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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]

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]

$⟨ S ⟩$ 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]

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).

[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]

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 $p$ (Three equivalent descriptions of an extraspecial $p$ -group).

Proof

technique · direct

1.1

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 $P_{i} \cap P_{j} = Z (P)$ .

F3L1

2.1

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 $[x_{i}, y_{i}] = z$ .

F1F2L2L5L10step 1.1

3.1

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.

F1L2step 1.1step 2.1

4.1

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.

F2F3L8step 2.1step 3.1

5.1

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.

F3L2step 3.1step 4.1

6.1

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 $B$ .

L3L4L9step 4.1step 5.1

7.1

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 $A \cap B = Z (P)$ .

L6L7L9step 5.1step 6.1

8.1

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

F4L5L10step 2.1step 7.1∎

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.

Depends on

Used by

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Sources

M. van Beek, Topics in Finite p-Groups, Proposition 2.41(iv) and Theorem 2.40(iv) (standard reference, not scraped)

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