p}$ is an automorphism of $A$ of order $p$ (Raising to the power $1 + p$ is an automorphism of order $p$ of a cyclic group of order $p^{2}$ ), so $α_{s^{i}}$ is the $i$ -th power of that automorphism and is an automorphism. If $s^{i} = s^{j}$ then $p$ divides $i - j$ , and the $p$ -th power of that automorphism is the identity, so $α_{s^{i}} = α_{s^{j}}$ . Finally $α_{s^{i}} \circ α_{s^{j}} = α_{s^{i + j}}$ , so $α$ is a homomorphism into $Aut (A)$ and hence an action by automorphisms (An action of a group $H$ on a group $N$ by automorphisms, Group isomorphisms, automorphisms and the set $Aut (G)$ ).

The modular group of order $p^{3}$ is the external semidirect product ( The external semidirect product $N ⋊_{α} H$ , The semidirect-product multiplication makes $N \times H$ a group)

$M_{p} : = A ⋊_{α} B .$

Writing $a$ and $s$ for the canonical images of the two generators (The canonical copy of $N$ is normal, the canonical copy of $H$ is a complement, and conjugation induces the action), the group is generated by $a$ and $s$ subject to

$a^{p^{2}} = 1, s^{p} = 1, s a s^{- 1} = a^{1 + p} .$

Remarks

Craven writes ${Mod}_{n} (p)$ for the analogous group of order $p^{n}$ and reserves $p_{+}^{1 + 2}$ for the exponent- $p$ group; only the case $n = 3$ is built here. Its relation uses the standard order- $p$ power automorphism $a \mapsto a^{1 + p}$ , the first nonidentity power automorphism congruent to the identity modulo $p$ .

At $p = 2$ the relation reads $s a s^{- 1} = a^{3} = a^{- 1}$ , so the action is inversion and $M_{2}$ is the generalized dihedral group of a cyclic group of order four ( $Dih (C_{n}) = C_{n} ⋊ C_{2}$ with inversion action has order $2 n$ and the dihedral relations).

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The modular group of order $p^{3}$ is extraspecial, of exponent $p^{2}$ when $p$ is odd

Statement

Let $p$ be a prime and let $M_{p} = A ⋊_{α} B$ be the modular group of order $p^{3}$ , with $A = ⟨ a ⟩$ of order $p^{2}$ , $B = ⟨ s ⟩$ of order $p$ and $s a s^{- 1} = a^{1 + p}$ . Then $∣ M_{p} ∣ = p^{3}$ , the group is nonabelian,

$Z (M_{p}) = [M_{p}, M_{p}] = ⟨ a^{p} ⟩$

has order $p$ , and $M_{p}$ is extraspecial of exponent $p^{2}$ . At $p = 2$ the group is the generalized dihedral group $Dih (C_{4})$ .

Facts & Assumptions

Given: A prime $p$ and the modular group $M_{p}$ with its generators $a$ and $s$ .

[F1]

The modular group of order $p^{3}$ is $M_{p} = A ⋊_{α} B$ with $A = ⟨ a ⟩$ of order $p^{2}$ , $B = ⟨ s ⟩$ of order $p$ , and $s a s^{- 1} = a^{1 + p}$ (The modular group of order $p^{3}$ as a semidirect product $C_{p^{2}} ⋊ C_{p}$ ).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F4]

For a finite group $G$ , $\exp (G) = \min {n \in N : n > 0 and g^{n} = e for every g \in G}$ (The exponent of a finite group).

[L1]

In $N ⋊_{α} H$ the sets $\overset{ˉ}{N} = {(n, 1)}$ and $\overset{ˉ}{H} = {(1, h)}$ are subgroups, $\overset{ˉ}{N}$ is normal, $\overset{ˉ}{N} \cap \overset{ˉ}{H} = {(1, 1)}$ , every element has a unique factorisation $(n, 1) (1, h)$ , and $(1, h) (n, 1) (1, h)^{- 1} = (α_{h} (n), 1)$ (The canonical copy of $N$ is normal, the canonical copy of $H$ is a complement, and conjugation induces the action).

[L2]

The class of $1 + p$ in $Z / p^{2}$ is a unit of multiplicative order $p$ , and $(1 + p)^{k} = 1 + k p$ (Raising to the power $1 + p$ is an automorphism of order $p$ of a cyclic group of order $p^{2}$ ).

[L3]

[L4]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L5]

If the quotient group $G / Z (G)$ is cyclic, then $G$ is abelian (If $G / Z (G)$ is cyclic, then $G$ is abelian).

[L6]

For $N ⊴ G$ , the quotient $G / N$ is abelian if and only if $[G, G] \subseteq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

[L7]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[L8]

For $n \geq 1$ , $Dih (C_{n}) = C_{n} ⋊ C_{2}$ where the nonidentity element of $C_{2}$ acts by inversion ( $Dih (C_{n}) = C_{n} ⋊ C_{2}$ with inversion action has order $2 n$ and the dihedral relations).

[L9]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L10]

Proof

technique · direct

1.1F1L1L9

Every element of $M_{p}$ has a unique factorisation $a^{i} s^{j}$ with $0 \leq i < p^{2}$ and $0 \leq j < p$ , so $∣ M_{p} ∣ = p^{2} \cdot p = p^{3}$ and $M_{p}$ is a finite $p$ -group generated by $a$ and $s$ .

1.2F1F3L2L10

The relation gives $[s, a] = s a s^{- 1} a^{- 1} = a^{1 + p} a^{- 1} = a^{p}$ , and $a^{p}$ has order $p$ because $a$ has order $p^{2}$ .

1.3F1F2L1L2L10

The element $a^{p}$ is central: it commutes with $a$ , and $s a^{p} s^{- 1} = (s a s^{- 1})^{p} = a^{(1 + p) p} = a^{p + p^{2}} = a^{p}$ .

2.1F1L2step 1.1

$M_{p}$ is nonabelian, since $s a s^{- 1} = a^{1 + p} \neq a$ : equality would give $a^{p} = 1$ , contradicting that $a$ has order $p^{2}$ .

2.2F3L6L9step 1.1step 1.2step 1.3

Modulo $⟨ a^{p} ⟩$ the images of $a$ and $s$ commute, by step 1.2, and they generate; so $M_{p} / ⟨ a^{p} ⟩$ is abelian and $[M_{p}, M_{p}] \leq ⟨ a^{p} ⟩$ . With $a^{p} = [s, a] \in [M_{p}, M_{p}]$ this gives $[M_{p}, M_{p}] = ⟨ a^{p} ⟩$ , of order $p$ .

3.1F2L4L5step 1.3step 2.1step 2.2

The centre contains $⟨ a^{p} ⟩$ and is not all of $M_{p}$ ; its order divides $p^{3}$ , and an order of $p^{2}$ would leave a quotient of order $p$ , necessarily cyclic, forcing $M_{p}$ abelian. So $Z (M_{p}) = ⟨ a^{p} ⟩$ has order $p$ and equals $[M_{p}, M_{p}]$ .

3.2F4L4L9step 1.1step 2.1

The exponent is $p^{2}$ : the element $a$ has order $p^{2}$ , so the exponent is a multiple of $p^{2}$ dividing $p^{3}$ ; it is not $p^{3}$ , because an element of order $p^{3}$ in a group of order $p^{3}$ would generate it and make it cyclic, hence abelian.

4.1L3L4L5L6L7step 2.1step 3.1

The quotient $M_{p} / Z (M_{p})$ has order $p^{2}$ , is abelian because $[M_{p}, M_{p}] = Z (M_{p})$ , and is not cyclic, so all of its nonidentity elements have order $p$ and it is elementary abelian; by the second description in the characterisation $M_{p}$ is extraspecial.

5.1F1L8L2step 1.1∎

At $p = 2$ the relation reads $s a s^{- 1} = a^{3} = a^{- 1}$ , so the action of $B$ on $A$ is inversion and $M_{2}$ is by definition the semidirect product of a cyclic group of order four by a cyclic group of order two acting by inversion, which is $Dih (C_{4})$ .

Remarks

The group is extraspecial at every prime and its exponent is $p^{2}$ at every prime; what changes at $p = 2$ is only that the resulting group already has a name, since inversion is the unique nontrivial power automorphism of a cyclic group of order four.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

$Dih (C_{4})$ and $Q_{8}$ are extraspecial of order $8$ , with six and two solutions of $x^{2} = 1$ respectively

Statement

Both $Dih (C_{4})$ and $Q_{8}$ are extraspecial groups of order $8$ : each is nonabelian, each has centre equal to its derived subgroup of order two, and each has elementary abelian central quotient. In $Dih (C_{4})$ there are exactly six solutions of $x^{2} = 1$ , and in $Q_{8}$ exactly two.

Facts & Assumptions

Given: The generalized dihedral group $Dih (C_{4}) = ⟨ r ⟩ ⋊ ⟨ s ⟩$ with $r$ of order four, and the quaternion group $Q_{8} = {1, - 1, i, - i, j, - j, k, - k}$ .

[F1]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F2]

[F3]

The generalized dihedral group of an abelian group $A$ is $Dih (A) = A ⋊ C_{2}$ with the nonidentity element of $C_{2}$ acting by inversion ( The generalized dihedral group $Dih (A) = A ⋊ C_{2}$ for an abelian group $A$ ).

[F4]

$Q_{8} : = {1, - 1, i, - i, j, - j, k, - k}$ inside the nonzero quaternions, where $i^{2} = j^{2} = k^{2} = - 1$ , $i j = k$ , $j k = i$ , $k i = j$ , $j i = - k$ , $k j = - i$ and $i k = - j$ (The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ).

[L1]

For $n \geq 1$ , $Dih (C_{n}) = C_{n} ⋊ C_{2}$ has order $2 n$ , and with $C_{n} = ⟨ r ⟩$ and $C_{2} = ⟨ s ⟩$ one has $r^{n} = s^{2} = 1$ , $s r s^{- 1} = r^{- 1}$ , and every element has a unique form $r^{i}$ or $r^{i} s$ with $0 \leq i < n$ ( $Dih (C_{n}) = C_{n} ⋊ C_{2}$ with inversion action has order $2 n$ and the dihedral relations).

[L2]

$Q_{8}$ is a subgroup of the nonzero quaternions with $∣ Q_{8} ∣ = 8$ ; the element $1$ is the only element of order $1$ , $- 1$ is the only element of order $2$ , and each of $\pm i, \pm j, \pm k$ has order $4$ ( $Q_{8}$ is a subgroup of $H^{\times}$ with eight elements, and $- 1$ is its only element of order $2$ ).

[L3]

[L4]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L5]

If the quotient group $G / Z (G)$ is cyclic, then $G$ is abelian (If $G / Z (G)$ is cyclic, then $G$ is abelian).

[L6]

For $N ⊴ G$ , the quotient $G / N$ is abelian if and only if $[G, G] \subseteq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

[L7]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[L8]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L9]

$ord (g)$ is the least $k \geq 1$ with $g^{k} = e$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Proof

technique · direct

1.1F1F3L1

In $D = Dih (C_{4})$ the relation $s r s^{- 1} = r^{- 1}$ gives $s r^{- 1} = r s$ , so $r (r^{i} s) r^{- 1} = r^{i + 1} s r^{- 1} = r^{i + 2} s$ , which differs from $r^{i} s$ because $r^{2} \neq 1$ ; and $r^{i}$ is central exactly when $r^{- i} = r^{i}$ , that is when $r^{2 i} = 1$ , that is when $i$ is $0$ or $2$ . So $Z (D) = {1, r^{2}}$ , of order two, and $D$ is nonabelian of order eight.

1.2F2L1L6L8

In $D$ one has $[s, r] = s r s^{- 1} r^{- 1} = r^{- 2} = r^{2}$ ; modulo $⟨ r^{2} ⟩$ the images of $r$ and $s$ commute and generate, so $[D, D] \leq ⟨ r^{2} ⟩$ and therefore $[D, D] = ⟨ r^{2} ⟩ = Z (D)$ .

1.3L1L9

In $D$ the squares are $1^{2} = 1$ , $(r^{2})^{2} = 1$ , $(r^{i} s)^{2} = r^{i} (s r^{i} s^{- 1}) s^{2} = r^{i} r^{- i} = 1$ for each of the four values of $i$ , while $r$ and $r^{3}$ have order four. So exactly six elements satisfy $x^{2} = 1$ .

1.4F1F4L2

In $Q_{8}$ the element $- 1$ is central, and $i$ is not, since $i j = k$ while $j i = - k$ and these are distinct elements of the eight-element set; the same computation excludes $\pm i, \pm j, \pm k$ , since $- x$ is central exactly when $x$ is. So $Z (Q_{8}) = {1, - 1}$ , of order two, and $Q_{8}$ is nonabelian of order eight.

1.5F2F4L2L6L8

In $Q_{8}$ one has $[i, j] = i j i^{- 1} j^{- 1} = k (- i) (- j) = (- j) (- j) = j^{2} = - 1$ ; modulo ${1, - 1}$ the images of $i$ and $j$ commute and generate, so $[Q_{8}, Q_{8}] = {1, - 1} = Z (Q_{8})$ .

1.6L2L9

In $Q_{8}$ the only solutions of $x^{2} = 1$ are $1$ and $- 1$ , because every other element has order four.

2.1L4L5L6L7step 1.1step 1.2step 1.4step 1.5

For each of the two groups the central quotient has order four by Lagrange, is abelian because the derived subgroup equals the centre, and is not cyclic, since a cyclic central quotient would force commutativity; an abelian group of order four that is not cyclic has all nonidentity elements of order two, hence is elementary abelian.

3.1L3step 1.1step 1.3step 1.4step 1.6step 2.1∎

Each group is therefore a nonabelian group of order $2^{3}$ with centre of order two and elementary abelian central quotient, so the second description in the characterisation makes both extraspecial; the solution counts are those of steps 1.3 and 1.6.

Remarks

The two solution counts are what separate the two groups: an isomorphism would carry solutions of $x^{2} = 1$ to solutions of $x^{2} = 1$ , and six is not two. Nothing about the centres or the derived subgroups distinguishes them, since those agree.

Both sources write the dihedral group of order eight as $D_{8}$ ; this library writes $D_{n}$ for the dihedral group of order $2 n$ , so the group here is $Dih (C_{4})$ , which is $D_{4}$ in that notation.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A nonabelian group of order $p^{3}$ is extraspecial

Statement

Let $p$ be a prime and let $P$ be a nonabelian group of order $p^{3}$ . Then $P$ is extraspecial: $Z (P) = [P, P] = Φ (P)$ has order $p$ and $P / Z (P)$ is elementary abelian of order $p^{2}$ .

Facts & Assumptions

Given: A prime $p$ and a nonabelian group $P$ with $∣ P ∣ = p^{3}$ .

[F1]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F2]

A finite $p$ -group is a finite group whose order has the form $∣ P ∣ = p^{n}$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L1]

If $P$ is a nontrivial finite $p$ -group then $p$ divides $∣ Z (P) ∣$ (Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$ ).

[L2]

If the quotient group $G / Z (G)$ is cyclic, then $G$ is abelian (If $G / Z (G)$ is cyclic, then $G$ is abelian).

[L3]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L4]

If $p$ is prime and $G$ is a group of order $p^{2}$ , then $G$ is abelian (Every group of order $p^{2}$ , for prime $p$ , is abelian).

[L5]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[L6]

[L7]

If $P$ is a finite $p$ -group and $H \leq P$ then $∣ H ∣ = p^{k}$ for some $k$ (Every subgroup of a finite $p$ -group has order a power of $p$ ).

Proof

technique · direct

1.1F1F2L1L3L7

$P$ is a nontrivial finite $p$ -group, so its centre has order a power of $p$ divisible by $p$ ; and $Z (P) \neq P$ because $P$ is nonabelian. So $∣ Z (P) ∣$ is $p$ or $p^{2}$ .

2.1L2L3step 1.1

If $∣ Z (P) ∣ = p^{2}$ then $P / Z (P)$ has order $p$ by Lagrange, hence is cyclic, and $P$ would be abelian. So $∣ Z (P) ∣ = p$ .

3.1L2L3L4L5step 2.1

By Lagrange $P / Z (P)$ has order $p^{2}$ , so it is abelian; it is not cyclic, since that would again force $P$ abelian; and an abelian group of order $p^{2}$ that is not cyclic has every nonidentity element of order $p$ , so it is elementary abelian.

4.1L6step 2.1step 3.1∎

So $P$ is a nonabelian finite $p$ -group with centre of order $p$ and elementary abelian central quotient, which is the second description in the characterisation; hence $P$ is extraspecial and $Z (P) = [P, P] = Φ (P)$ has order $p$ .

Remarks

Order $p^{3}$ is the smallest order at which a nonabelian $p$ -group exists, and the argument shows the extraspecial condition is automatic there. At larger orders it is not: a direct product of two nonabelian groups of order $p^{3}$ is nonabelian of order $p^{6}$ with centre of order $p^{2}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

For each prime there are exactly two nonabelian groups of order $p^{3}$ up to isomorphism

Statement

For every prime $p$ there are exactly two nonabelian groups of order $p^{3}$ up to isomorphism. For odd $p$ they are the Heisenberg group $H_{p}$ , of exponent $p$ , and the modular group $M_{p}$ , of exponent $p^{2}$ . For $p = 2$ they are $Dih (C_{4})$ and $Q_{8}$ .

Facts & Assumptions

Given: A prime $p$ and a nonabelian group $P$ with $∣ P ∣ = p^{3}$ .

[F1]

The Heisenberg group of order $p^{3}$ is the set $H_{p}$ of triples over $Z / p$ with $(a, b, c) (a^{'}, b^{'}, c^{'}) : = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'})$ (The Heisenberg group of order $p^{3}$ over $Z / p$ ).

[F2]

The modular group of order $p^{3}$ is $M_{p} = A ⋊_{α} B$ with $A = ⟨ a ⟩$ of order $p^{2}$ , $B = ⟨ s ⟩$ of order $p$ and $s a s^{- 1} = a^{1 + p}$ (The modular group of order $p^{3}$ as a semidirect product $C_{p^{2}} ⋊ C_{p}$ ).

[F3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F4]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F5]

For a finite group $G$ , $\exp (G) = \min {n \in N : n > 0 and g^{n} = e for every g \in G}$ (The exponent of a finite group).

[F6]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[F7]

[F8]

For a group $G$ and a prime $p$ , $G^{p} = ⟨ g^{p} : g \in G ⟩$ (The $p$ th-power subgroup $G^{p}$ ).

[L1]

A nonabelian group of order $p^{3}$ is extraspecial, with $Z (P) = [P, P] = Φ (P)$ of order $p$ and $P / Z (P)$ elementary abelian of order $p^{2}$ (A nonabelian group of order $p^{3}$ is extraspecial).

[L2]

[L3]

For every finite $p$ -group $P$ , $Φ (P) = P^{'} P^{p}$ ( $Φ (P) = P^{'} P^{p}$ for a finite $p$ -group).

[L4]

[L5]

[L6]

[L7]

$Z (H_{p}) = [H_{p}, H_{p}]$ is the third coordinate axis, of order $p$ ; $H_{p}$ is extraspecial; and for odd $p$ its exponent is $p$ (The Heisenberg group of order $p^{3}$ is extraspecial, and for odd $p$ it has exponent $p$ ).

[L8]

The Heisenberg multiplication is a group law with identity $(0, 0, 0)$ and inverse $(a, b, c)^{- 1} = (- a, - b, - c + a b)$ , the group is nonabelian of order $p^{3}$ , and $(1, 0, 0)^{a} = (a, 0, 0)$ , $(0, 1, 0)^{b} = (0, b, 0)$ , $(0, 0, 1)^{c} = (0, 0, c)$ (The Heisenberg multiplication is a group law, nonabelian, on a set of $p^{3}$ elements).

[L9]

$M_{p}$ is nonabelian of order $p^{3}$ , extraspecial of exponent $p^{2}$ , with $Z (M_{p}) = [M_{p}, M_{p}] = ⟨ a^{p} ⟩$ ; at $p = 2$ it is $Dih (C_{4})$ (The modular group of order $p^{3}$ is extraspecial, of exponent $p^{2}$ when $p$ is odd).

[L10]

$Dih (C_{4})$ and $Q_{8}$ are extraspecial of order $8$ , with exactly six and exactly two solutions of $x^{2} = 1$ respectively ( $Dih (C_{4})$ and $Q_{8}$ are extraspecial of order $8$ , with six and two solutions of $x^{2} = 1$ respectively).

[L11]

The conditions $N ⊴ G$ , $G = N H$ , $N \cap H = {1}$ hold if and only if conjugation restricts to an action $α : H \to Aut (N)$ and $(n, h) \mapsto n h$ is an isomorphism $N ⋊_{α} H \to G$ carrying the canonical factors onto $N$ and $H$ ( Recognition theorem: $G = N H$ with $N ⊴ G$ , $N \cap H = 1$ exactly realises an external semidirect product).

[L12]

For $n \geq 1$ , $Dih (C_{n}) = C_{n} ⋊ C_{2}$ with $r^{n} = s^{2} = 1$ and $s r s^{- 1} = r^{- 1}$ , of order $2 n$ ( $Dih (C_{n}) = C_{n} ⋊ C_{2}$ with inversion action has order $2 n$ and the dihedral relations).

[L13]

$Q_{8} = {1, - 1, i, - i, j, - j, k, - k}$ with $i^{2} = j^{2} = k^{2} = - 1$ , $i j = k$ and $j i = - k$ (The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ).

[L14]

$∣ Q_{8} ∣ = 8$ , the element $- 1$ is its only element of order two, and each of $\pm i, \pm j, \pm k$ has order four ( $Q_{8}$ is a subgroup of $H^{\times}$ with eight elements, and $- 1$ is its only element of order $2$ ).

[L15]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L16]

For all $m, n \in Z$ one has $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L17]

A cyclic group with a generator of finite order $n$ is isomorphic to $(Z / n, +)$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

Proof

technique · cases

1.1F3F4L1L2

$P$ is extraspecial: $Z (P) = [P, P] = Φ (P) = ⟨ z ⟩$ has order $p$ , and $V = P / Z (P)$ is elementary abelian of order $p^{2}$ .

1.2F8L1L3

Since $Φ (P) = [P, P] P^{p}$ contains $P^{p}$ , every $g \in P$ has $g^{p} \in Z (P)$ .

1.3F5F6F7L15L16

Every element order divides $p^{3}$ ; an element of order $p^{3}$ would generate $P$ and make it abelian, so every element has order dividing $p^{2}$ and $\exp (P)$ is $p$ or $p^{2}$ .

1.4F5L7L8L9

For odd $p$ the groups $H_{p}$ and $M_{p}$ are nonabelian of order $p^{3}$ with exponents $p$ and $p^{2}$ , so they are not isomorphic.

1.5L10

The groups $Dih (C_{4})$ and $Q_{8}$ are nonabelian of order eight and have six and two solutions of $x^{2} = 1$ , so they are not isomorphic.

1.6F7L16assume-case expp

First case: suppose $g^{p} = e$ for every $g \in P$ . At $p = 2$ this makes $P$ abelian, since $g h = (g h)^{- 1} = h^{- 1} g^{- 1} = h g$ ; so $p$ is odd here.

1.7F6F7L15L16assume-case exppsq

Second case: suppose instead that $P$ has an element $a$ of order $p^{2}$ . Then $⟨ a ⟩$ has order $p^{2}$ and index $p$ , and $a^{p}$ is a nonidentity element of $Z (P)$ , so $Z (P) = ⟨ a^{p} ⟩$ .

2.1F4F6L4step 1.1step 1.6

In the first case, nonabelianness gives $x, y$ with $[x, y] \neq e$ ; then $z : = [x, y]$ generates $Z (P)$ and $⟨ x, y ⟩$ has order $p^{3}$ , so $P = ⟨ x, y ⟩$ .

2.2F4F6L15step 1.7

In the second case, choose $b \notin ⟨ a ⟩$ ; then $⟨ a, b ⟩$ properly contains a subgroup of index $p$ and so equals $P$ , and $[b, a] \neq e$ since otherwise $P$ would be abelian.

3.1F3F6L1L15step 2.1

In the first case every element of $P$ is uniquely $x^{u} y^{v} z^{w}$ with $u, v, w$ in ${0, \dots, p - 1}$ : the images $\overset{ˉ}{x}, \overset{ˉ}{y}$ generate $V$ , so the products $x^{u} y^{v}$ meet every coset of $Z (P) = {z^{w}}$ , and there are exactly $p^{3}$ such triples of exponents.

3.2F4L5L16step 1.2step 1.7step 2.2

In the second case $[b, a]$ lies in $⟨ a^{p} ⟩$ and is not the identity, so $[b, a] = a^{p k}$ with $p ∤ k$ ; choosing $k^{'}$ with $k k^{'} \equiv 1 (m o d p)$ and replacing $b$ by $b^{k^{'}}$ , which still lies outside $⟨ a ⟩$ because its image in $V$ is nontrivial, gives $[b, a] = a^{p}$ , that is $b a b^{- 1} = a^{1 + p}$ . Moreover $b^{p} \in Z (P) = ⟨ a^{p} ⟩$ , say $b^{p} = a^{p m}$ .

4.1F4L5L16step 1.6step 3.1

In the first case the class-two identities give $y^{v} x^{u^{'}} = [y^{v}, x^{u^{'}}] x^{u^{'}} y^{v} = z^{- u^{'} v} x^{u^{'}} y^{v}$ , so $(x^{u} y^{v} z^{w}) (x^{u^{'}} y^{v^{'}} z^{w^{'}}) = x^{u + u^{'}} y^{v + v^{'}} z^{w + w^{'} - u^{'} v}$ , all exponents read modulo $p$ because $x^{p} = y^{p} = z^{p} = e$ .

4.2L6L16step 3.2assume-case odd

In the second case with $p$ odd, the $p$ -th power map is a homomorphism, so $c = b a^{- m}$ has $c^{p} = b^{p} (a^{- m})^{p} = e$ ; also $c \notin ⟨ a ⟩$ , and $c a c^{- 1} = b a b^{- 1} = a^{1 + p}$ .

4.3L16step 3.2assume-case two

In the second case with $p = 2$ , the element $b^{2}$ lies in ${e, a^{2}}$ , and the relation reads $b a b^{- 1} = a^{3} = a^{- 1}$ .

5.1F1F4L4L7L8step 3.1step 4.1

In $H_{p}$ put $X = (1, 0, 0)$ , $Y = (0, 1, 0)$ and $W = (0, 0, 1)$ . Then $X Y = (1, 1, 1)$ , $Y X = (1, 1, 0)$ and $(Y X)^{- 1} = (- 1, - 1, 1)$ , so $[X, Y] = (X Y) (Y X)^{- 1} = (0, 0, 1) = W$ , which generates $Z (H_{p})$ ; for odd $p$ every element of $H_{p}$ has $p$ -th power the identity, so steps 3.1 and 4.1 hold verbatim in $H_{p}$ with $X, Y, W$ in place of $x, y, z$ .

5.2F2F6L11L15L17step 4.2

In the second case with $p$ odd, $⟨ a ⟩$ is normal because $c$ conjugates it into itself and $a$ normalises it, $⟨ a ⟩ \cap ⟨ c ⟩$ is trivial because $c \notin ⟨ a ⟩$ and $⟨ c ⟩$ has prime order, and $∣ ⟨ a ⟩ ∣ ∣ ⟨ c ⟩ ∣ = p^{3}$ , so $P = ⟨ a ⟩ ⟨ c ⟩$ is an internal semidirect product whose conjugation action sends $a$ to $a^{1 + p}$ ; hence $P ≅ M_{p}$ .

5.3F6L11L12L15L17step 4.3

In the second case with $p = 2$ and $b^{2} = e$ : $⟨ a ⟩$ is normal of index two, $⟨ a ⟩ \cap ⟨ b ⟩$ is trivial, and $b$ acts on $⟨ a ⟩$ by inversion, so $P$ is the internal semidirect product of a cyclic group of order four by a cyclic group of order two acting by inversion, that is $P ≅ Dih (C_{4})$ .

5.4F6L13L14L15L16step 4.3

In the second case with $p = 2$ and $b^{2} = a^{2}$ : the eight elements $a^{u} b^{v}$ with $0 \leq u < 4$ and $v \in {0, 1}$ are distinct and exhaust $P$ , and the relations $a^{4} = e$ , $b^{2} = a^{2}$ and $b a = a^{- 1} b$ determine every product of two of them. The quaternion group satisfies the same three relations with $i$ for $a$ and $j$ for $b$ , since $i^{4} = 1$ , $j^{2} = - 1 = i^{2}$ and $j i = - k = i^{- 1} j$ , and $j \notin ⟨ i ⟩$ , so its eight elements have the same normal form; matching normal forms is therefore an isomorphism and $P ≅ Q_{8}$ .

6.1step 3.1step 4.1step 5.1

In the first case, matching normal forms gives a bijection $H_{p} \to P$ carrying $X^{u} Y^{v} W^{w}$ to $x^{u} y^{v} z^{w}$ , and both products are computed by the same rule, so it is an isomorphism and $P ≅ H_{p}$ .

7.1step 1.3step 1.4step 1.5step 5.2step 5.3step 5.4step 6.1cases-exhaustive∎

The two cases are exhaustive, and within the second the two parities are exhaustive; so for odd $p$ every nonabelian group of order $p^{3}$ is isomorphic to $H_{p}$ or to $M_{p}$ , and for $p = 2$ to $Dih (C_{4})$ or to $Q_{8}$ . With the two non-isomorphy statements this gives exactly two isomorphism classes at every prime.

Remarks

The parity of $p$ enters twice and in opposite directions. It rules out the exponent- $p$ case at $p = 2$ , where it forces commutativity; and it is what allows the correction of the second generator in the exponent- $p^{2}$ case, since that correction is made with the $p$ -th power homomorphism, which is available only for odd $p$ . At $p = 2$ the correction is not available and the two possible values of $b^{2}$ produce the two groups of order eight.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

$Q_{8} \circ Q_{8} ≅ Dih (C_{4}) \circ Dih (C_{4})$

Statement

Let $E = Q_{8} \circ Q_{8}$ be the central product of two copies of the quaternion group along the unique isomorphism between their centres. Then $E$ is also an internal central product of two subgroups isomorphic to $Dih (C_{4})$ meeting in $Z (E)$ , and therefore

$Q_{8} \circ Q_{8} ≅ Dih (C_{4}) \circ Dih (C_{4}) .$

Facts & Assumptions

Given: Two copies $E_{1}, E_{2}$ of $Q_{8}$ , the central product $E = E_{1} \circ_{α} E_{2}$ along the unique isomorphism $α$ of their centres, the canonical images $x_{1}, y_{1}$ of the generators $i, j$ of $E_{1}$ and $x_{2}, y_{2}$ of those of $E_{2}$ , and the common central image $z$ , so that $x_{i}^{2} = y_{i}^{2} = z$ , $z^{2} = 1$ and $y_{i} x_{i} y_{i}^{- 1} = x_{i}^{- 1}$ .

[F1]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F2]

Subgroups $G_{1}, \dots, G_{r}$ of $G$ form an internal central product when they generate $G$ and $[G_{i}, G_{j}] = 1$ for $i \neq j$ (Internal central products of a finite family of subgroups).

[F3]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[F4]

[L1]

$Q_{8} = {1, - 1, i, - i, j, - j, k, - k}$ with $i^{2} = j^{2} = - 1$ and $j i = - k = i^{- 1} j$ (The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ).

[L2]

[L3]

$Dih (C_{4})$ and $Q_{8}$ are extraspecial of order eight ( $Dih (C_{4})$ and $Q_{8}$ are extraspecial of order $8$ , with six and two solutions of $x^{2} = 1$ respectively).

[L4]

The canonical maps into a central product are injective homomorphisms whose images commute elementwise, generate the product, and meet in the image of the identified subgroup (The two canonical maps into a central product are injective homomorphisms whose images commute, generate it, and meet in the identified centre).

[L5]

A central product of two extraspecial $p$ -groups identified along their centres is extraspecial of order $∣ E_{1} ∣ ∣ E_{2} ∣ / p$ (A central product of extraspecial $p$ -groups identified along their centres is extraspecial).

[L6]

Subgroups form an internal central product of $G$ if and only if the multiplication map from their direct product is a surjective homomorphism; for two factors $G ≅ G_{1} \circ_{id} G_{2}$ along the identity of $G_{1} \cap G_{2}$ (Internal central products are the images of external ones).

[L7]

[L8]

The conditions $N ⊴ G$ , $G = N H$ , $N \cap H = {1}$ hold if and only if conjugation restricts to an action and $(n, h) \mapsto n h$ is an isomorphism $N ⋊_{α} H \to G$ ( Recognition theorem: $G = N H$ with $N ⊴ G$ , $N \cap H = 1$ exactly realises an external semidirect product).

[L9]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1F1L2L3L4L5

The two canonical images are isomorphic copies of $Q_{8}$ that commute elementwise, generate $E$ , and meet exactly in $⟨ z ⟩$ ; and $E$ is extraspecial of order $8 \cdot 8 / 2 = 32$ .

1.2F4L1L2

Each $x_{i}$ has order four, since $x_{i}^{2} = z$ and $z^{2} = 1$ with $z \neq 1$ ; likewise each $y_{i}$ .

2.1L1L4step 1.1step 1.2

Put $t_{1} = x_{2} y_{1}$ and $t_{2} = x_{1} y_{2}$ . Then $t_{1}^{2} = x_{2} y_{1} x_{2} y_{1} = x_{2}^{2} y_{1}^{2} = z \cdot z = 1$ , because $x_{2}$ commutes with $y_{1}$ ; likewise $t_{2}^{2} = 1$ .

2.2L1L4step 1.1

Also $t_{1} x_{1} t_{1}^{- 1} = x_{2} (y_{1} x_{1} y_{1}^{- 1}) x_{2}^{- 1} = y_{1} x_{1} y_{1}^{- 1} = x_{1}^{- 1}$ , because $x_{2}$ commutes with everything in the first image; likewise $t_{2} x_{2} t_{2}^{- 1} = x_{2}^{- 1}$ .

2.3F3L4step 1.1step 1.2

Neither $t_{1}$ lies in $⟨ x_{1} ⟩$ nor $t_{2}$ in $⟨ x_{2} ⟩$ : if $x_{2} y_{1}$ were a power of $x_{1}$ then $x_{2}$ would lie in the first canonical image, hence in $⟨ z ⟩$ , contradicting that $x_{2}$ has order four.

3.1F3L7L8L9step 2.1step 2.2step 2.3

Set $H_{1} = ⟨ x_{1}, t_{1} ⟩$ and $H_{2} = ⟨ x_{2}, t_{2} ⟩$ . In $H_{1}$ the cyclic subgroup $⟨ x_{1} ⟩$ of order four is normalised by $t_{1}$ and meets $⟨ t_{1} ⟩$ trivially, and $∣ ⟨ x_{1} ⟩ ∣ ∣ ⟨ t_{1} ⟩ ∣ = 8$ ; so $H_{1}$ is the internal semidirect product of a cyclic group of order four by a group of order two acting by inversion, that is $H_{1} ≅ Dih (C_{4})$ of order eight with $Z (H_{1}) = ⟨ x_{1}^{2} ⟩ = ⟨ z ⟩$ . The same holds for $H_{2}$ .

4.1F2F3L1L4step 1.1step 3.1

The four generators commute in pairs across the two subgroups: $x_{1}$ commutes with $x_{2}$ and with $t_{2} = x_{1} y_{2}$ ; $t_{1} = x_{2} y_{1}$ commutes with $x_{2}$ ; and $t_{1} t_{2} = x_{2} (y_{1} x_{1}) y_{2} = x_{1}^{- 1} x_{2} y_{1} y_{2}$ while $t_{2} t_{1} = x_{1} (y_{2} x_{2}) y_{1} = x_{2}^{- 1} x_{1} y_{1} y_{2}$ , and these agree because $x_{1}^{2} = z = x_{2}^{2}$ gives $x_{1}^{- 1} x_{2} = x_{2}^{- 1} x_{1}$ . Hence $[H_{1}, H_{2}] = 1$ .

4.2F3L4step 1.1step 3.1

The two subgroups generate $E$ : they contain $x_{1}, x_{2}, t_{1}, t_{2}$ , hence $y_{1} = x_{2}^{- 1} t_{1}$ and $y_{2} = x_{1}^{- 1} t_{2}$ , hence both canonical images, which generate $E$ .

5.1F2L6L9step 1.1step 3.1step 4.1step 4.2

So $H_{1}$ and $H_{2}$ form an internal central product of $E$ , and the recognition theorem gives $E ≅ H_{1} \circ_{id} H_{2}$ along the identity of $D = H_{1} \cap H_{2}$ . Comparing orders, $32 = ∣ E ∣ = 8 \cdot 8 / ∣ D ∣$ , so $∣ D ∣ = 2$ and $D = ⟨ z ⟩ = Z (H_{1}) = Z (H_{2})$ .

6.1F1L3step 3.1step 5.1∎

Since $H_{1}$ and $H_{2}$ are isomorphic to $Dih (C_{4})$ by isomorphisms carrying $D$ to the centre, $E$ is a central product of two copies of $Dih (C_{4})$ along the unique isomorphism of their centres, which is what was claimed.

Remarks

The identity $x_{1}^{2} = x_{2}^{2}$ is the whole of the computation in step 4.1, and it is where the quaternion hypothesis is spent: in a central product of two dihedral groups the corresponding squares are both trivial and the same computation succeeds for a different reason. What the statement records is that these two central products are the same group, so the number of quaternion factors in a decomposition is not an invariant of it.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A product formula for the number of square roots of the identity in a central product of extraspecial $2$ -groups

Statement

Let $P_{1}$ and $P_{2}$ be extraspecial $2$ -groups and let $P = P_{1} \circ_{α} P_{2}$ be the central product along an isomorphism $α$ of their centres. Write $t (G) = ∣ {g \in G : g^{2} = 1} ∣$ . Then

$t (P) = \frac{t (P_{1}) t (P_{2}) + (∣ P_{1} ∣ - t (P_{1})) (∣ P_{2} ∣ - t (P_{2}))}{2} .$

Facts & Assumptions

Given: Extraspecial $2$ -groups $P_{1}, P_{2}$ with $Z (P_{i}) = ⟨ z_{i} ⟩$ of order two, an isomorphism $α : Z (P_{1}) \to Z (P_{2})$ , and $P = P_{1} \circ_{α} P_{2}$ with quotient map $π : P_{1} \times P_{2} \to P$ .

[F1]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F2]

The external direct product $G \times H$ carries the componentwise operation (The external direct product $G \times H$ with componentwise multiplication).

[F3]

The quotient group $G / N$ has the left cosets $g N$ as elements, with product $(g N) (h N) : = g h N$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[F4]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[L1]

[L2]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[L3]

The subgroup $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ of $G \times H$ is central, hence normal (The identified subgroup used to form a central product is central, hence normal).

[L4]

[L5]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L6]

$∣ A ∣$ is the unique natural number $n$ with $A \approx n$ (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1F4L1L2

Since $P_{i} / Z (P_{i})$ is elementary abelian, $g^{2} \in Z (P_{i}) = {1, z_{i}}$ for every $g \in P_{i}$ ; so $P_{i}$ splits into the $t (P_{i})$ elements with $g^{2} = 1$ and the $∣ P_{i} ∣ - t (P_{i})$ elements with $g^{2} = z_{i}$ .

1.2F1F3L3L5L6

The quotient map $π$ is surjective with kernel $N = {(z, α (z)^{- 1}) : z \in Z (P_{1})}$ , which has two elements, so every element of $P$ has exactly two preimages in $P_{1} \times P_{2}$ .

1.3F2F3L4

Because the two canonical images commute, $π (g, h)^{2} = π (g^{2}, h^{2})$ for all $g \in P_{1}$ and $h \in P_{2}$ .

2.1F1F4step 1.1step 1.3

Hence $π (g, h)^{2} = 1$ exactly when $(g^{2}, h^{2}) \in N$ , that is exactly when $g^{2} = 1$ and $h^{2} = 1$ , or $g^{2} = z_{1}$ and $h^{2} = α (z_{1})^{- 1} = z_{2}$ .

3.1L6step 1.1step 2.1

The number of pairs $(g, h)$ with $π (g, h)^{2} = 1$ is therefore $t (P_{1}) t (P_{2}) + (∣ P_{1} ∣ - t (P_{1})) (∣ P_{2} ∣ - t (P_{2}))$ .

4.1L6step 1.2step 3.1∎

Each element of $P$ with square the identity is counted exactly twice in that total, so $t (P)$ is half of it, which is the displayed formula.

Remarks

The second summand is what makes the formula more than a product: an element of $P$ can square to the identity because both of its coordinates square to the identity, or because both square to the identified central element and those two squares cancel in the quotient.

Writing $∣ P_{i} ∣ = 2^{1 + 2 n_{i}}$ and $t (P_{i}) = 2^{2 n_{i}} + ε_{i} 2^{n_{i}}$ with $ε_{i} = \pm 1$ , the formula collapses to $t (P) = 2^{2 n} + ε_{1} ε_{2} 2^{n}$ with $n = n_{1} + n_{2}$ : the signs multiply.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

For each $n \geq 1$ there are exactly two extraspecial groups of order $2^{1 + 2 n}$

Statement

For each $n \geq 1$ there are exactly two extraspecial groups of order $2^{1 + 2 n}$ up to isomorphism. Writing $t (G)$ for the number of solutions of $x^{2} = 1$ in $G$ , one of them has $t = 2^{2 n} + 2^{n}$ and the other has $t = 2^{2 n} - 2^{n}$ , and an extraspecial group of that order is determined up to isomorphism by which of the two values it takes.

Facts & Assumptions

Given: An integer $n \geq 1$ and an extraspecial group $P$ of order $2^{1 + 2 n}$ with $Z (P) = ⟨ z ⟩$ .

[F1]

Subgroups $G_{1}, \dots, G_{r}$ of $G$ form an internal central product when they generate $G$ and $[G_{i}, G_{j}] = 1$ for $i \neq j$ (Internal central products of a finite family of subgroups).

[F2]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F4]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L1]

There are $n \geq 1$ subgroups $P_{1}, \dots, P_{n}$ of $P$ , each nonabelian of order $p^{3}$ with $Z (P_{i}) = Z (P)$ , which form an internal central product of $P$ ; such a family is admissible, $∣ P ∣ = p^{1 + 2 n}$ , and peeling one member leaves an extraspecial group of order $p^{1 + 2 (n - 1)}$ with the induced admissible family (Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$ ).

[L2]

For every prime $p$ there are exactly two nonabelian groups of order $p^{3}$ up to isomorphism; at $p = 2$ they are $Dih (C_{4})$ and $Q_{8}$ (For each prime there are exactly two nonabelian groups of order $p^{3}$ up to isomorphism).

[L3]

$Dih (C_{4})$ and $Q_{8}$ are extraspecial of order eight, with exactly six and exactly two solutions of $x^{2} = 1$ ( $Dih (C_{4})$ and $Q_{8}$ are extraspecial of order $8$ , with six and two solutions of $x^{2} = 1$ respectively).

[L4]

$Q_{8} \circ Q_{8}$ is an internal central product of two subgroups isomorphic to $Dih (C_{4})$ meeting in its centre ( $Q_{8} \circ Q_{8} ≅ Dih (C_{4}) \circ Dih (C_{4})$ ).

[L5]

For extraspecial $2$ -groups, $t (P_{1} \circ_{α} P_{2}) = (t (P_{1}) t (P_{2}) + (∣ P_{1} ∣ - t (P_{1})) (∣ P_{2} ∣ - t (P_{2}))) / 2$ (A product formula for the number of square roots of the identity in a central product of extraspecial $2$ -groups).

[L6]

[L7]

[L8]

A finite $p$ -group $P$ is extraspecial when it is nonabelian and $Z (P) = P^{'} = Φ (P)$ is elementary abelian of order $p$ (Special and extraspecial $p$ -groups).

[L9]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · induction

1.1L2L3L8base

At $n = 1$ an extraspecial group of order eight is nonabelian, hence isomorphic to $Dih (C_{4})$ or to $Q_{8}$ ; these have $t = 6 = 2^{2} + 2^{1}$ and $t = 2 = 2^{2} - 2^{1}$ , so there are exactly two and the value of $t$ tells them apart.

1.2ih

Assume, for every $m$ with $1 \leq m < n$ : both values $2^{2 m} + 2^{m}$ and $2^{2 m} - 2^{m}$ are realised; every extraspecial group of order $2^{1 + 2 m}$ has an admissible family with at most one quaternion member; if that number is $k$ then $t = 2^{2 m} + (- 1)^{k} 2^{m}$ ; and two such groups with equal $t$ are isomorphic.

1.3F2algebra

If $ϕ : G \to G^{'}$ and $ψ : H \to H^{'}$ are isomorphisms carrying the identified central subgroups to the identified central subgroups compatibly with the identifying isomorphisms, then $ϕ \times ψ$ carries $N$ onto $N^{'}$ and induces an isomorphism $G \circ_{α} H \to G^{'} \circ_{α^{'}} H^{'}$ ; when all four identified subgroups have order two the compatibility is automatic, since a group of order two has only one automorphism.

1.4F1L1L2

Let $P$ be extraspecial of order $2^{1 + 2 n}$ with $n \geq 2$ and take an admissible family $P_{1}, \dots, P_{n}$ ; each member is nonabelian of order eight, hence isomorphic to $Dih (C_{4})$ or to $Q_{8}$ .

2.1F1F3F4L4L6step 1.3step 1.4

If two members $P_{i}, P_{j}$ are isomorphic to $Q_{8}$ , then $R = ⟨ P_{i}, P_{j} ⟩$ is an internal central product of them, so $R ≅ Q_{8} \circ Q_{8}$ and $R$ is an internal central product of two subgroups isomorphic to $Dih (C_{4})$ with the same centre $Z (P)$ ; replacing $P_{i}, P_{j}$ by those two subgroups leaves an admissible family with two fewer quaternion members. Repeating, $P$ has an admissible family with $k \in {0, 1}$ quaternion members.

3.1F1L1L6L9step 2.1

Fix such a family. Since $n \geq 2$ and $k \leq 1$ , some member $P_{1}$ is isomorphic to $Dih (C_{4})$ ; peeling it leaves $C = ⟨ P_{2}, \dots, P_{n} ⟩$ , extraspecial of order $2^{1 + 2 (n - 1)}$ with an admissible family of $n - 1$ members of which $k$ are quaternion, and $P ≅ P_{1} \circ_{id} C$ along the identity of $Z (P)$ .

4.1L5L9step 1.2step 3.1

By the induction hypothesis $t (C) = 2^{2 (n - 1)} + (- 1)^{k} 2^{n - 1}$ , and $t (P_{1}) = 6 = 2^{2} + 2$ ; the counting formula then gives $t (P) = 2^{2 n} + (- 1)^{k} 2^{n}$ .

5.1F3step 1.2step 1.3step 3.1step 4.1

If $P$ and $P^{'}$ are extraspecial of order $2^{1 + 2 n}$ with $t (P) = t (P^{'})$ , their normalised families have the same $k$ by step 4.1, so peeling a dihedral member from each gives $C$ and $C^{'}$ extraspecial of order $2^{1 + 2 (n - 1)}$ with equal $t$ , hence isomorphic by the induction hypothesis, by an isomorphism carrying $Z (C) = Z (P)$ onto $Z (C^{'}) = Z (P^{'})$ ; the peeled members are isomorphic too, so $P ≅ P^{'}$ .

6.1L5L7step 1.1step 1.2step 4.1step 5.1discharge-induction∎

Both values are realised: if $C$ is extraspecial of order $2^{1 + 2 (n - 1)}$ then $Dih (C_{4}) \circ C$ is extraspecial of order $2^{1 + 2 n}$ with $t = 2^{2 n} + (- 1)^{k} 2^{n}$ where $t (C) = 2^{2 (n - 1)} + (- 1)^{k} 2^{n - 1}$ , so the two groups supplied by the induction hypothesis produce one group of each value. With step 5.1 this gives exactly two isomorphism classes at order $2^{1 + 2 n}$ and completes the induction.

Remarks

The quaternion factors are not an invariant of the group, only their parity is: two of them can always be traded for two dihedral factors, and it is exactly that trade which leaves the count $t$ unchanged, since the two signs multiply.

The count $t$ is an isomorphism invariant because an isomorphism carries solutions of $x^{2} = 1$ to solutions of $x^{2} = 1$ ; that is what makes the two classes provably distinct rather than merely differently presented.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

For odd $p$ , a central product of two modular groups of order $p^{3}$ is a central product of a modular group with a Heisenberg group

Statement

Let $p$ be an odd prime and let $E$ be a central product of two copies of the modular group $M_{p}$ along an isomorphism of their centres. Then $E$ is an internal central product of a subgroup isomorphic to the Heisenberg group $H_{p}$ and a subgroup isomorphic to $M_{p}$ ; consequently

$M_{p} \circ M_{p} ≅ H_{p} \circ M_{p} .$

Facts & Assumptions

Given: An odd prime $p$ , two copies of $M_{p}$ with generators $a_{i}$ of order $p^{2}$ and $s_{i}$ of order $p$ satisfying $s_{i} a_{i} s_{i}^{- 1} = a_{i}^{1 + p}$ , and the central product $E$ of the two along an isomorphism of their centres, with canonical images $x_{i}, y_{i}$ of $a_{i}, s_{i}$ and common central image $z$ .

[F1]

[F2]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F3]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F4]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F5]

For a finite group $G$ , $\exp (G) = \min {n \in N : n > 0 and g^{n} = e for every g \in G}$ (The exponent of a finite group).

[F6]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L1]

$M_{p}$ is nonabelian of order $p^{3}$ , extraspecial of exponent $p^{2}$ , with $Z (M_{p}) = [M_{p}, M_{p}] = ⟨ a^{p} ⟩$ (The modular group of order $p^{3}$ is extraspecial, of exponent $p^{2}$ when $p$ is odd).

[L2]

[L3]

[L4]

[L5]

An extraspecial $p$ -group is nilpotent of class exactly two, its derived subgroup satisfies $P^{'} = Z (P)$ and has order $p$ , and every nonidentity commutator has order $p$ (An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$ ).

[L6]

[L7]

[L8]

If $x, y \in P$ satisfy $[x, y] \neq e$ and $F = ⟨ x, y ⟩$ , then $P = F C_{P} (F)$ and $F \cap C_{P} (F) = Z (P)$ , and $C_{P} (F)$ is extraspecial of order $∣ P ∣ / p^{2}$ with centre $Z (P)$ when $∣ P ∣ > p^{3}$ (Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$ , The centralizer $C_{G} (H)$ of a subgroup).

[L9]

For every prime $p$ there are exactly two nonabelian groups of order $p^{3}$ up to isomorphism; for odd $p$ they are $H_{p}$ , of exponent $p$ , and $M_{p}$ , of exponent $p^{2}$ (For each prime there are exactly two nonabelian groups of order $p^{3}$ up to isomorphism).

[L10]

[L11]

Proof

technique · direct

1.1F2F4L1L2L3

$E$ is extraspecial of order $p^{5}$ , its centre is the common image $⟨ z ⟩$ of the two centres, and the two canonical images commute elementwise, are injective and generate $E$ .

1.2F1F2L1L11

The generator $a_{2}$ may be replaced by a power $a_{2}^{c}$ with $c$ not divisible by $p$ without changing the relations of the second copy, and such a replacement multiplies $x_{2}^{p}$ by $c$ in the exponent; choosing $c$ suitably we may assume $x_{1}^{p} = x_{2}^{p} = z$ .

2.1F1F3L1L3step 1.2

Each $x_{i}$ has order $p^{2}$ and each $y_{i}$ has order $p$ , and $[x_{2}, y_{2}]$ is the image of $[a_{2}, s_{2}] = a_{2}^{- p}$ , so $[x_{2}, y_{2}] = z^{- 1}$ .

2.2L4L5L11step 1.1step 1.2

Put $w = x_{2} x_{1}^{- 1}$ . The $p$ -th power map on $E$ is a homomorphism, because $p$ is odd and $[E, E] = Z (E)$ has order $p$ , so $w^{p} = x_{2}^{p} (x_{1}^{p})^{- 1} = z z^{- 1} = e$ .

3.1L3step 1.1step 2.1step 2.2

Moreover $w \neq e$ : otherwise $x_{1} = x_{2}$ would lie in both canonical images, hence in $⟨ z ⟩$ , contradicting that $x_{1}$ has order $p^{2}$ . So $w$ has order $p$ .

3.2F3L3L6step 1.1step 2.1

Since $x_{1}$ and $y_{2}$ lie in different canonical images they commute, so $[w, y_{2}] = [x_{2} x_{1}^{- 1}, y_{2}] = [x_{2}, y_{2}] [x_{1}^{- 1}, y_{2}] = [x_{2}, y_{2}] = z^{- 1} \neq e$ .

4.1F6L7step 3.2

Hence $F = ⟨ w, y_{2} ⟩$ is extraspecial of order $p^{3}$ with $Z (F) = Z (E)$ .

5.1F5F6L4step 2.2step 3.1step 4.1

The elements of $E$ whose $p$ -th power is the identity form the kernel of the $p$ -th power homomorphism, hence a subgroup; it contains $w$ , $y_{2}$ and $z$ , so it contains $F$ , and $F$ has exponent $p$ .

5.2L8step 1.1step 3.2step 4.1

By the splitting clause, $E = F C_{E} (F)$ with $F \cap C_{E} (F) = Z (E)$ , and $C = C_{E} (F)$ is extraspecial of order $p^{5} / p^{2} = p^{3}$ with $Z (C) = Z (E)$ ; so $F$ and $C$ form an internal central product of $E$ .

6.1L9step 4.1step 5.1

A nonabelian group of order $p^{3}$ and exponent $p$ is isomorphic to $H_{p}$ , since the other one has exponent $p^{2}$ ; so $F ≅ H_{p}$ .

6.2F5L4L9step 2.1step 5.1step 5.2

If $C$ had exponent $p$ then every element of $E = F C$ would be a product of two commuting elements of $p$ -th power the identity, so $E$ would have exponent $p$ , contradicting that $x_{1}$ has order $p^{2}$ . Hence $C$ has exponent $p^{2}$ and $C ≅ M_{p}$ .

7.1L10step 6.1step 5.2step 6.2∎

Therefore $E$ is an internal central product of $F ≅ H_{p}$ and $C ≅ M_{p}$ meeting in $Z (E)$ , and the recognition theorem identifies it with $H_{p} \circ M_{p}$ along the identity of that centre.

Remarks

The construction of the exponent- $p$ subgroup is where oddness of $p$ is spent: the element $w = x_{2} x_{1}^{- 1}$ has order $p$ only because the $p$ -th power map is a homomorphism, and at $p = 2$ that map is not one. The corresponding statement at $p = 2$ is the trade of two quaternion factors for two dihedral ones, which is a different computation with a different outcome.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

For odd $p$ and each $n \geq 1$ there are exactly two extraspecial groups of order $p^{1 + 2 n}$ , distinguished by their exponent

Statement

Let $p$ be an odd prime. For each $n \geq 1$ there are exactly two extraspecial groups of order $p^{1 + 2 n}$ up to isomorphism, and they are distinguished by their exponent: one has exponent $p$ and the other has exponent $p^{2}$ .

Facts & Assumptions

Given: An odd prime $p$ , an integer $n \geq 1$ , and an extraspecial group $P$ of order $p^{1 + 2 n}$ with $Z (P) = ⟨ z ⟩$ .

[F1]

Subgroups $G_{1}, \dots, G_{r}$ of $G$ form an internal central product when they generate $G$ and $[G_{i}, G_{j}] = 1$ for $i \neq j$ (Internal central products of a finite family of subgroups).

[F2]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F3]

For a finite group $G$ , $\exp (G) = \min {n \in N : n > 0 and g^{n} = e for every g \in G}$ (The exponent of a finite group).

[F4]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F5]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[F6]

For a group $G$ and a prime $p$ , $G^{p} = ⟨ g^{p} : g \in G ⟩$ (The $p$ th-power subgroup $G^{p}$ ).

[F7]

The Heisenberg group is $H_{p} = (Z / p)^{3}$ with $(a, b, c) (a^{'}, b^{'}, c^{'}) = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'})$ (The Heisenberg group of order $p^{3}$ over $Z / p$ ).

[L1]

[L2]

[L3]

For odd $p$ , a central product of two copies of $M_{p}$ along an isomorphism of their centres is an internal central product of a subgroup isomorphic to $H_{p}$ and a subgroup isomorphic to $M_{p}$ (For odd $p$ , a central product of two modular groups of order $p^{3}$ is a central product of a modular group with a Heisenberg group).

[L4]

[L5]

[L6]

[L7]

For every finite $p$ -group $P$ , $Φ (P) = P^{'} P^{p}$ ( $Φ (P) = P^{'} P^{p}$ for a finite $p$ -group).

[L8]

[L9]

[L10]

$H_{p}$ is extraspecial and, for odd $p$ , has exponent $p$ (The Heisenberg group of order $p^{3}$ is extraspecial, and for odd $p$ it has exponent $p$ ).

[L11]

$M_{p}$ is extraspecial of exponent $p^{2}$ (The modular group of order $p^{3}$ is extraspecial, of exponent $p^{2}$ when $p$ is odd).

[L12]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L13]

The canonical maps from both factors into a central product are injective homomorphisms; their images commute and generate the central product (The two canonical maps into a central product are injective homomorphisms whose images commute, generate it, and meet in the identified centre).

Proof

technique · induction

1.1L2L6L10L11base

At $n = 1$ an extraspecial group of order $p^{3}$ is nonabelian, hence isomorphic to $H_{p}$ or to $M_{p}$ ; their exponents are $p$ and $p^{2}$ , so there are exactly two and the exponent tells them apart.

1.2ih

Assume, for every $m$ with $1 \leq m < n$ : both exponents $p$ and $p^{2}$ are realised by extraspecial groups of order $p^{1 + 2 m}$ ; every such group has an admissible family with at most one modular member; its exponent is $p$ when that number is zero and $p^{2}$ when it is one; and two such groups with the same exponent are isomorphic.

1.3F2algebra

If $ϕ : G \to G^{'}$ and $ψ : H \to H^{'}$ are isomorphisms carrying the identified central subgroups onto the identified central subgroups compatibly with the identifying isomorphisms, then $ϕ \times ψ$ carries $N$ onto $N^{'}$ and induces an isomorphism $G \circ_{α} H \to G^{'} \circ_{α^{'}} H^{'}$ .

1.4F3F6L6L7

Every $g \in P$ has $g^{p} \in P^{p} \leq Φ (P) = Z (P)$ , so $g^{p^{2}} = (g^{p})^{p} = e$ and the exponent of $P$ divides $p^{2}$ .

1.5F1L1L2

Let $P$ be extraspecial of order $p^{1 + 2 n}$ with $n \geq 2$ and take an admissible family $P_{1}, \dots, P_{n}$ ; each member is nonabelian of order $p^{3}$ , hence isomorphic to $H_{p}$ or to $M_{p}$ .

2.1F1F4F5L3L8step 1.3step 1.5

If two members $P_{i}, P_{j}$ are isomorphic to $M_{p}$ , then $R = ⟨ P_{i}, P_{j} ⟩$ is an internal central product of them, so $R ≅ M_{p} \circ M_{p}$ and $R$ is an internal central product of a subgroup isomorphic to $H_{p}$ and one isomorphic to $M_{p}$ , both with centre $Z (P)$ ; replacing $P_{i}, P_{j}$ by those two leaves an admissible family with one fewer modular member. Repeating, $P$ has an admissible family with $k \in {0, 1}$ modular members.

3.1F3F5L4L5L10L11step 1.4step 2.1

If $k = 0$ every member has exponent $p$ ; since the members commute and generate $P$ and the $p$ -th power map is a homomorphism, every element of $P$ is a product of elements of the members and has $p$ -th power the identity, so $\exp (P) = p$ . If $k = 1$ the modular member contains an element of order $p^{2}$ , so $\exp (P)$ is a multiple of $p^{2}$ , and by step 1.4 it equals $p^{2}$ . Thus the exponent determines $k$ .

3.2F1L1L8L12step 2.1

Since $n \geq 2$ and $k \leq 1$ , some member $P_{1}$ is isomorphic to $H_{p}$ ; peeling it leaves $C = ⟨ P_{2}, \dots, P_{n} ⟩$ , extraspecial of order $p^{1 + 2 (n - 1)}$ with an admissible family of $n - 1$ members of which $k$ are modular, and $P ≅ P_{1} \circ_{id} C$ along the identity of $Z (P)$ .

4.1F4F7L10step 1.2step 1.3step 3.1step 3.2

If $P$ and $P^{'}$ are extraspecial of order $p^{1 + 2 n}$ with the same exponent, their normalised families have the same $k$ by step 3.1, so the peeled subgroups $C$ and $C^{'}$ have the same exponent and are isomorphic by the induction hypothesis. Any such isomorphism restricts to an isomorphism $Z (C) \to Z (C^{'})$ . For the peeled Heisenberg factors, the maps $(a, b, c) \mapsto (r a, b, r c)$ with $r \in F_{p}^{\times}$ preserve the multiplication of [F7] and induce every automorphism of their order- $p$ centres; choose one whose central restriction makes the two factor isomorphisms compatible. Step 1.3 then induces $P ≅ P^{'}$ .

5.1L4L9L10L13step 1.1step 1.2step 1.4step 3.1step 4.1discharge-induction∎

Both exponents are realised: if $C$ is extraspecial of order $p^{1 + 2 (n - 1)}$ then $H_{p} \circ C$ is extraspecial of order $p^{1 + 2 n}$ . When $C$ has exponent $p$ , the commuting generating images of [L13] and [L4] show that the product has exponent $p$ ; when $C$ has exponent $p^{2}$ , its injective canonical image from [L13] still contains an element of order $p^{2}$ , while step 1.4 bounds the product exponent by $p^{2}$ . Applying this to the two groups supplied by the induction hypothesis gives one group of each exponent. With step 4.1 this gives exactly two isomorphism classes at order $p^{1 + 2 n}$ and completes the induction.

Remarks

The modular factors are not an invariant of the group and, unlike the quaternion factors at $p = 2$ , not even their parity is: two of them can be traded for one Heisenberg factor and one modular factor, so the count drops by one rather than by two. What survives is the presence or absence of an element of order $p^{2}$ , which is the exponent.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Plus and minus type of an extraspecial $p$ -group

Definition

Let $p$ be a prime and $n \geq 1$ . By For odd $p$ and each $n \geq 1$ there are exactly two extraspecial groups of order $p^{1 + 2 n}$ , distinguished by their exponent for odd $p$ and by For each $n \geq 1$ there are exactly two extraspecial groups of order $2^{1 + 2 n}$ for $p = 2$ , there are exactly two extraspecial groups of order $p^{1 + 2 n}$ up to isomorphism. They are named as follows.

For odd $p$ , write $p_{+}^{1 + 2 n}$ for the one of exponent $p$ and $p_{-}^{1 + 2 n}$ for the one of exponent $p^{2}$ (The exponent of a finite group). At $n = 1$ these are the Heisenberg group and the modular group (The Heisenberg group of order $p^{3}$ is extraspecial, and for odd $p$ it has exponent $p$ , The modular group of order $p^{3}$ is extraspecial, of exponent $p^{2}$ when $p$ is odd).

For $p = 2$ , write $2_{+}^{1 + 2 n}$ for the one with $2^{2 n} + 2^{n}$ solutions of $x^{2} = 1$ and $2_{-}^{1 + 2 n}$ for the one with $2^{2 n} - 2^{n}$ solutions. At $n = 1$ these are $Dih (C_{4})$ and $Q_{8}$ ( $Dih (C_{4})$ and $Q_{8}$ are extraspecial of order $8$ , with six and two solutions of $x^{2} = 1$ respectively).

The two names are well defined because in each case the two classification theorems supply exactly two isomorphism classes and an invariant that separates them, so the label is a property of the isomorphism class and not of a presentation.

Remarks

The two source conventions differ in scope and are both recorded here. van Beek's Definition 2.38 writes $p_{\pm}^{1 + 2 n}$ for every prime, the sign being read off an iterated central product, which is the convention taken above. Craven's Definition 3.3 introduces $p_{+}^{1 + 2}$ only for the odd exponent- $p$ group of order $p^{3}$ and names the others by their constructions. Where the two overlap they agree, and the convention in force here is van Beek's.

At $p = 2$ the signs multiply under central products, as the counting formula shows. At odd $p$ the notation records exponent instead: a central product is of plus type precisely when every order- $p^{3}$ factor is Heisenberg. If a modular factor occurs, the absorption lemma reduces all modular factors to one, so the product is of minus type. Thus the two uses of the signs agree on the basic plus factors but obey different product rules.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

An extraspecial group of odd order has exponent $p$ or $p^{2}$ , and an extraspecial $2$ -group has exponent $4$

Statement

Let $P$ be an extraspecial $p$ -group of order $p^{1 + 2 n}$ . If $p$ is odd then $\exp (P) = p$ when $P$ is of plus type and $\exp (P) = p^{2}$ when $P$ is of minus type. If $p = 2$ then $\exp (P) = 4$ , whichever type $P$ is.

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ of order $p^{1 + 2 n}$ with $n \geq 1$ .

[F1]

For a finite group $G$ , $\exp (G) = \min {n \in N : n > 0 and g^{n} = e for every g \in G}$ (The exponent of a finite group).

[F2]

For odd $p$ , $p_{+}^{1 + 2 n}$ is the extraspecial group of that order with exponent $p$ and $p_{-}^{1 + 2 n}$ the one with exponent $p^{2}$ ; at $p = 2$ the two are named by their number of solutions of $x^{2} = 1$ (Plus and minus type of an extraspecial $p$ -group).

[F3]

[F4]

For a group $G$ and a prime $p$ , $G^{p} = ⟨ g^{p} : g \in G ⟩$ (The $p$ th-power subgroup $G^{p}$ ).

[L1]

For odd $p$ there are exactly two extraspecial groups of order $p^{1 + 2 n}$ , distinguished by their exponent, which is $p$ for one and $p^{2}$ for the other (For odd $p$ and each $n \geq 1$ there are exactly two extraspecial groups of order $p^{1 + 2 n}$ , distinguished by their exponent).

[L2]

For each $n \geq 1$ there are exactly two extraspecial groups of order $2^{1 + 2 n}$ , separated by the number of solutions of $x^{2} = 1$ (For each $n \geq 1$ there are exactly two extraspecial groups of order $2^{1 + 2 n}$ ).

[L3]

There are $n \geq 1$ subgroups of $P$ , each nonabelian of order $p^{3}$ with centre $Z (P)$ , forming an internal central product of $P$ (Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$ ).

[L4]

For every prime $p$ there are exactly two nonabelian groups of order $p^{3}$ ; at $p = 2$ they are $Dih (C_{4})$ and $Q_{8}$ (For each prime there are exactly two nonabelian groups of order $p^{3}$ up to isomorphism).

[L5]

In $Dih (C_{4})$ the rotation has order four, and in $Q_{8}$ the element $i$ has order four ( $Dih (C_{4})$ and $Q_{8}$ are extraspecial of order $8$ , with six and two solutions of $x^{2} = 1$ respectively, $Q_{8}$ is a subgroup of $H^{\times}$ with eight elements, and $- 1$ is its only element of order $2$ ).

[L6]

[L7]

For every finite $p$ -group $P$ , $Φ (P) = P^{'} P^{p}$ ( $Φ (P) = P^{'} P^{p}$ for a finite $p$ -group).

Proof

technique · direct

1.1F1F4L6L7

Every $g \in P$ has $g^{p} \in P^{p} \leq Φ (P) = Z (P)$ , a group of order $p$ , so $g^{p^{2}} = (g^{p})^{p} = e$ and $\exp (P)$ divides $p^{2}$ .

1.2F2L1

For odd $p$ the classification names the two isomorphism classes by their exponents, which are $p$ and $p^{2}$ , and the plus and minus labels are those names.

2.1F1F3L2L3L4L5step 1.1∎

At $p = 2$ , take an internal central product decomposition into subgroups of order eight; each is nonabelian, hence isomorphic to $Dih (C_{4})$ or to $Q_{8}$ , and each contains an element of order four. So $\exp (P)$ is a multiple of four, and by step 1.1 it divides four; hence $\exp (P) = 4$ for both types.

Remarks

At $p = 2$ the exponent does not separate the two types, and that is why the classification there uses the number of solutions of $x^{2} = 1$ instead. The two invariants are not interchangeable: for odd $p$ the exponent separates the types and the number of solutions of $x^{p} = 1$ does so as well, while at $p = 2$ only the second does.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The square map of an extraspecial $2$ -group relative to a chosen generator of its centre

Definition

Let $P$ be an extraspecial $2$ -group (Special and extraspecial $p$ -groups) and fix the generator $z$ of its centre, so $Z (P) = {1, z}$ with $z^{2} = 1$ (The center $Z (G)$ of a group, The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity). Write $V = P / Z (P)$ with its canonical $F_{2}$ -vector-space structure (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , An elementary abelian $p$ -group has a canonical $F_{p}$ -vector-space structure, For every prime $p$ , the two operations on $Z / p$ make it a field), and let $b_{z}$ be the commutator pairing (The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre).

The square map of $P$ relative to $z$ is

$q_{z} : V \to F_{2}, x^{2} = z^{q_{z} (\overset{ˉ}{x})} .$

Why the exponent exists and is unique. The quotient $P / Z (P)$ is elementary abelian (Three equivalent descriptions of an extraspecial $p$ -group), so $x^{2} \in Z (P) = {1, z}$ for every $x \in P$ ; and $z^{0} = 1 \neq z = z^{1}$ , so exactly one class in $F_{2}$ records which of the two values $x^{2}$ takes.

That the value depends only on the coset $\overset{ˉ}{x}$ , and the identity relating it to the commutator pairing, are proved in The square map is well defined on the central quotient and satisfies $q (\overset{ˉ}{x} \overset{ˉ}{y}) = q (\overset{ˉ}{x}) + q (\overset{ˉ}{y}) + b (\overset{ˉ}{x}, \overset{ˉ}{y})$ ↗.

Remarks

The map is not a homomorphism to $F_{2}$ : it fails additivity by exactly the value of the commutator pairing, and that failure is the whole content of the identity proved for it. Where the pairing vanishes the map is additive, and it is on such subspaces that the counting of elementary abelian subgroups is done.

The map is defined only at $p = 2$ . For odd $p$ the corresponding assignment $x \mapsto x^{p}$ also lands in the centre, but it is a homomorphism by the class-two power formula (For an odd prime $p$ , the $p$ -th power map is a homomorphism on a finite group whose derived subgroup is central of exponent dividing $p$ ) and carries no extra information beyond the type.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The square map is well defined on the central quotient and satisfies $q (\overset{ˉ}{x} \overset{ˉ}{y}) = q (\overset{ˉ}{x}) + q (\overset{ˉ}{y}) + b (\overset{ˉ}{x}, \overset{ˉ}{y})$

Statement

Let $P$ be an extraspecial $2$ -group with $Z (P) = ⟨ z ⟩$ and $V = P / Z (P)$ , and let $q = q_{z}$ be its square map and $b = b_{z}$ its commutator pairing. Then $q$ is well defined on $V$ and

$q (\overset{ˉ}{x} \overset{ˉ}{y}) = q (\overset{ˉ}{x}) + q (\overset{ˉ}{y}) + b (\overset{ˉ}{x}, \overset{ˉ}{y}) for all \overset{ˉ}{x}, \overset{ˉ}{y} \in V .$

Moreover $q (\overset{ˉ}{x}) = 0$ exactly when the elements of the coset $\overset{ˉ}{x}$ satisfy $x^{2} = 1$ ; a subgroup $U \leq V$ on which $q$ vanishes satisfies $U \leq U^{⊥}$ and has elementary abelian preimage in $P$ of order $2 ∣ U ∣$ ; and every maximal elementary abelian subgroup of $P$ contains $Z (P)$ .

Facts & Assumptions

Given: An extraspecial $2$ -group $P$ with $Z (P) = ⟨ z ⟩$ of order two, the quotient $V = P / Z (P)$ , the commutator pairing $b$ and the square map $q$ .

[F1]

The square map of $P$ relative to $z$ is $q_{z} : V \to F_{2}$ determined by $x^{2} = z^{q_{z} (\overset{ˉ}{x})}$ (The square map of an extraspecial $2$ -group relative to a chosen generator of its centre).

[F2]

The commutator pairing of $P$ relative to $z$ is $b_{z} : V \times V \to F_{p}$ determined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ (The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre).

[F3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F4]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ (Elementary abelian $p$ -groups).

[F5]

[L1]

[L2]

If $[G, G] \leq Z (G)$ then $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ for every $n \in N$ (In a group with central derived subgroup, $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ ).

[L3]

An extraspecial $p$ -group is nilpotent of class exactly two, its derived subgroup satisfies $P^{'} = Z (P)$ and has order $p$ (An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$ ).

[L4]

[L5]

[L6]

The quotient group $G / N$ has the left cosets $g N$ as elements (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Proof

technique · direct

1.1F1F3L1L6

Since $P / Z (P)$ is elementary abelian, $x^{2} \in Z (P) = {1, z}$ for every $x \in P$ , and $z^{2} = 1$ ; so $(x z)^{2} = x^{2} z^{2} = x^{2}$ and the value of $q$ depends only on the coset $\overset{ˉ}{x}$ , which makes $q$ well defined on $V$ .

2.1F1F5step 1.1

By definition $q (\overset{ˉ}{x}) = 0$ exactly when $x^{2} = 1$ , and since $q$ is well defined this holds for one element of the coset exactly when it holds for both.

2.2F1F2L2L3L4L5step 1.1

The class-two power formula at exponent two gives $(x y)^{2} = [y, x]^{(\binom{2}{2})} x^{2} y^{2} = [y, x] x^{2} y^{2}$ , and $[y, x] = z^{b (\overset{ˉ}{y}, \overset{ˉ}{x})}$ . Over $F_{2}$ one has $- 1 = 1$ , so $b (\overset{ˉ}{y}, \overset{ˉ}{x}) = - b (\overset{ˉ}{x}, \overset{ˉ}{y}) = b (\overset{ˉ}{x}, \overset{ˉ}{y})$ ; hence $z^{q (\overset{ˉ}{x} \overset{ˉ}{y})} = z^{b (\overset{ˉ}{x}, \overset{ˉ}{y}) + q (\overset{ˉ}{x}) + q (\overset{ˉ}{y})}$ and the displayed identity holds.

3.1F2step 2.2

Let $U \leq V$ satisfy $q (u) = 0$ for every $u \in U$ . Then for $u, u^{'} \in U$ the identity gives $0 = q (u u^{'}) = q (u) + q (u^{'}) + b (u, u^{'}) = b (u, u^{'})$ , so $U \leq U^{⊥}$ .

4.1F3F4L6step 2.1step 3.1

Let $E$ be the preimage of such a $U$ in $P$ . It contains $Z (P)$ , has order $2 ∣ U ∣$ , is abelian because $b$ vanishes on $U$ , and every one of its elements $x$ satisfies $x^{2} = 1$ by step 2.1 together with $z^{2} = 1$ ; so $E$ is elementary abelian of order $2 ∣ U ∣$ .

5.1F3F4step 2.1step 4.1∎

If $E$ is elementary abelian and does not contain $Z (P)$ , then $E \cap Z (P)$ is trivial and $E Z (P)$ is an abelian subgroup all of whose elements square to the identity, properly containing $E$ ; so a maximal elementary abelian subgroup contains $Z (P)$ .

Remarks

The identity is not additivity, and the correction term is where the two isomorphism types differ: on a subspace where $b$ vanishes the map $q$ is additive and its zero set is a subspace, and it is precisely the size of the largest such subspace that separates the two extraspecial groups of a given order.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

An automorphism of an extraspecial $p$ -group acting trivially on its Frattini quotient is inner

Statement

Let $P$ be an extraspecial $p$ -group of order $p^{1 + 2 n}$ and let $ρ_{P} : Aut (P) \to {Aut}_{F_{p}} (P / Φ (P))$ be the induced-action homomorphism. Then $\ker ρ_{P} = Inn (P)$ : an automorphism of $P$ acting trivially on the Frattini quotient is inner.

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ of order $p^{1 + 2 n}$ .

[F1]

$Inn (G) : = {c_{g} : g \in G}$ with $c_{g} (x) = g x g^{- 1}$ (Inner automorphisms and $Inn (G)$ ).

[F2]

For a finite $p$ -group $P$ , the generator rank $d (P)$ is the common size of a basis of $P / Φ (P)$ (The generator rank $d (P)$ of a finite $p$ -group).

[F3]

$Φ (G)$ is the intersection of the maximal proper subgroups of $G$ (The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group).

[F4]

An isomorphism is a bijective group homomorphism, and $Aut (G)$ is the set of automorphisms of $G$ (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[L1]

[L2]

An extraspecial $p$ -group has $∣ P ∣ = p^{1 + 2 n}$ with $n \geq 1$ and $∣ P / Z (P) ∣ = p^{2 n}$ (An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$ ).

[L3]

An extraspecial $p$ -group of order $p^{1 + 2 n}$ has $Φ (P) = Z (P)$ of order $p$ and generator rank $d (P) = 2 n$ , and every minimal generating set has $2 n$ elements (An extraspecial $p$ -group of order $p^{1 + 2 n}$ has generator rank $2 n$ ).

[L4]

Every automorphism of a finite $p$ -group induces an $F_{p}$ -linear automorphism of $P / Φ (P)$ , and these form a homomorphism $ρ_{P}$ (Automorphisms act linearly on the Frattini quotient).

[L5]

A subset $X$ of a finite $p$ -group $P$ is a minimal generating set if and only if the quotient map restricts to a bijection from $X$ onto a basis of $P / Φ (P)$ (Burnside Basis Theorem).

[L6]

$G / Z (G) ≅ Inn (G)$ ( $G / Z (G) ≅ Inn (G)$ ).

[L7]

[L8]

$∣ A ∣$ is the unique natural number $n$ with $A \approx n$ (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1F2F3L1L2L3

The Frattini subgroup is $Φ (P) = Z (P)$ of order $p$ , the Frattini quotient has order $p^{2 n}$ , and every minimal generating set of $P$ has exactly $2 n$ elements.

2.1F4L5step 1.1

Fix a minimal generating set $X = {g_{1}, \dots, g_{2 n}}$ , which exists because the Burnside basis theorem matches minimal generating sets with bases of the Frattini quotient. An automorphism of $P$ is determined by its values on $X$ , since $X$ generates $P$ .

2.2F1L1L4step 1.1

Every inner automorphism lies in $\ker ρ_{P}$ : for $g, x \in P$ one has $c_{g} (x) x^{- 1} = g x g^{- 1} x^{- 1} \in [P, P] = Φ (P)$ , so $c_{g}$ fixes every coset of $Φ (P)$ .

2.3F1L2L6step 1.1

The inner automorphism group has order $∣ P / Z (P) ∣ = p^{2 n}$ .

3.1F3L4L7L8step 1.1step 2.1

If $θ$ lies in $\ker ρ_{P}$ then $θ (g_{i}) Φ (P) = g_{i} Φ (P)$ for each $i$ , so $θ (g_{i}) = g_{i} u_{i}$ with $u_{i} \in Φ (P) = Z (P)$ , a set of $p$ elements. Hence $θ$ is determined by the tuple $(u_{1}, \dots, u_{2 n})$ , and there are at most $p^{2 n}$ such tuples, so $∣ \ker ρ_{P} ∣ \leq p^{2 n}$ .

4.1L8step 3.1step 2.2step 2.3∎

So $Inn (P)$ is a subset of $\ker ρ_{P}$ of size $p^{2 n}$ , while $\ker ρ_{P}$ has at most $p^{2 n}$ elements; hence the two coincide.

Remarks

The equality is forced by two counts that happen to agree, and each uses the extraspecial hypothesis: the upper bound uses $Φ (P) = Z (P)$ of order $p$ together with the generator rank $2 n$ , and the lower bound uses $∣ P / Z (P) ∣ = p^{2 n}$ . For a general finite $p$ -group the kernel can be larger than the inner automorphism group; equality is not asserted or excluded without additional hypotheses.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

An automorphism fixing the centre pointwise induces a pairing-preserving automorphism of the central quotient, with kernel the inner automorphisms

Statement

Let $P$ be an extraspecial $p$ -group with $Z (P) = ⟨ z ⟩$ and commutator pairing $b_{z}$ , and let $α \in Aut (P)$ fix $Z (P)$ pointwise.

Then $α$ induces an automorphism $\overset{ˉ}{α}$ of $P / Z (P)$ satisfying

$b_{z} (\overset{ˉ}{α} (\overset{ˉ}{x}), \overset{ˉ}{α} (\overset{ˉ}{y})) = b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$

for all $\overset{ˉ}{x}, \overset{ˉ}{y} \in P / Z (P)$ . The kernel of the action of the centre-fixing automorphism subgroup on $P / Z (P)$ is $Inn (P)$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ with $Z (P) = ⟨ z ⟩$ , its commutator pairing $b_{z}$ , and an automorphism $α$ fixing $Z (P)$ pointwise.

[F1]

For an extraspecial $p$ -group $P$ with $Z (P) = ⟨ z ⟩$ , the commutator pairing is the map $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) \in Z / p$ determined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ (The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre).

[L1]

[L2]

Every extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$ (An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$ ).

[L3]

For an extraspecial $p$ -group of order $p^{1 + 2 n}$ , an automorphism acting trivially on its Frattini quotient is inner (An automorphism of an extraspecial $p$ -group acting trivially on its Frattini quotient is inner).

[L4]

Group isomorphisms, automorphisms and the set $Aut (G)$ . (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[L5]

Inner automorphisms and $Inn (G)$ . (Inner automorphisms and $Inn (G)$ ).

[L6]

$[g, h] : = g h g^{- 1} h^{- 1} .$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

Proof

technique · direct

1.1F1L4L6

Because $α$ fixes $Z (P)$ pointwise, it sends each coset $x Z (P)$ to $α (x) Z (P)$ ; this is well defined, so $α$ induces an automorphism $\overset{ˉ}{α}$ of $P / Z (P)$ . Also $[α (x), α (y)] = α ([x, y]) = α (z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}) = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ , so $b_{z} (\overset{ˉ}{α} (\overset{ˉ}{x}), \overset{ˉ}{α} (\overset{ˉ}{y})) = b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ for all $\overset{ˉ}{x}, \overset{ˉ}{y} \in P / Z (P)$ .

2.1L1L2L3L5L6step 1.1∎

If $\overset{ˉ}{α}$ is the identity on $P / Z (P)$ , then $α$ acts trivially on the central quotient. Since $P$ is extraspecial, [L1] gives $Φ (P) = Z (P)$ and [L2] supplies the order hypothesis of [L3], so $α$ acts trivially on the Frattini quotient and is inner. Conversely, every inner automorphism $c_{g} (x) = g x g^{- 1}$ acts trivially on $P / Z (P)$ because $g x g^{- 1} x^{- 1} = [g, x] \in [P, P] = Z (P)$ by [L1] and [L6]. Hence the kernel of the action on $P / Z (P)$ is exactly $Inn (P)$ .

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The maximal elementary abelian subgroups of the two extraspecial groups of order $2^{1 + 2 n}$ have orders $2^{n + 1}$ and $2^{n}$

Statement

The maximal elementary abelian subgroups of the two extraspecial groups of order $2^{1 + 2 n}$ have orders $2^{n + 1}$ and $2^{n}$ .

Facts & Assumptions

Given: The hypotheses of the Statement.

[L1]

For an extraspecial $2$ -group $P$ of order $2^{1 + 2 n}$ with $Z (P) = ⟨ z ⟩$ , the square map is the function $q$ determined by $x^{2} = z^{q (\overset{ˉ}{x})}$ , with values in $F_{2}$ (The square map of an extraspecial $2$ -group relative to a chosen generator of its centre).

[L2]

The square map is well defined on the central quotient and satisfies $q (\overset{ˉ}{x} \overset{ˉ}{y}) = q (\overset{ˉ}{x}) + q (\overset{ˉ}{y}) + b (\overset{ˉ}{x}, \overset{ˉ}{y})$ (The square map is well defined on the central quotient and satisfies $q (\overset{ˉ}{x} \overset{ˉ}{y}) = q (\overset{ˉ}{x}) + q (\overset{ˉ}{y}) + b (\overset{ˉ}{x}, \overset{ˉ}{y})$ ).

[L3]

For a subgroup $\overset{ˉ}{A}$ of $P / Z (P)$ one has $∣ \overset{ˉ}{A} ∣ ∣ {\overset{ˉ}{A}}^{⊥} ∣ = ∣ P / Z (P) ∣$ (A subgroup of the central quotient and its orthogonal complement have orders multiplying to the order of the quotient).

[L4]

For each $n \geq 1$ there are exactly two extraspecial groups of order $2^{1 + 2 n}$ up to isomorphism, with $2^{2 n} + 2^{n}$ and $2^{2 n} - 2^{n}$ solutions of $x^{2} = 1$ (For each $n \geq 1$ there are exactly two extraspecial groups of order $2^{1 + 2 n}$ ).

[L5]

If $P_{1}$ and $P_{2}$ are extraspecial $2$ -groups with $t_{i}$ solutions of $x^{2} = 1$ , then $P_{1} \circ P_{2}$ has $(t_{1} t_{2} + (∣ P_{1} ∣ - t_{1}) (∣ P_{2} ∣ - t_{2})) / 2$ such solutions (A product formula for the number of square roots of the identity in a central product of extraspecial $2$ -groups).

[L6]

Every maximal abelian subgroup of an extraspecial $p$ -group of order $p^{1 + 2 n}$ has order $p^{1 + n}$ (In an extraspecial $p$ -group of order $p^{1 + 2 n}$ every maximal abelian subgroup has order $p^{1 + n}$ ).

[L7]

An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$ (An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$ ).

[L8]

The commutator pairing is independent of the coset representatives, is $F_{p}$ -bilinear on $P / Z (P)$ , and is alternating (The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating).

[L9]

The commutator pairing of an extraspecial $p$ -group has trivial radical (The commutator pairing of an extraspecial $p$ -group has trivial radical).

[L10]

[L11]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ ; the trivial group is permitted (,, ). (Elementary abelian $p$ -groups).

[L12]

The set $S$ is independent when $\prod_{s \in S} s^{a_{s}} = e$ with finite support forces every $a_{s} = 0$ . A basis of an elementary abelian $p$ -group is an independent spanning subset for its canonical $F_{p}$ -linear structure. ( $F_{p}$ -spanning sets, independence, and bases in an elementary abelian $p$ -group).

[L13]

Every finite elementary abelian $p$ -group has a basis; every independent subset extends to a basis, every spanning subset contains a basis, and all bases have the same finite size. (Finite elementary abelian $p$ -groups have bases, basis extension, and a well-defined dimension).

[L14]

$∣ G ∣ = [G : H] ∣ H ∣ .$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L15]

$⟨ S ⟩; : =; ⋂ {H; :; H \leq G and S \subseteq H} .$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L16]

$Z (G) : = {z \in G : z g = g z for every g \in G} .$ (The center $Z (G)$ of a group).

[L17]

An extraspecial $2$ -group of order $2^{1 + 2 n}$ is of plus type when it has $2^{2 n} + 2^{n}$ solutions of $x^{2} = 1$ and of minus type when it has $2^{2 n} - 2^{n}$ such solutions; for odd $p$ , plus and minus mean exponent $p$ and $p^{2}$ respectively (Plus and minus type of an extraspecial $p$ -group).

[L18]

A central product of extraspecial $p$ -groups identified along their centres is extraspecial (A central product of extraspecial $p$ -groups identified along their centres is extraspecial).

[L19]

Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$ pairwise intersecting in its centre (Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$ ).

Proof

technique · direct

1.1L1L2L7L11L14L16

A maximal elementary abelian subgroup $E$ contains $Z (P)$ , so $\overset{ˉ}{E} \leq P / Z (P)$ has order $2^{k}$ with $∣ E ∣ = 2^{k + 1}$ , and $q$ vanishes on $\overset{ˉ}{E}$ .

1.2L1L2L8L10L12L13

On a coset $\overset{ˉ}{E} \overset{ˉ}{v}$ with $\overset{ˉ}{v} \notin {\overset{ˉ}{E}}^{⊥}$ the map $\overset{ˉ}{u} \mapsto b (\overset{ˉ}{u}, \overset{ˉ}{v})$ is a nonzero $F_{2}$ -linear functional on $\overset{ˉ}{E}$ , so its kernel has index two and exactly $2^{k - 1}$ of the coset's classes have $q = 0$ .

2.1L2L10L11L15step 1.1

Maximality of $E$ says no $\overset{ˉ}{v} \notin \overset{ˉ}{E}$ orthogonal to $\overset{ˉ}{E}$ has $q (\overset{ˉ}{v}) = 0$ : the polar identity would make $q$ vanish on $\overset{ˉ}{E} ⟨ \overset{ˉ}{v} ⟩$ , whose preimage is a strictly larger elementary abelian subgroup.

2.2L1step 1.1

On the coset $\overset{ˉ}{E}$ itself $q$ vanishes identically, contributing $2^{k}$ classes.

3.1L1L2step 2.1

On a coset $\overset{ˉ}{E} \overset{ˉ}{v}$ with $\overset{ˉ}{v} \in {\overset{ˉ}{E}}^{⊥} ∖ \overset{ˉ}{E}$ the correction term vanishes, so $q$ is constantly $q (\overset{ˉ}{v}) = 1$ by step 2.1 and the coset contributes no class.

4.1L1L2L3L7L9L14step 2.2step 3.1step 1.2algebra

With $∣ {\overset{ˉ}{E}}^{⊥} ∣ = 2^{2 n - k}$ there are $2^{2 n - 2 k}$ cosets inside the complement and $2^{2 n - k} - 2^{2 n - 2 k}$ outside, so the classes with $q = 0$ number $2^{k} + (2^{2 n - k} - 2^{2 n - 2 k}) 2^{k - 1} = 2^{2 n - 1} + 2^{k} - 2^{2 n - k - 1}$ , and the elements with $x^{2} = 1$ number twice that.

5.1L4L5step 4.1algebra

The classification counts those elements as $2^{2 n} + 2^{n}$ in the plus case and $2^{2 n} - 2^{n}$ in the minus case, so $2^{k} - 2^{2 n - k - 1} = \pm 2^{n - 1}$ .

6.1L11L13L17L18L19step 2.1step 5.1algebra

The left side is strictly increasing in $k$ , so $k = n$ in the plus case and $k = n - 1$ in the minus case are the only solutions; both are attained, giving $∣ E ∣ = 2^{n + 1}$ and $∣ E ∣ = 2^{n}$ for EVERY maximal elementary abelian subgroup.

7.1L6step 1.1∎

Independently, $E$ is abelian, so the maximal-abelian bound gives $∣ E ∣ \leq 2^{n + 1}$ and hence $k \leq n$ without the monotonicity argument; this is the free half of the plus case.

5 · Examples, counterexamples and false statements

None yet.

Extraspecial pp-Groups and Central Products

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Commutator identities in a group whose derived subgroup is central

Statement

Facts & Assumptions

Proof

Remarks

In a group with central derived subgroup, (xy)n=[y,x](n2)xnyn(xy)n=[y,x](2n​)xnyn

Statement

Facts & Assumptions

Proof

Remarks

For an odd prime pp, the pp-th power map is a homomorphism on a finite group whose derived subgroup is central of exponent dividing pp

Statement

Facts & Assumptions

Proof

Remarks

Special and extraspecial pp-groups

Definition

Remarks

Three equivalent descriptions of an extraspecial pp-group

Statement

Facts & Assumptions

Proof

Remarks

An extraspecial pp-group is nilpotent of class exactly two and its derived subgroup has order pp

Statement

Facts & Assumptions

Proof

Remarks

The centre of an extraspecial pp-group has no complement

Statement

Facts & Assumptions

Proof

Remarks

Every conjugacy class of an extraspecial pp-group outside the centre has exactly pp elements

Statement

Facts & Assumptions

Proof

Remarks

A noncentral element of an extraspecial pp-group has centraliser of index pp

Statement

Facts & Assumptions

Proof

Remarks

The central product G∘αHG∘α​H of two groups along an isomorphism of central subgroups

Definition

Remarks

The identified subgroup used to form a central product is central, hence normal

Statement

Facts & Assumptions

Proof

Remarks

The two canonical maps into a central product are injective homomorphisms whose images commute, generate it, and meet in the identified centre

Statement

Facts & Assumptions

Proof

Remarks

Order, centre and derived subgroup of a central product

Statement

Facts & Assumptions

Proof

Remarks

Homomorphisms out of a central product

Statement

Facts & Assumptions

Proof

Remarks

Internal central products of a finite family of subgroups

Definition

Remarks

Internal central products are the images of external ones

Statement

Facts & Assumptions

Proof

Remarks

A central product of extraspecial pp-groups identified along their centres is extraspecial

Extraspecial $p$ -Groups and Central Products

In a group with central derived subgroup, $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$

For an odd prime $p$ , the $p$ -th power map is a homomorphism on a finite group whose derived subgroup is central of exponent dividing $p$

Special and extraspecial $p$ -groups

Three equivalent descriptions of an extraspecial $p$ -group

An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$

The centre of an extraspecial $p$ -group has no complement

Every conjugacy class of an extraspecial $p$ -group outside the centre has exactly $p$ elements

A noncentral element of an extraspecial $p$ -group has centraliser of index $p$

The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups

A central product of extraspecial $p$ -groups identified along their centres is extraspecial

The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre

The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating

The commutator pairing of an extraspecial $p$ -group has trivial radical

Two elements of an extraspecial $p$ -group with nontrivial commutator generate an extraspecial subgroup of order $p^{3}$

Every extraspecial $p$ -group is an internal central product of nonabelian subgroups of order $p^{3}$

An extraspecial $p$ -group has order $p^{1 + 2 n}$ for some $n \geq 1$

An extraspecial $p$ -group of order $p^{1 + 2 n}$ has generator rank $2 n$

In an extraspecial $p$ -group of order $p^{1 + 2 n}$ every maximal abelian subgroup has order $p^{1 + n}$

An extraspecial $p$ -group is the product of two maximal abelian subgroups meeting in its centre

The Heisenberg group of order $p^{3}$ over $Z / p$

The Heisenberg multiplication is a group law, nonabelian, on a set of $p^{3}$ elements

The Heisenberg group of order $p^{3}$ is extraspecial, and for odd $p$ it has exponent $p$

Raising to the power $1 + p$ is an automorphism of order $p$ of a cyclic group of order $p^{2}$