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Sylow's Theorems, $p$ -Groups and Nilpotent Groups

1 · Prerequisites

2 · Summary

Finite group actions, orbit-stabilizer, the fixed-point congruence for finite $p$ -groups, and the prime factorization of $∣ G ∣$ provide the counting framework for Sylow theory. The published nilpotence criteria and semidirect-product constructions supply the group-theoretic background used for normalizers, products of normal subgroups, and the concrete finite-group applications.

Sylow subgroups, their number, $p$ -cores, the Fitting subgroup, and the Frattini subgroup are defined first. Sylow I establishes existence, Sylow II gives containment and conjugacy, and Sylow III gives the normalizer index, congruence, and divisibility restrictions. These results lead to normalizer and Frattini arguments, Sylow characterizations of finite nilpotence, and applications to finite groups of specified orders.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Sylow $p$ -subgroups of a finite group

Definition

Let $G$ be a finite group, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $a \in N$ and $p ∤ m$ . A subgroup $P \leq G$ is a Sylow $p$ -subgroup when $∣ P ∣ = p^{a}$ . Equivalently, its order is the largest power of $p$ dividing $∣ G ∣$ . This is a property of a subgroup and does not presume that such a subgroup exists; existence is proved in Sylow I: every finite group has a Sylow $p$ -subgroup.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

The number $n_{p} (G)$ of Sylow $p$ -subgroups

Definition

For a finite group $G$ and a prime $p$ , let ${Syl}_{p} (G)$ be the set of Sylow $p$ -subgroups (Sylow $p$ -subgroups of a finite group). Define $n_{p} (G) : = ∣ {Syl}_{p} (G) ∣ .$ This cardinal is defined even before existence is proved because ${Syl}_{p} (G)$ is a subset of the finite power set of $G$ ; Sylow I: every finite group has a Sylow $p$ -subgroup later shows it is nonzero.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Maximal proper subgroups

Definition

A subgroup $M < G$ is maximal proper when there is no subgroup $H$ with $M < H < G$ . Equivalently, every subgroup containing $M$ is either $M$ or $G$ . The word maximal refers to inclusion among proper subgroups, not to cardinality.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-17Open item page →

$p$ -complements in finite groups

Definition

Let $G$ be finite and let $p$ be prime. A $p$ -complement is a subgroup $H \leq G$ whose order is prime to $p$ and whose index $[G : H]$ is a power of $p$ (The coset set $G / H$ and the index $[G : H]$ of a subgroup). Equivalently, when $∣ G ∣ = p^{a} m$ with $p ∤ m$ , a $p$ -complement has order $m$ . The definition asserts no existence.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

The $p$ -core $O_{p} (G)$ as the largest normal $p$ -subgroup

Definition

For a finite group $G$ and a prime $p$ , the $p$ -core $O_{p} (G)$ is the subgroup generated by all normal $p$ -subgroups of $G$ . There are finitely many such subgroups. If $A$ and $B$ are two of them, then $A B$ is a subgroup by If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ and is normal because $g A B g^{- 1} = A B$ for every $g \in G$ . For a fixed $a b \in A B$ , the fibres of the multiplication map $A \times B \to A B$ are exactly the pairs $(a x, x^{- 1} b)$ with $x \in A \cap B$ , so

∣ A B ∣ = \frac{∣ A ∣ ∣ B ∣}{∣ A \cap B ∣} .

Lagrange's theorem Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ makes $∣ A \cap B ∣$ a power of $p$ , and therefore $A B$ is again a normal $p$ -subgroup. Induction shows that the product of all normal $p$ -subgroups is a normal $p$ -subgroup. It contains every such subgroup, so it is the unique largest normal $p$ -subgroup; this product is $O_{p} (G)$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-17Open item page →

The Fitting subgroup $F (G) = \prod_{p} O_{p} (G)$ of a finite group

Definition

For a finite group $G$ , the Fitting subgroup is $F (G) : = \prod_{p ∣ ∣ G ∣} O_{p} (G),$ the product of its $p$ -cores (The $p$ -core $O_{p} (G)$ as the largest normal $p$ -subgroup). If $A, B ⊴ G$ , then $A B$ is a subgroup, is normal because $g A B g^{- 1} = A B$ , and satisfies $A B = B A$ : indeed $a b = (a b a^{- 1}) a \in B A$ , and the reverse inclusion is symmetric. Induction therefore shows that the finite product of the normal factors $O_{p} (G)$ is a normal subgroup independent of their order. For the trivial group the product is empty and equals $1$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group

Definition

For a finite group $G$ , the Frattini subgroup is $Φ (G) : = ⋂ {M \leq G : M is maximal proper} .$ If $G = 1$ , the family is empty and its intersection inside $G$ is $G$ itself. Thus $Φ (1) = 1$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

The centralizer $C_{G} (H)$ of a subgroup

Definition

Let $G$ be a group (Group and abelian group) and let $H \leq G$ . The centralizer of $H$ in $G$ is

$C_{G} (H) : = {g \in G : g h = h g for every h \in H} .$

Equivalently, $C_{G} (H) = ⋂_{h \in H} C_{G} (h)$ , the intersection of the centralizers of the individual elements of $H$ (The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

Why it is a subgroup. The identity satisfies $e h = h = h e$ for every $h \in H$ , so $e \in C_{G} (H)$ . If $g_{1}, g_{2} \in C_{G} (H)$ and $h \in H$ , then

$(g_{1} g_{2}) h = g_{1} (g_{2} h) = g_{1} (h g_{2}) = (g_{1} h) g_{2} = (h g_{1}) g_{2} = h (g_{1} g_{2}),$

so $g_{1} g_{2} \in C_{G} (H)$ . If $g \in C_{G} (H)$ and $h \in H$ , then multiplying $g h = h g$ by $g^{- 1}$ on both sides gives $h g^{- 1} = g^{- 1} h$ , so $g^{- 1} \in C_{G} (H)$ . Hence $C_{G} (H) \leq G$ .

Two special cases are used without further comment. Taking $H = G$ gives $C_{G} (G) = Z (G)$ , the center (The center $Z (G)$ of a group). Intersecting with $H$ gives $C_{G} (H) \cap H = Z (H)$ , since an element of $H$ lies in $C_{G} (H)$ exactly when it commutes with every element of $H$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

The centralizer of a normal subgroup is normal

Statement

If $N ⊴ G$ , then $C_{G} (N) ⊴ G$ .

Facts & Assumptions

Given: A group $G$ and a normal subgroup $N ⊴ G$ .

[L1]

$C_{G} (N) = {g \in G : g n = n g for every n \in N}$ , and it is a subgroup of $G$ (The centralizer $C_{G} (H)$ of a subgroup).

[L2]

$N ⊴ G$ means $g N g^{- 1} = N$ for every $g \in G$ (Normal subgroup: invariance under conjugation).

Proof

technique · direct

1.1

Fix $g \in G$ , $x \in C_{G} (N)$ and $n \in N$ . By [L2] the element $g^{- 1} n g$ lies in $N$ , so $x (g^{- 1} n g) = (g^{- 1} n g) x$ by [L1].

L1L2given

2.1

Multiplying that identity by $g$ on the left and by $g^{- 1}$ on the right gives $(g x g^{- 1}) n = g x (g^{- 1} n g) g^{- 1} = g (g^{- 1} n g) x g^{- 1} = n (g x g^{- 1})$ , and since $n \in N$ was arbitrary, $g x g^{- 1} \in C_{G} (N)$ by [L1].

L1step 1.1algebra

3.1

Hence $g C_{G} (N) g^{- 1} \subseteq C_{G} (N)$ for every $g \in G$ . Applying this inclusion to $g^{- 1}$ and conjugating by $g$ gives $C_{G} (N) \subseteq g C_{G} (N) g^{- 1}$ , so $g C_{G} (N) g^{- 1} = C_{G} (N)$ for every $g \in G$ , which is normality by [L2]. This proves the stated claim.

L1L2step 2.1∎

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

A nontrivial normal subgroup of a solvable group contains a nontrivial abelian subgroup normal in the whole group

Statement

Let $Q$ be a solvable group and let $N ⊴ Q$ with $N \neq 1$ . Then $N$ contains a subgroup $A \neq 1$ that is abelian and normal in $Q$ .

Facts & Assumptions

Given: A solvable group $Q$ and a normal subgroup $N ⊴ Q$ with $N \neq 1$ .

[L1]

The derived series of a group $N$ is $N^{(0)} = N$ , $N^{(r + 1)} = [N^{(r)}, N^{(r)}]$ ; $N$ is solvable when $N^{(n)} = 1$ for some $n \in N$ , and its derived length is the least such $n$ . The trivial group has derived length $0$ (The derived series, solvable groups, and derived length).

[L2]

Every subgroup and every quotient of a solvable group is solvable; no finiteness hypothesis is required (Subgroups and quotients of solvable groups are solvable).

[L3]

For every group $G$ , the derived subgroup $G^{'} = [G, G]$ is characteristic, hence normal (The derived subgroup is characteristic and the abelianization is universal).

[L4]

If $K char H$ and $H char G$ , then $K char G$ (Characteristic subgroups are normal, and characteristicity is transitive).

[L5]

If $K$ is characteristic in $N$ and $N ⊴ G$ , then $K ⊴ G$ (If $K$ is characteristic in $N$ and $N$ is normal in $G$ , then $K$ is normal in $G$ ).

[L6]

Let $N ⊴ G$ . Then $G / N$ is abelian if and only if $[G, G] \subseteq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

Proof

technique · direct

1.1

$N$ is a subgroup of the solvable group $Q$ , so $N$ is solvable by [L2]. Let $n$ be its derived length, so $N^{(n)} = 1$ and $N^{(r)} \neq 1$ for $r < n$ by the leastness in [L1]. Since $N \neq 1$ and the trivial group is the only group of derived length $0$ , we have $n \geq 1$ .

L1L2given

2.1

Put $A : = N^{(n - 1)}$ . Then $A \neq 1$ by the leastness in step 1.1, and $A \leq N$ because each term of the derived series lies in the preceding one by [L1].

L1step 1.1

2.2

$[A, A] = N^{(n)} = 1$ by [L1] and step 1.1, so $A / 1$ is abelian by [L6] applied to the trivial normal subgroup of $A$ ; that is, $A$ is abelian.

L1L6step 1.1

3.1

Each term of the derived series is characteristic in the preceding term by [L3], so iterating [L4] along $N^{(n - 1)} char \dots char N^{(0)} = N$ makes $A$ characteristic in $N$ .

L1L3L4step 2.1

4.1

$A$ is characteristic in $N$ and $N ⊴ Q$ , so $A ⊴ Q$ by [L5]. With steps 2.1 and 2.2, $A$ is a nontrivial abelian subgroup of $N$ that is normal in $Q$ . This proves the stated claim.

L5givenstep 2.1step 2.2step 3.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

If $∣ G ∣ = p^{a} m$ with $p ∤ m$ , then $v_{p} (\binom{p^{a} m}{p^{a}}) = 0$

Statement

Let $p$ be prime and let $a, m \in N$ satisfy $p ∤ m$ . Then $v_{p} ⁣ ((\binom{p^{a} m}{p^{a}})) = 0.$ The valuation is applied only to nonzero integers. See The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a prime $p$ and a nonzero integer $x$ , the $p$ -adic valuation is $v_{p} (x) : = \max {k \in N : p^{k} ∣ x};$ the value $v_{p} (0)$ is undefined. (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).

[L2]

For a prime $p$ and nonzero integers $x, y$ , the product $x y$ is nonzero and $v_{p} (x y) = v_{p} (x) + v_{p} (y) .$ ( $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ for nonzero integers $a, b$ , and $v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)}$ whenever $a$ , $b$ and $a + b$ are all nonzero).

[L3]

For a finite set $A$ and $k \in N$ put $[A]^{k} : = {S \subseteq A : ∣ S ∣ = k},$ the set of $k$ -element subsets of $A$ . Every $S \subseteq A$ is finite (thm-subset-of-a-finite-set), so the condition $∣ S ∣ = k$ makes sense for every subset. (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L4]

Let $n, k \in N$ with $k \leq n$ . Then $(\binom{n}{k}) k! (n - k)! = n!,$ and consequently $(\binom{n}{k}) k! = n^{\underline{k}}$ . ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L5]

The factorial and falling factorial satisfy $0! = 1, (n + 1)! = n! (n + 1), n^{\underline{0}} = 1, n^{\underline{k + 1}} = n^{\underline{k}} (n - k),$ and for $k \leq n$ , $n^{\underline{k}}$ is the product $n (n - 1) \dots (n - k + 1)$ . (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

Proof

technique · direct

1.1

We apply the valuation to the identity $(\binom{p^{a} m}{p^{a}}) (p^{a})! = (p^{a} m)^{\underline{p^{a}}}$ .

L1L2L3L4L5givenalgebra

2.1

If $0 < j < p^{a}$ , write $j = p^{r} u$ with $p ∤ u$ ; then $r < a$ and $p^{a} m - j = p^{r} (p^{a - r} m - u)$ , whose parenthesized factor is prime to $p$ . Hence $v_{p} (p^{a} m - j) = v_{p} (j)$ . The factor $p^{a} m$ contributes $a + v_{p} (m) = a$ , exactly matching the contribution of $p^{a}$ in $(p^{a})!$ , so all valuations cancel.

step 1.1givenalgebra

3.1

If $a = 0$ , the binomial coefficient is $(\binom{m}{1}) = m$ , which is nonzero and prime to $p$ . No valuation of zero occurs in either case. This proves the stated claim.

step 2.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Sylow I: every finite group has a Sylow $p$ -subgroup

Statement

Let $G$ be finite, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then $G$ has a subgroup of order $p^{a}$ , hence a Sylow $p$ -subgroup (Sylow $p$ -subgroups of a finite group). See Sylow $p$ -subgroups of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be a finite group, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $a \in N$ and $p ∤ m$ . A subgroup $P \leq G$ is a Sylow $p$ -subgroup when $∣ P ∣ = p^{a}$ . Equivalently, its order is the largest power of $p$ dividing $∣ G ∣$ . This is a property of a subgroup and does not presume that such a subgroup exists; existence is proved in thm-sylow-first-theorem. (Sylow $p$ -subgroups of a finite group).

[L2]

Let $p$ be prime and let $a, m \in N$ satisfy $p ∤ m$ . Then $v_{p} ⁣ ((\binom{p^{a} m}{p^{a}})) = 0.$ The valuation is applied only to nonzero integers. (If $∣ G ∣ = p^{a} m$ with $p ∤ m$ , then $v_{p} (\binom{p^{a} m}{p^{a}}) = 0$ ).

[L3]

Let $G$ act on $X$ and let $x \in X$ . The rule $Φ : G / G_{x} ⟶ G \cdot x, Φ (g G_{x}) = g \cdot x,$ is well-defined and bijective. Thus every orbit is naturally in bijection with the left cosets of its stabilizer. (Orbit-stabiliser: $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a well-defined bijection).

[L4]

For an action of $G$ on $X$ and $x \in X$ , $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite. In particular, if $G$ is finite, then $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣ .$ . (Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$ ).

[L5]

Let $G$ be a finite group and $H \leq G$ . Then $∣ G ∣ = [G : H] ∣ H ∣ .$ Consequently, under the canonical embedding $ι : N \to Z$ , $∣ H ∣$ divides $∣ G ∣$ . (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L6]

Let $P$ be a property of naturals such that for every $n \in N$ , if $P (m)$ holds for all $m < n$ then $P (n)$ . Then $P (n)$ holds for all $n \in N$ . (At $n = 0$ the hypothesis is vacuous, so $P (0)$ is forced.). (Strong (complete) induction).

[L7]

For a left action of $G$ on $X$ , the relation $x \sim y$ defined by $y = g \cdot x$ for some $g \in G$ is an equivalence relation whose class at $x$ is $G \cdot x$ , and the distinct orbits partition $X$ (The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set).

Proof

technique · direct

1.1

Argue by strong induction [L6] on $∣ G ∣$ , the induction statement being that every finite group of order $n$ has a subgroup of order $p^{a}$ whenever $n = p^{a} m$ with $p ∤ m$ . Let $Ω$ be the set of subsets of $G$ of size $p^{a}$ and let $G$ act on $Ω$ by left translation, $g \cdot A : = g A$ ; this is an action, and $∣ g A ∣ = ∣ A ∣$ because left translation is a bijection of $G$ . Counting subsets gives $∣ Ω ∣ = (\binom{p^{a} m}{p^{a}})$ , so [L2] yields $v_{p} (∣ Ω ∣) = 0$ , that is $p ∤ ∣ Ω ∣$ .

L2L6givenalgebra

2.1

By [L7] the orbits partition $Ω$ , so $∣ Ω ∣$ is the sum of the orbit sizes. Were $p$ to divide every orbit size it would divide $∣ Ω ∣$ , so some orbit $O$ has $p ∤ ∣ O ∣$ . Choose $A \in O$ and put $H = G_{A} = {g \in G : g A = A}$ . The bijection of [L3] between $G / H$ and $O$ gives $∣ O ∣ = [G : H]$ , so [L4] gives $p^{a} m = ∣ G ∣ = ∣ H ∣ ∣ O ∣$ ; since $p ∤ ∣ O ∣$ , the full power $p^{a}$ divides $∣ H ∣$ .

step 1.1L3L4L7choose

3.1

Suppose $H = G$ . Then $g A = A$ for every $g \in G$ , so for any $x \in A$ the set $A$ contains $G x = G$ , whence $A = G$ and $p^{a} = ∣ A ∣ = ∣ G ∣ = p^{a} m$ . Thus $m = 1$ and $G$ itself is a subgroup of order $p^{a}$ .

step 2.1givenalgebra

3.2

Suppose instead $H \neq G$ , so $∣ H ∣ < ∣ G ∣$ . By [L5], $∣ H ∣$ divides $p^{a} m$ ; writing $∣ H ∣ = p^{b} m^{'}$ with $p ∤ m^{'}$ , step 2.1 gives $b \geq a$ , while $p^{b} ∣ p^{a} m$ with $p ∤ m$ gives $b \leq a$ . Hence $∣ H ∣ = p^{a} m^{'}$ with $p ∤ m^{'}$ , and the induction hypothesis applied to $H$ supplies a subgroup of $H$ of order $p^{a}$ , which is a subgroup of $G$ .

step 2.1L5L6givenalgebra

4.1

Steps 3.1 and 3.2 are exhaustive, so $G$ has a subgroup $P$ of order $p^{a}$ , and $∣ P ∣$ is the largest power of $p$ dividing $∣ G ∣$ , so $P$ is a Sylow $p$ -subgroup by [L1]. At $a = 0$ the argument returns the trivial subgroup, of order $p^{0} = 1$ ; for the trivial group $∣ G ∣ = 1$ this is $G$ itself, and $m = 1$ is the case settled in step 3.1.

L1step 3.1step 3.2given∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class

Statement

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ . In particular, for every Sylow $p$ -subgroup $Q$ there is $g \in G$ with $Q = g P g^{- 1}$ , so the Sylow $p$ -subgroups form one conjugacy class. See Sylow I: every finite group has a Sylow $p$ -subgroup.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be finite, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then $G$ has a subgroup of order $p^{a}$ , hence a Sylow $p$ -subgroup (def-sylow-p-subgroup). (Sylow I: every finite group has a Sylow $p$ -subgroup).

[L2]

If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p) .$ . (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L3]

For a finite group $G$ and a prime $p$ , let ${Syl}_{p} (G)$ be the set of Sylow $p$ -subgroups (def-sylow-p-subgroup). Define $n_{p} (G) : = ∣ {Syl}_{p} (G) ∣ .$ This cardinal is defined even before existence is proved because ${Syl}_{p} (G)$ is a subset of the finite power set of $G$ ; thm-sylow-first-theorem later shows it is nonzero. (The number $n_{p} (G)$ of Sylow $p$ -subgroups).

[L4]

Let $H \leq G$ be a subgroup (def-subgroup). The normalizer of $H$ in $G$ is $N_{G} (H) : = {g \in G : g H g^{- 1} = H} .$ Thus $g \in N_{G} (H)$ exactly when the conjugation automorphism $c_{g}$ preserves $H$ setwise (thm-conjugation-is-an-automorphism). The subgroup property is proved in lem-centralizers-and-normalizers-are-subgroups. (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

[L5]

Proof

technique · direct

1.1

Let a finite $p$ -subgroup $H$ act by left multiplication on $G / P$ .

L1L2L3L4L5givenalgebra

2.1

Its cardinality is prime to $p$ , so the fixed-point congruence gives a fixed coset $g P$ , and the fixed-coset condition is exactly $H \leq g P g^{- 1}$ .

step 1.1givenalgebra

3.1

Taking $H$ Sylow, Lagrange's theorem turns containment into equality; applying this to any Sylow $p$ -subgroup $Q$ gives $Q = g P g^{- 1}$ .

step 2.1givenalgebra

4.1

If $H = {1}$ , then $X^{H} = X$ and step 2.1 returns any coset, with ${1} \leq g P g^{- 1}$ for every $g$ . If $p ∤ ∣ G ∣$ , then $a = 0$ , so $P = {1}$ is the Sylow $p$ -subgroup and the only $p$ -subgroup of $G$ is ${1}$ itself, which is already equal to it. This proves the stated claim.

step 3.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Sylow III*: $n_{p} (G) = [G : N_{G} (P)]$

Statement

If $P$ is a Sylow $p$ -subgroup of a finite group $G$ , then $n_{p} (G) = [G : N_{G} (P)] .$ See Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ . In particular, for every Sylow $p$ -subgroup $Q$ there is $g \in G$ with $Q = g P g^{- 1}$ , so the Sylow $p$ -subgroups form one conjugacy class. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L2]

[L3]

Let $H \leq G$ . The rule $G / N_{G} (H) ⟶ {g H g^{- 1} : g \in G}, g N_{G} (H) ⟼ g H g^{- 1},$ is a well-defined bijection. If $G$ is finite, the number of distinct conjugates of $H$ is $[G : N_{G} (H)]$ . (The conjugates of $H$ are in bijection with $G / N_{G} (H)$ and, for finite $G$ , number $[G : N_{G} (H)]$ ).

Proof

technique · direct

1.1

Conjugation is transitive on the Sylow $p$ -subgroups by Sylow II, and the stabilizer of $P$ is exactly $N_{G} (P)$ .

L1L2L3givenalgebra

2.1

By step 1.1 the set of Sylow $p$ -subgroups is exactly the conjugacy class of $P$ , so [L3] counts it as $[G : N_{G} (P)]$ , and [L2] identifies that count with $n_{p} (G)$ . If $P$ is the unique Sylow $p$ -subgroup, then $N_{G} (P) = G$ and both sides are $1$ ; if $p ∤ ∣ G ∣$ , then $P = {1}$ , again with $N_{G} (P) = G$ and both sides $1$ . This proves the stated claim.

step 1.1L2L3givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$

Statement

Let $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then the number of Sylow $p$ -subgroups satisfies $n_{p} (G) \equiv 1 (m o d p), n_{p} (G) ∣ m .$ See Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ . In particular, for every Sylow $p$ -subgroup $Q$ there is $g \in G$ with $Q = g P g^{- 1}$ , so the Sylow $p$ -subgroups form one conjugacy class. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L2]

If $P$ is a Sylow $p$ -subgroup of a finite group $G$ , then $n_{p} (G) = [G : N_{G} (P)] .$ . (Sylow III*: $n_{p} (G) = [G : N_{G} (P)]$ ).

[L3]

[L4]

[L5]

Let $G$ be finite, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then $G$ has a subgroup of order $p^{a}$ , hence a Sylow $p$ -subgroup (Sylow I: every finite group has a Sylow $p$ -subgroup).

Proof

technique · direct

1.1

By [L5] the set ${Syl}_{p} (G)$ of [L4] is nonempty, so fix $P \in {Syl}_{p} (G)$ and let it act by conjugation on that set; the action is well defined because a conjugate of a Sylow $p$ -subgroup again has order $p^{a}$ .

L4L5givenalgebra

2.1

Suppose $P$ fixes $Q \in {Syl}_{p} (G)$ , so $P \leq N_{G} (Q)$ , and also $Q \leq N_{G} (Q)$ . Both have order $p^{a}$ , the full power of $p$ dividing $∣ G ∣$ and hence dividing $∣ N_{G} (Q) ∣$ , so both are Sylow $p$ -subgroups of $N_{G} (Q)$ ; by [L1] applied inside $N_{G} (Q)$ they are conjugate there, and $Q ⊴ N_{G} (Q)$ makes every such conjugate equal to $Q$ , so $P = Q$ . Thus $P$ itself is the only fixed point, and since $P$ is a finite $p$ -group acting on the finite set ${Syl}_{p} (G)$ , [L3] gives $n_{p} (G) \equiv 1 (m o d p)$ .

step 1.1L1L3L4givenalgebra

3.1

By [L2], $n_{p} (G) = [G : N_{G} (P)]$ . Since $P \leq N_{G} (P) \leq G$ , the index tower gives $[G : P] = [G : N_{G} (P)] [N_{G} (P) : P]$ , and $[G : P] = m$ because $∣ P ∣ = p^{a}$ ; hence $n_{p} (G)$ divides $m$ . This proves the stated claim.

step 1.1step 2.1L2givenalgebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

A Sylow $p$ -subgroup is normal if and only if it is unique

Statement

A Sylow $p$ -subgroup of a finite group is normal if and only if it is the unique Sylow $p$ -subgroup. See Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ . In particular, for every Sylow $p$ -subgroup $Q$ there is $g \in G$ with $Q = g P g^{- 1}$ , so the Sylow $p$ -subgroups form one conjugacy class. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L2]

[L3]

Let $G$ be a group and let $N \leq G$ be a subgroup (def-subgroup). For $g \in G$ , write $g N g^{- 1} : = {g n g^{- 1} : n \in N} .$ The subgroup $N$ is normal in $G$ when $g N g^{- 1} = N for every g \in G .$ In that case write $N ⊴ G$ . Equivalently, every inner conjugation of $G$ maps $N$ onto itself. The connection with equality of the left and right cosets of def-coset is proved in thm-normal-subgroup-characterisations. (Normal subgroup: invariance under conjugation).

Proof

technique · direct

1.1

A normal Sylow subgroup is fixed by every conjugation, and Sylow II says every Sylow subgroup is one of its conjugates.

L1L2L3givenalgebra

2.1

Conversely, uniqueness makes the subgroup conjugation-invariant. This proves the stated claim.

step 1.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

A subgroup containing the normalizer of a Sylow subgroup is self-normalizing

Statement

Let $P$ be a Sylow $p$ -subgroup of a finite group $G$ . If $N_{G} (P) \leq H \leq G$ , then $N_{G} (H) = H$ . In particular, $N_{G} (N_{G} (P)) = N_{G} (P)$ . See Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ . In particular, for every Sylow $p$ -subgroup $Q$ there is $g \in G$ with $Q = g P g^{- 1}$ , so the Sylow $p$ -subgroups form one conjugacy class. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L2]

Proof

technique · direct

1.1

For $N_{G} (P) \leq H \leq G$ and $x \in N_{G} (H)$ , the groups $P$ and $x P x^{- 1}$ are Sylow in $H$ .

L1L2givenalgebra

2.1

Conjugate them inside $H$ ; the resulting element puts $x$ in $H$ .

step 1.1givenalgebra

3.1

Specialize to $H = N_{G} (P)$ to obtain $N_{G} (N_{G} (P)) = N_{G} (P)$ . This proves the stated claim.

step 2.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Frattini argument: if $N ⊴ G$ and $P$ is Sylow in $N$ , then $G = N N_{G} (P)$

Statement

If $N ⊴ G$ is finite and $P$ is a Sylow $p$ -subgroup of $N$ , then $G = N N_{G} (P) .$ See Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ . In particular, for every Sylow $p$ -subgroup $Q$ there is $g \in G$ with $Q = g P g^{- 1}$ , so the Sylow $p$ -subgroups form one conjugacy class. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L2]

[L3]

If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ . Here $H N : = {h n : h \in H, n \in N}$ . (If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ ).

Proof

technique · direct

1.1

For each $g \in G$ , normality makes $g P g^{- 1}$ Sylow in $N$ .

L1L2L3givenalgebra

2.1

Sylow II inside $N$ supplies $n \in N$ conjugating it back to $P$ , so $n g \in N_{G} (P)$ and $g \in N N_{G} (P)$ .

step 1.1givenalgebra

3.1

The reverse containment is immediate. This proves the stated claim.

step 2.1givenalgebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

A normal Sylow subgroup of a normal subgroup is normal in the whole group

Statement

If $N ⊴ G$ and $P$ is a normal Sylow $p$ -subgroup of $N$ , then $P ⊴ G$ . See A Sylow $p$ -subgroup is normal if and only if it is unique.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A Sylow $p$ -subgroup of a finite group is normal if and only if it is the unique Sylow $p$ -subgroup. (A Sylow $p$ -subgroup is normal if and only if it is unique).

[L2]

If $K$ is characteristic in $N$ and $N ⊴ G$ , then $K ⊴ G$ . (If $K$ is characteristic in $N$ and $N$ is normal in $G$ , then $K$ is normal in $G$ ).

Proof

technique · direct

1.1

A unique Sylow subgroup is preserved by every automorphism of its ambient normal subgroup, so it is characteristic there; characteristic-in-normal gives normality in the whole group.

L1L2givenalgebra

2.1

Step 1.1 uses only that $P$ is the unique Sylow $p$ -subgroup of $N$ , so the degenerate cases are covered as well: if $p ∤ ∣ N ∣$ then $P = {1}$ , which is normal in $G$ ; if $N = {1}$ then $P = {1}$ likewise; and if $N = G$ the asserted normality is the hypothesis itself. This proves the stated claim.

step 1.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Every proper subgroup of a finite nilpotent group is properly contained in its normalizer

Statement

Every proper subgroup of a finite nilpotent group is properly contained in its normalizer. See Nilpotence via central series, the upper central series, and the lower central series.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a group $G$ and $c \in N$ , the following are equivalent: 1. $G$ has a central series $1 = H_{0} \leq \dots \leq H_{c} = G$ ; 2. $Z_{c} (G) = G$ ; 3. $γ_{c + 1} (G) = 1$ . Hence $G$ is nilpotent exactly when its lower central series reaches $1$ , and the least such $c$ is its nilpotency class. (Nilpotence via central series, the upper central series, and the lower central series).

[L2]

Proof

technique · direct

1.1

We use a central series and take the first term not contained in the proper subgroup $H$ .

L1L2givenalgebra

2.1

The preceding term lies in $H$ , so an element newly appearing at that stage normalizes $H$ modulo the preceding term but is not in $H$ .

step 1.1givenalgebra

3.1

Both boundary cases are admitted and hold. If $G$ has nilpotency class zero then $G = 1$ , which has no proper subgroup, so the claim is vacuously true and step 1.1 is never entered. If $H = 1$ and $G \neq 1$ , then $N_{G} (H) = G$ , which properly contains $H$ . This proves the stated claim.

step 1.1step 2.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Maximal subgroups of finite nilpotent groups are normal of prime index

Statement

Every maximal proper subgroup of a finite nilpotent group is normal and has prime index. See Maximal proper subgroups.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A subgroup $M < G$ is maximal proper when there is no subgroup $H$ with $M < H < G$ . Equivalently, every subgroup containing $M$ is either $M$ or $G$ . The word maximal refers to inclusion among proper subgroups, not to cardinality. (Maximal proper subgroups).

[L2]

Every proper subgroup of a finite nilpotent group is properly contained in its normalizer. (Every proper subgroup of a finite nilpotent group is properly contained in its normalizer).

[L3]

Let $G$ be a finite group and let $p$ be prime. If $p ∣ ∣ G ∣$ , then $G$ contains an element of order $p$ . (Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ ).

[L4]

Every subgroup and every quotient of a nilpotent group is nilpotent. Every finite direct product of nilpotent groups is nilpotent; the class of a subgroup or quotient is at most the class of the original group, and the class of a nonempty finite product is at most the maximum of the factor classes. The empty product is the trivial group of class zero. (Subgroups, quotients, and finite direct products of nilpotent groups are nilpotent).

Proof

technique · direct

1.1

The normalizer condition and maximality force $N_{G} (M) = G$ , hence $M ⊴ G$ .

L1L2L3L4givenalgebra

2.1

The quotient has no nontrivial proper subgroup; Cauchy's theorem then forces its nontrivial order to be prime.

step 1.1givenalgebra

3.1

A maximal subgroup is proper by definition, so $M \neq G$ and the quotient $G / M$ of step 2.1 is nontrivial; its order is therefore a genuine prime rather than $1$ , and step 1.1 has already made $M$ normal. This proves the stated claim.

step 1.1step 2.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Every nontrivial finite $p$ -group has a normal subgroup of index $p$

Statement

Every nontrivial finite $p$ -group has a normal subgroup of index $p$ . See Every finite $p$ -group is nilpotent.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Every finite $p$ -group is nilpotent. The trivial group is included and has nilpotency class zero. (Every finite $p$ -group is nilpotent).

[L2]

Every maximal proper subgroup of a finite nilpotent group is normal and has prime index. (Maximal subgroups of finite nilpotent groups are normal of prime index).

[L3]

Proof

technique · direct

1.1

In the finite nonempty collection of proper subgroups choose one maximal by inclusion.

L1L2L3givenalgebra

2.1

The nilpotent maximal-subgroup theorem makes its index prime, while Lagrange makes that index divide a power of $p$ , so it is $p$ .

step 1.1givenalgebra

3.1

The maximal-subgroup theorem also makes $M$ normal, so $M$ is the required normal subgroup of index $p$ . This proves the stated claim.

step 2.1givenalgebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Distinct normal Sylow subgroups centralize one another

Statement

Normal Sylow subgroups for distinct primes centralize one another. See Sylow $p$ -subgroups of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

[L2]

For subgroups $A, B \leq G$ , their subgroup commutator is $[A, B] = ⟨ [a, b] : a \in A, b \in B ⟩,$ where $[a, b] = a b a^{- 1} b^{- 1}$ (def-commutator-and-commutator-subgroup, def-generated-subgroup). The lower central series is $γ_{1} (G) = G, γ_{r + 1} (G) = [G, γ_{r} (G)] (r \geq 1) .$ Each $γ_{r} (G)$ is characteristic in $G$ , and the series descends because $[G, N] \leq N$ whenever $N ⊴ G$ . (Subgroup commutators and the lower central series).

[L3]

[L4]

Proof

technique · direct

1.1

For normal Sylow $p$ - and $q$ -subgroups with $p \neq q$ , every commutator lies in their intersection.

L1L2L3L4givenalgebra

2.1

Lagrange makes that intersection trivial because its order divides coprime prime powers. This proves the stated claim.

step 1.1givenalgebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

A finite group is nilpotent if and only if all Sylow subgroups are normal, if and only if it is their internal direct product

Statement

For a finite group $G$ , the following are equivalent: $G$ is nilpotent; every Sylow subgroup is normal; and $G$ is the internal direct product of its Sylow subgroups. See Every proper subgroup of a finite nilpotent group is properly contained in its normalizer.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Every proper subgroup of a finite nilpotent group is properly contained in its normalizer. (Every proper subgroup of a finite nilpotent group is properly contained in its normalizer).

[L2]

Let $P$ be a Sylow $p$ -subgroup of a finite group $G$ . If $N_{G} (P) \leq H \leq G$ , then $N_{G} (H) = H$ . In particular, $N_{G} (N_{G} (P)) = N_{G} (P)$ . (A subgroup containing the normalizer of a Sylow subgroup is self-normalizing).

[L3]

[L4]

Normal Sylow subgroups for distinct primes centralize one another. (Distinct normal Sylow subgroups centralize one another).

[L5]

Let $N_{0}, \dots, N_{r - 1} ⊴ G$ . The following are equivalent: the $N_{i}$ form an internal direct product of $G$ ; every $g \in G$ has a unique expression $g = n_{0} \dots n_{r - 1}$ with $n_{i} \in N_{i}$ ; and the multiplication map $μ : \prod_{i < r} N_{i} \to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

[L6]

[L7]

Every finite $p$ -group is nilpotent. The trivial group is included and has nilpotency class zero. (Every finite $p$ -group is nilpotent).

Proof

technique · direct

1.1

If $G$ is nilpotent and $P$ is Sylow, self-normalization of $N_{G} (P)$ contradicts the normalizer condition unless $N_{G} (P) = G$ .

L1L2L3L4L5L6L7givenalgebra

2.1

Normal Sylow subgroups for distinct primes commute and have trivial intersections; their product has order $∣ G ∣$ , so it is the internal direct product.

step 1.1givenalgebra

3.1

Conversely each factor is a finite $p$ -group and hence nilpotent, and a finite direct product of nilpotent groups is nilpotent.

step 2.1givenalgebra

4.1

The two degenerate cases are admitted by [L5] and hold. For $G = 1$ the family of Sylow subgroups is empty, the empty internal direct product is the trivial group, and $G$ is nilpotent of class zero by [L7]. If $∣ G ∣$ is a power of a single prime, the family has one member, namely $G$ itself, the one-factor internal direct product is $G$ , and [L7] again makes $G$ nilpotent. This proves the stated claim.

L5L7step 3.1givenalgebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

A finite product of normal $p$ -subgroups is a normal $p$ -subgroup

Statement

A finite product of normal $p$ -subgroups of a group is a normal $p$ -subgroup. The empty product is the trivial subgroup. See A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $p$ be a prime natural number (def-prime). A finite $p$ -group is a finite group $P$ (def-group, def-finite-cardinality) whose order has the form $∣ P ∣ = p^{n}$ for some $n \in N$ , with natural exponentiation as in def-nat-power. The case $n = 0$ permits the trivial group. A finite $p$ -group is nontrivial exactly when $n \geq 1$ . (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L2]

If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ . Here $H N : = {h n : h \in H, n \in N}$ . (If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ ).

[L3]

If $H$ is a subgroup of a finite group $G$ , then $∣ G ∣ = [G : H] ∣ H ∣$ ; in particular $∣ H ∣$ divides $∣ G ∣$ . (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

Induct on the number of factors; the empty product is the trivial normal $p$ -subgroup.

L1L2givenalgebra

2.1

Let $A$ and $B$ be normal $p$ -subgroups. By [L2], $A B$ is a subgroup, and $g A B g^{- 1} = A B$ for every ambient-group element $g$ , so $A B$ is normal. The multiplication map $A \times B \to A B$ is surjective. For a fixed factorization $a b$ , all its preimages are exactly $(a x, x^{- 1} b)$ with $x \in A \cap B$ . Thus every fibre has $∣ A \cap B ∣$ elements and $∣ A B ∣ = ∣ A ∣ ∣ B ∣ / ∣ A \cap B ∣$ . By [L3], $∣ A \cap B ∣$ is a power of $p$ , so $∣ A B ∣$ is a power of $p$ .

step 1.1L2L3givenalgebra

3.1

Applying step 2.1 repeatedly proves the result for every finite product, including one factor and repeated factors.

step 1.1step 2.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group

Statement

For every finite group $G$ , $F (G)$ is nilpotent and normal, and every normal nilpotent subgroup of $G$ is contained in $F (G)$ . See The Fitting subgroup $F (G) = \prod_{p} O_{p} (G)$ of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a finite group $G$ , the Fitting subgroup is $F (G) : = \prod_{p ∣ ∣ G ∣} O_{p} (G),$ the product of its $p$ -cores (def-p-core-of-a-finite-group). The factors are normal, so their finite product is a normal subgroup and does not depend on the order of multiplication. For the trivial group the product is empty and equals $1$ . (The Fitting subgroup $F (G) = \prod_{p} O_{p} (G)$ of a finite group).

[L2]

For a finite group $G$ , the following are equivalent: $G$ is nilpotent; every Sylow subgroup is normal; and $G$ is the internal direct product of its Sylow subgroups. (A finite group is nilpotent if and only if all Sylow subgroups are normal, if and only if it is their internal direct product).

[L3]

Let $N_{0}, \dots, N_{r - 1} ⊴ G$ . They form an internal direct product of $G$ if and only if every $g \in G$ has a unique expression $g = n_{0} \dots n_{r - 1}$ with $n_{i} \in N_{i}$ , equivalently the multiplication map $\prod_{i < r} N_{i} \to G$ is an isomorphism. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

[L4]

If $K$ is characteristic in $N$ and $N ⊴ G$ , then $K ⊴ G$ . (If $K$ is characteristic in $N$ and $N$ is normal in $G$ , then $K$ is normal in $G$ ).

[L5]

The $p$ -core $O_{p} (G)$ is the unique largest normal $p$ -subgroup of the finite group $G$ . (The $p$ -core $O_{p} (G)$ as the largest normal $p$ -subgroup).

[L6]

Every $p$ -subgroup of a finite group is contained in a Sylow $p$ -subgroup, and all Sylow $p$ -subgroups are conjugate. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L7]

The order of a subgroup of a finite group divides the order of the group. (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

For distinct primes $p$ and $q$ , normality puts every commutator of an element of $O_{p} (G)$ with an element of $O_{q} (G)$ in $O_{p} (G) \cap O_{q} (G)$ . By [L7], the order of this intersection divides powers of both $p$ and $q$ , so the intersection is trivial and the two $p$ -cores centralize one another. If a product $\prod_{p} x_{p}$ with $x_{p} \in O_{p} (G)$ equals $1$ , then for each $p$ the element $x_{p}$ lies both in the $p$ -group $O_{p} (G)$ and in the commuting product of the other prime-power groups; its order divides two coprime numbers and is therefore $1$ . Thus product expressions are unique, so [L3] identifies $F (G)$ with the internal direct product of the $p$ -cores. The direct-product order shows that $O_{p} (G)$ is the Sylow $p$ -subgroup of $F (G)$ ; [L2] makes $F (G)$ nilpotent, and [L1] gives its normality in $G$ .

L1L2L3L5L7givenalgebra

2.1

If $N ⊴ G$ is nilpotent, [L2] makes each Sylow subgroup normal in $N$ , and conjugacy [L6] makes it unique. Automorphisms preserve orders, so this unique Sylow subgroup is characteristic in $N$ ; [L4] makes it normal in $G$ , and the maximality clause [L5] puts it in the corresponding $p$ -core.

L2L4L5L6step 1.1givenalgebra

3.1

By [L2], $N$ is the product of its Sylow subgroups, and step 2.1 puts every factor in the corresponding $p$ -core. Hence $N \leq F (G)$ .

L1L2step 2.1given

4.1

For $G = 1$ the family of $p$ -cores is empty and $F (G) = 1$ , which is nilpotent and contains every normal nilpotent subgroup. This proves the stated claim.

L1step 1.1step 3.1givenalgebra∎

TheoremStatement: AI-adaptedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Philip Hall: in a finite solvable group the Fitting subgroup contains its own centralizer

Statement

Let $G$ be a finite solvable group. Then $C_{G} (F (G)) \leq F (G)$ .

Facts & Assumptions

Given: A finite solvable group $G$ . Write $F : = F (G)$ and $C : = C_{G} (F)$ .

[L1]

For a finite group $G$ the Fitting subgroup is $F (G) = \prod_{p ∣ ∣ G ∣} O_{p} (G)$ ; and if $A, B ⊴ G$ , then $A B$ is a subgroup, is normal, and satisfies $A B = B A$ (The Fitting subgroup $F (G) = \prod_{p} O_{p} (G)$ of a finite group).

[L2]

For every finite group $G$ , $F (G)$ is nilpotent and normal, and every normal nilpotent subgroup of $G$ is contained in $F (G)$ (The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group).

[L3]

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

[L4]

If $N ⊴ G$ , then $C_{G} (N) ⊴ G$ (The centralizer of a normal subgroup is normal).

[L5]

Every subgroup and every quotient of a solvable group is solvable; no finiteness hypothesis is required (Subgroups and quotients of solvable groups are solvable).

[L6]

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

[L7]

Let $Q$ be a solvable group and let $N ⊴ Q$ with $N \neq 1$ . Then $N$ contains a subgroup $A \neq 1$ that is abelian and normal in $Q$ (A nontrivial normal subgroup of a solvable group contains a nontrivial abelian subgroup normal in the whole group).

[L8]

Let $A, B, C \leq G$ with $A \leq C$ . If $A B$ is a subgroup of $G$ , then $A (B \cap C) = A B \cap C$ (Dedekind's modular law for subgroup products).

[L9]

Let $N ⊴ G$ . Then $G / N$ is abelian if and only if $[G, G] \subseteq N$ ( $G / N$ is abelian if and only if $[G, G] \subseteq N$ ).

[L10]

Let $G$ be a group and let $N$ be a nonempty family of normal subgroups of $G$ . Then $⋂_{N \in N} N$ is a normal subgroup of $G$ (The intersection of a nonempty family of normal subgroups is normal).

[L11]

A group $G$ is nilpotent if $Z_{c} (G) = G$ for some $c \in N$ , where $(Z_{r} (G))$ is its upper central series (Nilpotent groups and nilpotency class).

[L12]

The upper central series begins with $Z_{0} (G) = 1$ and satisfies $Z_{r + 1} (G) / Z_{r} (G) = Z (G / Z_{r} (G))$ ; in particular $Z_{1} (G) = Z (G)$ (The upper central series).

[L13]

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

[L14]

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

Proof

technique · contradiction

1.1

By [L2], $F ⊴ G$ and every normal nilpotent subgroup of $G$ is contained in $F$ .

L2given

1.2

Assume towards a contradiction that $C \leq̸ F$ .

assume-contragiven

2.1

$F$ is normal in $G$ , so $C = C_{G} (F) ⊴ G$ by [L4].

L4step 1.1

3.1

$C$ and $F$ are both normal in $G$ , so by [L1] the product $C F$ is a subgroup of $G$ , is normal in $G$ , and satisfies $C F = F C$ . Since the identity lies in $C$ , also $F \leq C F$ .

L1step 1.1step 2.1

4.1

$G$ is solvable, so its quotient $G / F$ is solvable by [L5]; and $C F ⊴ G$ with $F \leq C F$ , so $C F / F ⊴ G / F$ by [L6]. By step 1.2 some $c \in C$ lies outside $F$ , so $c F \neq F$ and $C F / F \neq 1$ .

L5L6givenstep 1.2step 3.1

5.1

Apply [L7] to the solvable group $G / F$ and its nontrivial normal subgroup $C F / F$ : there is a subgroup $\overset{ˉ}{A} \neq 1$ of $C F / F$ that is abelian and normal in $G / F$ . Let $A$ be its preimage under $G \to G / F$ . By [L6], $F \leq A \leq C F$ , $A ⊴ G$ , and $A / F = \overset{ˉ}{A}$ is nontrivial and abelian.

L6L7step 4.1

6.1

$F ⊴ A$ and $A / F$ is abelian, so $[A, A] \leq F$ by [L9].

L9step 5.1

6.2

Put $D : = C \cap A$ . Both $C$ and $A$ are normal in $G$ , so $D ⊴ G$ by [L10].

L10step 2.1step 5.1

7.1

Apply [L8] with its $A : = F$ , its $B : = C$ and its $C : = A$ . Its hypotheses hold: $F \leq A$ by step 5.1, and $F C = C F$ is a subgroup by step 3.1. Hence $F (C \cap A) = F C \cap A$ , and $A \leq C F = F C$ makes the right-hand side equal to $A$ . Therefore $A = F D$ .

L8step 3.1step 5.1step 6.2

7.2

$D \leq A$ , so $[D, D] \leq [A, A] \leq F$ by step 6.1. Also $D \leq C = C_{G} (F)$ , so by [L3] every element of $D$ commutes with every element of $F$ , in particular with every element of $[D, D]$ . Since $[D, D] \leq D$ , this says $[D, D] \leq Z (D)$ by [L13].

L3L13step 6.1step 6.2

8.1

$Z (D) ⊴ D$ by [L14], and $[D, D] \leq Z (D)$ , so $D / Z (D)$ is abelian by [L9].

L9L14step 7.2

9.1

By [L12], $Z_{1} (D) = Z (D)$ and $Z_{2} (D) / Z_{1} (D) = Z ⁣ (D / Z_{1} (D))$ . Step 8.1 makes $D / Z_{1} (D)$ abelian, and the center of an abelian group is the whole group by [L13], so $Z_{2} (D) / Z_{1} (D) = D / Z_{1} (D)$ and hence $Z_{2} (D) = D$ . By [L11], $D$ is nilpotent.

L11L12L13step 8.1

10.1

$D$ is a normal nilpotent subgroup of $G$ by steps 6.2 and 9.1, so $D \leq F$ by step 1.1.

step 1.1step 6.2step 9.1

11.1

Substituting into step 7.1 gives $A = F D \leq F$ , so $A / F$ is trivial, contradicting step 5.1. The assumption of step 1.2 is therefore untenable, and $C_{G} (F (G)) \leq F (G)$ . This proves the stated claim.

discharge-contradiction: step 1.2step 5.1step 7.1step 10.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

The Frattini subgroup consists exactly of the nongenerators of a finite group

Statement

For a finite group $G$ , an element $x$ lies in $Φ (G)$ if and only if, for every subset $S \subseteq G$ , $⟨ S, x ⟩ = G$ implies $⟨ S ⟩ = G$ . See The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a finite group $G$ , the Frattini subgroup is $Φ (G) : = ⋂ {M \leq G : M is maximal proper} .$ If $G = 1$ , the family is empty and its intersection inside $G$ is $G$ itself. Thus $Φ (1) = 1$ . (The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group).

[L2]

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

Proof

technique · direct

1.1

If $x \notin Φ (G)$ , a maximal subgroup omitting $x$ shows that adjoining $x$ can enlarge a generating set.

L1L2givenalgebra

2.1

Conversely, if $⟨ S, x ⟩ = G$ while $⟨ S ⟩ \neq G$ , extend the latter finite subgroup to a maximal subgroup; it contains $S$ but cannot contain $x$ .

step 1.1givenalgebra

3.1

We treat the trivial group, whose empty intersection convention gives $Φ (1) = 1$ . This proves the stated claim.

step 2.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Nilpotence lifts over the Frattini subgroup of a finite group

Statement

Let $G$ be finite and let $Φ (G) \leq N ⊴ G$ . Then $N$ is nilpotent if and only if $N / Φ (G)$ is nilpotent. In particular, $G$ is nilpotent if and only if $G / Φ (G)$ is nilpotent. See The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

[L2]

If $N ⊴ G$ is finite and $P$ is a Sylow $p$ -subgroup of $N$ , then $G = N N_{G} (P) .$ . (Frattini argument: if $N ⊴ G$ and $P$ is Sylow in $N$ , then $G = N N_{G} (P)$ ).

[L3]

[L4]

For $K ⊴ G$ , subgroups of $G / K$ correspond to subgroups of $G$ containing $K$ , and the correspondence preserves normality. (Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

[L5]

Every subgroup and every quotient of a nilpotent group is nilpotent, and every finite direct product of nilpotent groups is nilpotent. (Subgroups, quotients, and finite direct products of nilpotent groups are nilpotent).

[L6]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ ; in particular all Sylow $p$ -subgroups are conjugate. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L7]

If $K$ is characteristic in $N$ and $N ⊴ G$ , then $K ⊴ G$ . (If $K$ is characteristic in $N$ and $N$ is normal in $G$ , then $K$ is normal in $G$ ).

Proof

technique · direct

1.1

If $N$ is nilpotent, then its quotient $N / Φ (G)$ is nilpotent by [L5].

L1L5given

1.2

Conversely, assume that $N / Φ (G)$ is nilpotent, and let $P$ be a Sylow $p$ -subgroup of $N$ . Every automorphism of $G$ permutes its maximal proper subgroups, so [L1] makes $Φ (G)$ characteristic and hence normal in $G$ . If $S$ is Sylow in $Φ (G)$ , then [L6] gives $n \in N$ with $S \leq n P n^{- 1}$ ; normality of $Φ (G)$ gives $n^{- 1} S n \leq P \cap Φ (G)$ . Thus $P \cap Φ (G)$ contains a Sylow $p$ -subgroup of $Φ (G)$ and, being a $p$ -subgroup, is itself Sylow there. Consequently $P Φ (G) / Φ (G)$ is a Sylow $p$ -subgroup of $N / Φ (G)$ .

L1L3L4L6givenalgebra

2.1

By nilpotence and [L3], this quotient Sylow subgroup is normal; conjugacy [L6] makes it unique, hence characteristic in $N / Φ (G)$ . Since $N / Φ (G) ⊴ G / Φ (G)$ , [L7] and [L4] give $P Φ (G) ⊴ G$ .

step 1.2L3L4L6L7givenalgebra

3.1

Because $P \cap Φ (G)$ is Sylow in $Φ (G)$ , the product-order formula shows that $P$ is Sylow in $P Φ (G)$ . Apply the Frattini argument [L2] to the normal subgroup $P Φ (G)$ : $G = P Φ (G) N_{G} (P) = Φ (G) N_{G} (P) .$ If $N_{G} (P)$ were proper, finiteness would place it in a maximal subgroup $M$ ; [L1] gives $Φ (G) \leq M$ , contradicting the displayed equality. Hence $N_{G} (P) = G$ .

L1L2step 1.2step 2.1givenalgebra

4.1

Thus every Sylow subgroup of $N$ is normal, so [L3] makes $N$ nilpotent. Together with step 1.1 this proves both directions.

L3step 1.1step 3.1given

5.1

Taking $N = G$ gives the asserted special case. When $G = 1$ , one has $N = Φ (G) = 1$ , and both groups in the equivalence are trivial and nilpotent.

L1step 4.1given∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

The Frattini subgroup of a finite group is nilpotent

Statement

The Frattini subgroup of every finite group is nilpotent. See Nilpotence lifts over the Frattini subgroup of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be finite and let $Φ (G) \leq N ⊴ G$ . Then $N$ is nilpotent if and only if $N / Φ (G)$ is nilpotent. In particular, $G$ is nilpotent if and only if $G / Φ (G)$ is nilpotent. (Nilpotence lifts over the Frattini subgroup of a finite group).

Proof

technique · direct

1.1

Every automorphism of $G$ permutes the maximal proper subgroups, so their intersection $Φ (G)$ is characteristic and therefore normal; the lifting theorem applies with $N = Φ (G)$ , whose quotient by $Φ (G)$ is the trivial nilpotent group.

L1givenalgebra

2.1

The trivial group is admitted: it has no maximal proper subgroup, so the defining family is empty and its intersection inside $G$ is $G$ itself, giving $Φ (1) = 1$ , which is nilpotent of class zero. This proves the stated claim.

step 1.1givenalgebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

The Frattini subgroup is contained in the Fitting subgroup

Statement

For every finite group $G$ , $Φ (G) \leq F (G)$ . See The Frattini subgroup of a finite group is nilpotent.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

The Frattini subgroup of every finite group is nilpotent. (The Frattini subgroup of a finite group is nilpotent).

[L2]

For every finite group $G$ , $F (G)$ is nilpotent and normal, and every normal nilpotent subgroup of $G$ is contained in $F (G)$ . (The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group).

Proof

technique · direct

1.1

The Frattini subgroup is characteristic, hence normal, and is nilpotent; maximality of the Fitting subgroup gives the inclusion.

L1L2givenalgebra

2.1

No finiteness beyond that of the Statement is used, and the degenerate case is consistent: for $G = {1}$ the family of maximal proper subgroups is empty, so $Φ (G) = {1} = F (G)$ and the inclusion holds with equality. This proves the stated claim.

step 1.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

$F (G / Φ (G)) = F (G) / Φ (G)$ for every finite group

Statement

For every finite group $G$ , $F (G / Φ (G)) = F (G) / Φ (G) .$ See The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

[L2]

For every finite group $G$ , $Φ (G) \leq F (G)$ . (The Frattini subgroup is contained in the Fitting subgroup).

[L3]

[L4]

Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved. 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 $K \leq G / N$ ; they preserve normality. (Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

Proof

technique · direct

1.1

The image $F (G) / Φ (G)$ is normal and nilpotent, giving one inclusion.

L1L2L3L4givenalgebra

2.1

For the reverse inclusion, pull $F (G / Φ (G))$ back to a normal subgroup $N$ of $G$ ; the lifting theorem makes $N$ nilpotent, so $N \leq F (G)$ .

step 1.1givenalgebra

3.1

If $G / Φ (G)$ is trivial, then $Φ (G) = G$ , and $Φ (G) \leq F (G) \leq G$ from [L2] forces $F (G) = G$ ; both sides of the identity are then the trivial group. Together with the two inclusions of steps 1.1 and 2.1 this gives equality in every case. This proves the stated claim.

step 1.1step 2.1L2givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Sylow and maximal-subgroup characterizations of finite nilpotence

Statement

For a finite group $G$ , the following are equivalent: $G$ is nilpotent; every Sylow subgroup is normal; $G$ is the internal direct product of its Sylow subgroups; and every maximal subgroup of $G$ is normal. See A finite group is nilpotent if and only if all Sylow subgroups are normal, if and only if it is their internal direct product.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

[L2]

Every maximal proper subgroup of a finite nilpotent group is normal and has prime index. (Maximal subgroups of finite nilpotent groups are normal of prime index).

[L3]

[L4]

[L5]

[L6]

If a prime $p$ divides the order of a finite group $H$ , then $H$ contains an element, and hence a subgroup, of order $p$ . (Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ ).

[L7]

Every finite $p$ -group is nilpotent, including the trivial group. (Every finite $p$ -group is nilpotent).

[L8]

[L9]

For $K ⊴ G$ , subgroups of $G / K$ correspond to subgroups of $G$ containing $K$ , and the correspondence preserves inclusion and normality. (Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

Proof

technique · direct

1.1

By [L1], the first three conditions are equivalent, and [L2] proves that any of them implies normality of every maximal subgroup.

L1L2given

2.1

Conversely, assume every maximal subgroup $M$ is normal. The diagonal map $G / Φ (G) \to \prod_{M} G / M$ is injective because its kernel is the intersection [L4]. By [L9] and maximality, each nontrivial quotient $G / M$ has no nontrivial proper subgroup. If $p$ divides its order, [L6] supplies a subgroup of order $p$ , which must be all of $G / M$ . Thus each factor has prime order and is nilpotent by [L7].

L4L6L7L9step 1.1givenalgebra

3.1

The finite product in step 2.1 is nilpotent and so is its subgroup $G / Φ (G)$ by [L8]. The lifting theorem [L3] now makes $G$ nilpotent.

L3L8step 2.1

4.1

This proves the reverse implication and hence all four equivalences.

step 1.1step 3.1

5.1

If the family of maximal subgroups is empty, finiteness forces $G = 1$ ; the diagonal target is then the empty product $1$ , and every condition holds.

L1L4L5L7step 4.1given∎

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

For prime $p$ , $∣ Aut ((Z / p) \times (Z / p)) ∣ = (p^{2} - 1) (p^{2} - p)$

Statement

For every prime $p$ , $∣ Aut ((Z / p) \times (Z / p)) ∣ = (p^{2} - 1) (p^{2} - p) .$ See Group isomorphisms, automorphisms and the set $Aut (G)$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

An isomorphism is a bijective group homomorphism; an automorphism is an isomorphism from a group to itself, and $Aut (G) : = {f : G \to G : f is an automorphism} .$ (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[L2]

Let $G$ and $H$ be groups. Their external direct product has underlying set $G \times H : = {(g, h) : g \in G, h \in H}$ and componentwise operation $(g, h) (g^{'}, h^{'}) : = (g g^{'}, h h^{'}) .$ The fact that this operation makes $G \times H$ a group, with the indicated identity and inverses, is proved in thm-external-direct-product-is-a-group. Until that result is used, this definition introduces only the set and its componentwise binary operation. (The external direct product $G \times H$ with componentwise multiplication).

[L3]

For groups $G$ and $H$ , the componentwise operation of def-external-direct-product-of-groups makes $G \times H$ a group. Its identity is $(e_{G}, e_{H})$ , and $(g, h)^{- 1} = (g^{- 1}, h^{- 1}) .$ Moreover the coordinate maps $π_{G} (g, h) = g$ and $π_{H} (g, h) = h$ are group homomorphisms. ( $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

[L4]

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

[L5]

Let $n$ be a positive integer. Every class in $Z / n$ (def-integers-modulo-n) contains exactly one integer $r$ with $0 \leq r < n$ . Consequently the map $r ⟼ [r]_{n} (0 \leq r < n)$ is a bijection from the von Neumann natural $n$ to $Z / n$ , and $∣ Z / n ∣ = n$ . This includes $n = 1$ , where the only representative is $0$ . For $n = 0$ , the map $a \mapsto [a]_{0}$ is a bijection $Z \to Z / 0$ . (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[L6]

If $A$ and $B$ are finite then $A \times B$ is finite and $∣ A \times B ∣ = ∣ A ∣ \cdot ∣ B ∣$ (def-finite-cardinality). 2. Let $m \in N$ and let $A_{0}, \dots, A_{m - 1}$ be finite sets. Write $\prod_{i < m} A_{i} : = {f : f is a function with domain m and f (i) \in A_{i} for every i < m} .$ Then $\prod_{i < m} A_{i}$ is finite and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ , the right-hand product being the $N$ -valued one of def-nat-finite-sum-and-product. (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ).

[L7]

Let $A$ be a finite set (def-countable) and let $B \subseteq A$ . Then: 1. $B$ is finite; 2. $∣ B ∣ \leq ∣ A ∣$ (def-finite-cardinality); 3. $∣ B ∣ = ∣ A ∣$ if and only if $B = A$ ; 4. every injection $f : A \to A$ is a bijection, and every surjection $f : A \to A$ is a bijection. (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

[L8]

Proof

technique · direct

1.1

In the additive group $E_{p} = (Z / p) \times (Z / p)$ , a homomorphism $E_{p} \to E_{p}$ is determined by the images $u$ and $v$ of the two coordinate generators, because every element has a unique coordinate expression.

L1L2L3L4L5L6L7L8givenalgebra

2.1

If $u = 0$ or $v \in ⟨ u ⟩$ , the image is a proper cyclic subgroup. If $u \neq 0$ and $v \notin ⟨ u ⟩$ , the $p$ elements of each coset $j v + ⟨ u ⟩$ are disjoint as $j$ varies, so $u, v$ generate all $p^{2}$ elements and the homomorphism is bijective.

step 1.1givenalgebra

3.1

There are $p^{2} - 1$ choices for nonzero $u$ . Its cyclic subgroup has exactly $p$ elements, leaving $p^{2} - p$ choices for $v$ ; multiplication gives $(p^{2} - 1) (p^{2} - p)$ automorphisms.

step 2.1givenalgebra

4.1

When $p = 2$ , the same count gives $(4 - 1) (4 - 2) = 6$ ; the argument is entirely in coordinates and makes no matrix-group identification. This proves the stated claim.

step 1.1step 2.1step 3.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Every group of order $p^{2} q$ for distinct primes has a normal Sylow subgroup

Statement

Every group of order $p^{2} q$ for distinct primes has a normal Sylow subgroup. See Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then the number of Sylow $p$ -subgroups satisfies $n_{p} (G) \equiv 1 (m o d p), n_{p} (G) ∣ m .$ . (Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$ ).

[L2]

A Sylow $p$ -subgroup of a finite group is normal if and only if it is the unique Sylow $p$ -subgroup. (A Sylow $p$ -subgroup is normal if and only if it is unique).

Proof

technique · direct

1.1

If $p > q$ , the restrictions $n_{p} ∣ q$ and $n_{p} \equiv 1 (m o d p)$ force $n_{p} = 1$ .

L1L2givenalgebra

2.1

If $q > p$ , then $n_{q} \in {1, p^{2}}$ ; the second value forces $q ∣ (p - 1) (p + 1)$ , hence the sole exceptional pair $(p, q) = (2, 3)$ .

step 1.1givenalgebra

3.1

For order $12$ , if the four Sylow $3$ -subgroups are nonnormal, their eight nonidentity elements leave exactly four elements, so every Sylow $2$ -subgroup is that same four-element complement and is normal.

step 2.1givenalgebra

4.1

The two orderings exhaust the hypothesis, since $p$ and $q$ are distinct: step 1.1 settles $p > q$ with a normal Sylow $p$ -subgroup, and step 2.1 settles $q > p$ with a normal Sylow $q$ -subgroup except at $(p, q) = (2, 3)$ , which step 3.1 settles with a normal Sylow $2$ -subgroup. Every case therefore produces a normal Sylow subgroup, and [L2] turns each uniqueness count $n = 1$ into normality. This proves the stated claim.

step 1.1step 2.1step 3.1L2givenalgebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

No group of order $p^{2} q$ for distinct primes is simple

Statement

No group of order $p^{2} q$ for distinct primes is simple. See Every group of order $p^{2} q$ for distinct primes has a normal Sylow subgroup.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Every group of order $p^{2} q$ for distinct primes has a normal Sylow subgroup. (Every group of order $p^{2} q$ for distinct primes has a normal Sylow subgroup).

[L2]

A group $G$ is simple if $G \neq {1}$ and its only normal subgroups are ${1}$ and $G$ , where normality is as in def-normal-subgroup. (Simple groups).

Proof

technique · direct

1.1

The normal Sylow subgroup supplied by the theorem has prime-power order strictly between one and $p^{2} q$ , so it is a nontrivial proper normal subgroup.

L1L2givenalgebra

2.1

Both cases of step 1.1 are genuinely nontrivial and proper: a normal Sylow $p$ -subgroup has order $p^{2}$ with $1 < p^{2} < p^{2} q$ because $q > 1$ , and a normal Sylow $q$ -subgroup has order $q$ with $1 < q < p^{2} q$ because $p^{2} > 1$ . Either one is therefore a normal subgroup other than ${1}$ and $G$ , which is what [L2] requires for $G$ to fail simplicity. This proves the stated claim.

step 1.1L2givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Every group of order $30$ has normal Sylow $3$ - and $5$ -subgroups and is not simple

Statement

Every group of order $30$ has normal Sylow $3$ - and $5$ -subgroups and is not simple. See Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

[L2]

For $H \leq G$ , left multiplication gives a transitive action on $G / H$ and a homomorphism $ρ : G \to Sym (G / H)$ with $\ker ρ = {Core}_{G} (H) .$ (Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$ ).

[L3]

For every natural $n$ , the function $sgn : S_{n} \to {+ 1, - 1}$ is a group homomorphism. It is surjective exactly when $n \geq 2$ ; for $n = 0$ and $n = 1$ its image is ${1}$ . (The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ ).

[L4]

A cycle of length $k$ has sign $(- 1)^{k - 1}$ . If $σ \in S_{n}$ and $c (σ)$ is the number of cycles after every fixed point is included as a one-cycle, then $sgn (σ) = (- 1)^{n - c (σ)} .$ . (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

[L5]

The image of a group homomorphism is a subgroup and its kernel is a normal subgroup. For every group homomorphism $f : G \to H$ , one has $im f \leq H$ and $\ker f ⊴ G$ . (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

[L6]

First isomorphism theorem for groups: $G / \ker f ≅ im f$ . 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]

Let $N ⊴ G$ . If $[G : N]$ is finite, then the quotient group $G / N$ is finite and $∣ G / N ∣ = [G : N] .$ In particular, if $G$ is finite, then $∣ G / N ∣ = \frac{∣ G ∣}{∣ N ∣} .$ . (If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

[L8]

Let $p < q$ be primes. - If $p ∤ (q - 1)$ , every group of order $p q$ is cyclic. - If $p ∣ (q - 1)$ , there are exactly two isomorphism classes of groups of order $p q$ : the cyclic group $C_{p q}$ and one nonabelian semidirect product $C_{q} ⋊ C_{p}$ . (Classification of groups of order $p q$ for primes $p < q$ ).

[L9]

If $N ⊴ G$ and $P$ is a normal Sylow $p$ -subgroup of $N$ , then $P ⊴ G$ . (A normal Sylow subgroup of a normal subgroup is normal in the whole group).

[L10]

A group $G$ is simple if $G \neq {1}$ and its only normal subgroups are ${1}$ and $G$ , where normality is as in def-normal-subgroup. (Simple groups).

Proof

technique · direct

1.1

Choose an involution by Cauchy's theorem and use the left regular permutation representation.

L1L2L3L4L5L6L7L8L9L10givenalgebra

2.1

Left multiplication by the involution is a product of fifteen transpositions, so composing with sign gives a surjection to $\pm 1$ whose normal kernel has order $15$ .

step 1.1givenalgebra

3.1

The published order- $p q$ classification makes that kernel cyclic.

step 2.1givenalgebra

4.1

Its Sylow $3$ - and $5$ -subgroups are normal in the kernel and hence normal in $G$ .

step 3.1givenalgebra

5.1

Either is a nontrivial proper normal subgroup. This proves the stated claim.

step 4.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Every group of order $105$ has normal Sylow $5$ - and $7$ -subgroups and is not simple

Statement

Every group of order $105$ has normal Sylow $5$ - and $7$ -subgroups and is not simple. See Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

[L2]

A Sylow $p$ -subgroup of a finite group is normal if and only if it is the unique Sylow $p$ -subgroup. (A Sylow $p$ -subgroup is normal if and only if it is unique).

[L3]

If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ . Here $H N : = {h n : h \in H, n \in N}$ . (If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ ).

[L4]

Let $H \leq G$ . Left multiplication defines a transitive action on $G / H$ , and the corresponding homomorphism $ρ : G \to Sym (G / H)$ has $\ker ρ = {Core}_{G} (H) .$ (Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$ ).

[L5]

[L6]

[L7]

[L8]

[L9]

If $N ⊴ G$ and $P$ is a normal Sylow $p$ -subgroup of $N$ , then $P ⊴ G$ . (A normal Sylow subgroup of a normal subgroup is normal in the whole group).

[L10]

A group $G$ is simple if $G \neq {1}$ and its only normal subgroups are ${1}$ and $G$ , where normality is as in def-normal-subgroup. (Simple groups).

[L11]

If $H \leq G$ and $G$ is finite, then $∣ G ∣ = [G : H] ∣ H ∣$ ; in particular $∣ H ∣$ divides $∣ G ∣$ . (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L12]

A set of cardinality $n$ has exactly $n!$ bijections to itself; in particular the symmetric group on three points has order $3! = 6$ . (A finite set $A$ with $∣ A ∣ = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

[L13]

For $H \leq G$ , the core $K = {Core}_{G} (H)$ is a normal subgroup of $G$ , satisfies $K \leq H$ , and contains every normal subgroup of $G$ that is contained in $H$ . ( ${Core}_{G} (H)$ is the largest normal subgroup of $G$ contained in $H$ ).

Proof

technique · direct

1.1

The nonunique counts $n_{5} = 21$ and $n_{7} = 15$ would already contribute $84 + 90$ distinct nonidentity elements, so at least one of the two Sylow subgroups is normal.

L1L2L3L4L5L6L7L8L9L10givenalgebra

2.1

Let $P$ be the normal Sylow subgroup supplied by step 1.1 and let $Q$ be a Sylow subgroup for the other prime. By [L3], $H = P Q$ is a subgroup. Its multiplication map $P \times Q \to P Q$ has fibres indexed by $P \cap Q = 1$ , so $∣ H ∣ = ∣ P ∣ ∣ Q ∣ = 35$ .

L3step 1.1givenalgebra

3.1

The action on the three left cosets has kernel $K = {Core}_{G} (H)$ , and $K \leq H$ by [L13]. By [L6], $G / K$ is isomorphic to a subgroup of the symmetric group on three points, so [L11] and [L12] give $[G : K] ∣ 6$ . Also [L7] and [L11] give $[G : K] ∣ 105$ , while $K \leq H$ gives $3 = [G : H] ∣ [G : K]$ . Hence $[G : K] = 3 = [G : H]$ , so $K = H$ and $H ⊴ G$ .

L4L5L6L7L11L12L13step 2.1givenalgebra

4.1

The order- $p q$ classification [L8] makes $H$ cyclic, and its Sylow $5$ - and $7$ -subgroups are normal in $H$ and therefore normal in $G$ by [L9]. Either is a nontrivial proper normal subgroup of $G$ , so [L10] also shows that $G$ is not simple.

L8L9L10step 3.1given∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

There are exactly two isomorphism classes of groups of order $105$

Statement

Up to isomorphism, the groups of order $105$ are the cyclic group $C_{105}$ and the direct product $C_{5} \times (C_{7} ⋊ C_{3})$ , where the action of $C_{3}$ on $C_{7}$ is nontrivial. In particular, there are exactly two isomorphism classes. See Every group of order $105$ has normal Sylow $5$ - and $7$ -subgroups and is not simple.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Every group of order $105$ has normal Sylow $5$ - and $7$ -subgroups and is not simple. (Every group of order $105$ has normal Sylow $5$ - and $7$ -subgroups and is not simple).

[L2]

[L3]

If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ . Here $H N : = {h n : h \in H, n \in N}$ . (If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ ).

[L4]

Let $G$ be a finite group such that the positive integer $∣ G ∣$ is prime. Then every $g \neq e$ has order $∣ G ∣$ , satisfies $⟨ g ⟩ = G$ , and hence generates $G$ . In particular, $G$ is cyclic. (A finite group of prime order is cyclic and every nonidentity element generates it).

[L5]

Conjugation $x \mapsto g x g^{- 1}$ is an automorphism. For each $g \in G$ , the map $c_{g} : G \to G$ , $c_{g} (x) = g x g^{- 1}$ , is an automorphism. (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

[L6]

The automorphisms of a group form a group under composition. (The automorphisms of a group form a group under composition).

[L7]

[L8]

For every $n \geq 1$ , $Aut (C_{n}) ≅ (Z / n)^{\times} .$ If $C_{n} = ⟨ g ⟩$ , the unit class $[a]$ corresponds to the automorphism $g \mapsto g^{a}$ . ( $Aut (C_{n}) ≅ (Z / n Z)^{\times}$ ).

[L9]

Euler's totient satisfies $φ (1) = 1$ . If $p$ is prime (def-prime), then $φ (p) = p - 1.$ . ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ).

[L10]

[L11]

Let $N, H \leq G$ . The conditions $N ⊴ G, G = N H, N \cap H = {1}$ hold if and only if both of the following hold: conjugation $α_{h} (n) = h n h^{- 1}$ restricts to an action $α : H \to Aut (N)$ , and the resulting map $Φ : N ⋊_{α} H ⟶ G, (n, h) ⟼ n h$ is an isomorphism 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]

The canonical factors of $N ⋊_{α} H$ form an internal direct product if and only if $α_{h} = {id}_{N}$ for every $h \in H$ . In that case $N ⋊_{α} H$ is the external direct product $N \times H$ . (The canonical semidirect decomposition is an internal direct product if and only if the defining action is trivial).

[L13]

[L14]

If $g \in G$ and $h \in H$ have finite orders $m, n \geq 1$ , then in the external direct product $ord (g, h) = lcm (m, n) .$ (If $g$ and $h$ have finite orders $m$ and $n$ , then $ι (ord (g, h)) = lcm (ι (m), ι (n))$ in $G \times H$ ).

[L15]

If $G = ⟨ g ⟩$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G ≅ (Z, +)$ ; - if $g$ has finite order $n$ , necessarily $n \geq 1$ , then $G ≅ (Z / n, +)$ . (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

Proof

technique · direct

1.1

Let $N_{5}, N_{7}$ be the normal Sylow subgroups and let $P$ be Sylow of order $3$ . Since $N_{7}$ is normal, $H = N_{7} P$ is a subgroup of order $21$ ; similarly $G = N_{5} H$ and $N_{5} \cap H = 1$ .

L1L2L3L4L5L6L7L8L9L10L11L12L13L14L15givenalgebra

2.1

Conjugation gives a homomorphism $H \to Aut (N_{5})$ . Its image order divides $∣ H ∣ = 21$ and $∣ Aut (C_{5}) ∣ = 4$ , so the image is trivial.

step 1.1givenalgebra

3.1

Thus $H$ centralizes $N_{5}$ , and the internal product is $G ≅ C_{5} \times H$ .

step 1.1step 2.1givenalgebra

4.1

The order- $p q$ classification makes $H$ either $C_{21}$ or the unique nonabelian $C_{7} ⋊ C_{3}$ . In the first case, generators of $C_{5}$ and $C_{21}$ combine to an element of order $lcm (5, 21) = 105$ , so $C_{5} \times H ≅ C_{105}$ .

step 3.1givenalgebra

5.1

The two resulting groups are distinguished by abelianness, and exhaustiveness of the order- $21$ classification leaves no third case. This proves the stated claim.

step 3.1step 4.1givenalgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Every group of order $45$ is abelian

Statement

Every group of order $45$ is abelian. See Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

[L2]

A Sylow $p$ -subgroup of a finite group is normal if and only if it is the unique Sylow $p$ -subgroup. (A Sylow $p$ -subgroup is normal if and only if it is unique).

[L3]

Normal Sylow subgroups for distinct primes centralize one another. (Distinct normal Sylow subgroups centralize one another).

[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]

Proof

technique · direct

1.1

Sylow III forces both the order-nine and order-five Sylow subgroups to be unique.

L1L2L3L4L5givenalgebra

2.1

They commute, their product is the whole group, and both factors are abelian, so the internal direct product is abelian. This proves the stated claim.

step 1.1givenalgebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

No group of order $p q$ for distinct primes is simple

Statement

No group of order $p q$ for distinct primes is simple. See If $p < q$ are primes and $∣ G ∣ = p q$ , then $G$ has a normal subgroup of order $q$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $p < q$ be primes. Every group $G$ of order $p q$ has a normal subgroup of order $q$ . (If $p < q$ are primes and $∣ G ∣ = p q$ , then $G$ has a normal subgroup of order $q$ ).

[L2]

A group $G$ is simple if $G \neq {1}$ and its only normal subgroups are ${1}$ and $G$ , where normality is as in def-normal-subgroup. (Simple groups).

Proof

technique · direct

1.1

We order the primes as $p < q$ .

L1L2givenalgebra

2.1

The published order- $p q$ lemma supplies a normal subgroup of order $q$ , which is nontrivial and proper. This proves the stated claim.

step 1.1givenalgebra∎

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-17Open item page →

False statement: every divisor of the order of a finite group occurs as a subgroup order

Statement

False claim: every divisor of the order of a finite group occurs as a subgroup order. See Sylow I: every finite group has a Sylow $p$ -subgroup.

Facts & Assumptions

Given: The hypotheses and objects in the false claim.

[L1]

[L2]

Let $G$ be finite abelian and let $d$ be a positive divisor of $∣ G ∣$ . Then $G$ has a subgroup of order $d$ . (Converse of Lagrange for finite abelian groups: every divisor occurs as a subgroup order).

[L3]

For $n \in N$ , the alternating group is the kernel of the sign homomorphism, $A_{n} : = \ker (sgn : S_{n} \to {+ 1, - 1}) = {σ \in S_{n} : sgn (σ) = 1} .$ Thus $A_{n}$ consists exactly of the even permutations. The subgroup and normality assertions implicit in the word “group” follow from thm-image-subgroup-and-kernel-normal. (The alternating group $A_{n} = \ker (sgn)$ of even permutations).

[L4]

Refutation

technique · direct

1.1

The divisor $6$ of $∣ A_{4} ∣ = 12$ is not the order of a subgroup. Indeed, a hypothetical subgroup $H$ of order $6$ would have index $2$ and hence be normal. Sylow I applied inside $H$ gives an element of order $3$ .

L1L2L3L4givenalgebra

2.1

Conjugating that $3$ -cycle in $A_{4}$ , and also conjugating its inverse, puts all eight $3$ -cycles in the normal subgroup $H$ . Together with the identity this gives more than six elements, a contradiction. Thus the general converse fails although the cited abelian and prime-power special cases remain valid. This proves the stated claim.

step 1.1givenalgebra∎

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

False statement: all subgroups of the same $p$ -power order are conjugate

Statement

False claim: all subgroups of the same $p$ -power order are conjugate. See Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class.

Facts & Assumptions

Given: The hypotheses and objects in the false claim.

[L1]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ . In particular, for every Sylow $p$ -subgroup $Q$ there is $g \in G$ with $Q = g P g^{- 1}$ , so the Sylow $p$ -subgroups form one conjugacy class. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L2]

Let $n \in N$ , so that $n = {0, 1, \dots, n - 1}$ (def-natural-numbers). The symmetric group on $n$ letters is $S_{n} : = Sym (n) = Sym ({0, 1, \dots, n - 1}),$ the group of all bijections of $n$ under composition (def-symmetric-group), with the composition convention. (The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

[L3]

For $σ, τ \in S_{n}$ , there is a $g \in S_{n}$ with $τ = g σ g^{- 1}$ if and only if $σ$ and $τ$ have the same cycle type, including their numbers of fixed points. (Two elements of $S_{n}$ are conjugate if and only if they have the same cycle type).

Refutation

technique · direct

1.1

In the $S_{4} = Sym ({0, 1, 2, 3})$ of [L2], the order-two subgroups generated by $(0 1)$ and $(0 1) (2 3)$ cannot be conjugate because their nonidentity generators have different cycle types.

L1L2L3givenalgebra

2.1

Two subgroups of order $2$ are conjugate exactly when their nonidentity elements are, so step 1.1 exhibits two subgroups of the same $2$ -power order that are not conjugate, refuting the universal claim. No conflict with [L1] arises: both have order $2$ while the Sylow $2$ -subgroups of $S_{4}$ have order $8$ , and [L1] asserts conjugacy only among subgroups of that maximal $p$ -power order. This proves the stated claim.

step 1.1L1L3givenalgebra∎

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-17Open item page →

False statement: one unique Sylow subgroup forces the whole group to be a direct product

Statement

False claim: one unique Sylow subgroup forces the whole group to be a direct product. See A Sylow $p$ -subgroup is normal if and only if it is unique.

Facts & Assumptions

Given: The hypotheses and objects in the false claim.

[L1]

A Sylow $p$ -subgroup of a finite group is normal if and only if it is the unique Sylow $p$ -subgroup. (A Sylow $p$ -subgroup is normal if and only if it is unique).

[L2]

[L3]

Refutation

technique · direct

1.1

Let $C_{2}$ act on $C_{3}$ by inversion. The resulting semidirect product $C_{3} ⋊ C_{2}$ has order $6$ and is nonabelian.

L1L2L3givenalgebra

2.1

Its canonical $C_{3}$ is normal and therefore the unique Sylow $3$ -subgroup. If the product were direct, the two factors would commute, contradicting the inversion action. This proves the stated claim.

step 1.1givenalgebra∎

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-17Open item page →

False statement: every group of order $42$ has a normal Sylow $2$ -subgroup

Statement

False claim: every group of order $42$ has a normal Sylow $2$ -subgroup. See Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$ .

Facts & Assumptions

Given: The hypotheses and objects in the false claim.

[L1]

[L2]

Let $N$ and $H$ be groups (def-group), and let $α : H \to Aut (N)$ be an action by automorphisms (def-action-by-automorphisms). The external semidirect product $N ⋊_{α} H$ is the set $N \times H$ with multiplication. ( The external semidirect product $N ⋊_{α} H$ ).

[L3]

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

Refutation

technique · direct

1.1

We construct the affine group $F_{7} ⋊ F_{7}^{\times}$ of order $42$ .

L1L2L3givenalgebra

2.1

Its seven involutions $x \mapsto - x + b$ generate seven Sylow $2$ -subgroups, so none is normal.

step 1.1givenalgebra

3.1

The translations $x \mapsto x + b$ form a subgroup of order $7$ , and it is normal because $(x \mapsto a x + c)$ conjugates $x \mapsto x + b$ to $x \mapsto x + a b$ , again a translation; so the group does have a normal Sylow $7$ -subgroup, and it is only the Sylow $2$ -subgroups that fail to be normal. The count $n_{2} = 7$ of step 2.1 is consistent with [L1], since $7 \equiv 1 (m o d 2)$ and $7 ∣ 21$ . This proves the stated claim.

step 2.1L1givenalgebra∎

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-17Open item page →

False statement: finite nilpotent groups and finite solvable groups are the same

Statement

False claim: finite nilpotent groups and finite solvable groups are the same. See Nilpotent groups, and in particular finite $p$ -groups, are solvable.

Facts & Assumptions

Given: The hypotheses and objects in the false claim.

[L1]

Every nilpotent group is solvable. Consequently every finite $p$ -group is solvable. (Nilpotent groups, and in particular finite $p$ -groups, are solvable).

[L2]

The derived series of a group $G$ is defined recursively by $G^{(0)} = G, G^{(r + 1)} = [G^{(r)}, G^{(r)}] .$ Each term is characteristic, hence normal, in the preceding term by thm-derived-subgroup-is-characteristic-and-abelianization-is-universal. (The derived series, solvable groups, and derived length).

[L3]

[L4]

For a finite group $G$ , the following are equivalent: $G$ is nilpotent; every Sylow subgroup is normal; $G$ is the internal direct product of its Sylow subgroups; and every maximal subgroup of $G$ is normal. (Sylow and maximal-subgroup characterizations of finite nilpotence).

Refutation

technique · direct

1.1

One inclusion does hold: [L1] states that every nilpotent group is solvable, so every finite nilpotent group is solvable and only the converse can fail. Refuting the claim therefore requires a finite solvable group that is not nilpotent.

L1given

2.1

For the converse, take $S_{3}$ of [L3] and compute its derived series of [L2]: $S_{3}^{'} = A_{3}$ and $A_{3}^{'} = 1$ , so $S_{3}$ is solvable. Its three Sylow $2$ -subgroups are the subgroups generated by the transpositions, which are not normal, so the maximal-subgroup and Sylow clauses of [L4] deny that $S_{3}$ is nilpotent. A finite solvable group that is not nilpotent refutes the claim. This proves the stated claim.

step 1.1L2L3L4givenalgebra∎

Sources

Standard references

Recommended treatments; not extraction sources.

Sylow's Theorems, pp-Groups and Nilpotent Groups

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Sylow pp-subgroups of a finite group

Definition

The number np(G)np​(G) of Sylow pp-subgroups

Definition

Maximal proper subgroups

Definition

pp-complements in finite groups

Definition

The pp-core Op(G)Op​(G) as the largest normal pp-subgroup

Definition

The Fitting subgroup F(G)=∏pOp(G)F(G)=∏p​Op​(G) of a finite group

Definition

The Frattini subgroup Φ(G)Φ(G) as the intersection of the maximal subgroups of a finite group

Definition

The centralizer CG(H)CG​(H) of a subgroup

Definition

The centralizer of a normal subgroup is normal

Statement

Facts & Assumptions

Proof

A nontrivial normal subgroup of a solvable group contains a nontrivial abelian subgroup normal in the whole group

Statement

Facts & Assumptions

Proof

If ∣G∣=pam∣G∣=pam with p∤mp∤m, then vp(pampa)=0vp​(papam​)=0

Statement

Facts & Assumptions

Proof

Sylow I: every finite group has a Sylow pp-subgroup

Statement

Facts & Assumptions

Proof

Sylow II: in a finite group every pp-subgroup lies in a conjugate of any Sylow pp-subgroup, and the Sylow pp-subgroups form a single conjugacy class

Statement

Facts & Assumptions

Proof

Sylow III*: np(G)=[G:NG(P)]np​(G)=[G:NG​(P)]

Statement

Facts & Assumptions

Proof

Sylow III: np≡1(modp)np​≡1(modp) and np∣mnp​∣m when ∣G∣=pam∣G∣=pam with p∤mp∤m

Statement

Facts & Assumptions

Proof

A Sylow pp-subgroup is normal if and only if it is unique

Statement

Facts & Assumptions

Proof

A subgroup containing the normalizer of a Sylow subgroup is self-normalizing

Statement

Facts & Assumptions

Proof

Frattini argument: if N⊴GN⊴G and PP is Sylow in NN, then G=NNG(P)G=NNG​(P)

Statement

Facts & Assumptions

Proof

A normal Sylow subgroup of a normal subgroup is normal in the whole group

Statement

Facts & Assumptions

Proof

Every proper subgroup of a finite nilpotent group is properly contained in its normalizer

Statement

Facts & Assumptions

Proof

Maximal subgroups of finite nilpotent groups are normal of prime index

Statement

Facts & Assumptions

Proof

Every nontrivial finite pp-group has a normal subgroup of index pp

Statement

Facts & Assumptions

Proof

Distinct normal Sylow subgroups centralize one another

Statement

Facts & Assumptions

Sylow's Theorems, $p$ -Groups and Nilpotent Groups

Sylow $p$ -subgroups of a finite group

The number $n_{p} (G)$ of Sylow $p$ -subgroups

$p$ -complements in finite groups

The $p$ -core $O_{p} (G)$ as the largest normal $p$ -subgroup

The Fitting subgroup $F (G) = \prod_{p} O_{p} (G)$ of a finite group

The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group

The centralizer $C_{G} (H)$ of a subgroup

If $∣ G ∣ = p^{a} m$ with $p ∤ m$ , then $v_{p} (\binom{p^{a} m}{p^{a}}) = 0$

Sylow I: every finite group has a Sylow $p$ -subgroup

Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class

Sylow III*: $n_{p} (G) = [G : N_{G} (P)]$

Sylow III: $n_{p} \equiv 1 (m o d p)$ and $n_{p} ∣ m$ when $∣ G ∣ = p^{a} m$ with $p ∤ m$

A Sylow $p$ -subgroup is normal if and only if it is unique

Frattini argument: if $N ⊴ G$ and $P$ is Sylow in $N$ , then $G = N N_{G} (P)$

Every nontrivial finite $p$ -group has a normal subgroup of index $p$

A finite product of normal $p$ -subgroups is a normal $p$ -subgroup

$F (G / Φ (G)) = F (G) / Φ (G)$ for every finite group

For prime $p$ , $∣ Aut ((Z / p) \times (Z / p)) ∣ = (p^{2} - 1) (p^{2} - p)$

Every group of order $p^{2} q$ for distinct primes has a normal Sylow subgroup

No group of order $p^{2} q$ for distinct primes is simple

Every group of order $30$ has normal Sylow $3$ - and $5$ -subgroups and is not simple

Every group of order $105$ has normal Sylow $5$ - and $7$ -subgroups and is not simple

There are exactly two isomorphism classes of groups of order $105$

Every group of order $45$ is abelian

No group of order $p q$ for distinct primes is simple