LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27

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Normal p subgroup has proper commutator in a p group

Statement

Let $P$ be a finite $p$ -group and let $Q ⊴ P$ be a nontrivial normal subgroup (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ , Normal subgroup: invariance under conjugation). Then

$[Q, P] < Q,$

where $[Q, P] = ⟨ [q, u] : q \in Q, u \in P ⟩$ is the subgroup commutator of Subgroup commutators and the lower central series and Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ . Moreover $[Q, P] ⊴ Q$ , so the quotient $Q / [Q, P]$ is defined (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Facts & Assumptions

Given: A finite $p$ -group $P$ and a nontrivial normal subgroup $Q ⊴ P$ .

[F1]

$Q \cap Z (P) \neq {1}$ : a nontrivial normal subgroup of a finite $p$ -group meets the center nontrivially (Every nontrivial normal subgroup of a finite $p$ -group meets the center nontrivially, The center $Z (G)$ of a group).

[F2]

Commutators: $[a, b] = a b a^{- 1} b^{- 1}$ , and $[A, B] = ⟨ [a, b] : a \in A, b \in B ⟩$ ; if $A \leq B \leq G$ and $D \leq G$ then every generator $[a, d]$ of $[A, D]$ is a generator of $[B, D]$ , so $[A, D] \leq [B, D]$ . If $N ⊴ G$ , $A \leq N$ and $B \leq G$ , then $[A, B] \leq N$ : for $a \in A$ , $b \in B$ one has $b a b^{- 1} \in N$ , so $[a, b] = a (b a b^{- 1})^{- 1} \in N$ (Subgroup commutators and the lower central series, Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, Normal subgroup: invariance under conjugation, In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential, Subgroup).

[F3]

If $M ⊴ G$ , the quotient map $π : G \to G / M$ , $π (g) = g M$ , is a surjective group homomorphism, $π (A) = A M / M$ for every subgroup $A \leq G$ , and images of generated subgroups are generated by the images of the generators (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , Homomorphisms respect commutator subgroups and derived series, The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[F4]

Order facts: for $M ⊴ Q$ one has $∣ Q / M ∣ = [Q : M] = ∣ Q ∣ / ∣ M ∣$ ; subgroups of finite $p$ -groups are finite $p$ -groups; $[Q : M]$ divides $∣ Q ∣$ , so it is a power of $p$ when $Q$ is (If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ , Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , Every subgroup of a finite $p$ -group has order a power of $p$ , A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[F5]

Strong induction on the positive integer $∣ Q ∣$ (Strong (complete) induction).

Proof

technique · direct

1.1

We prove by strong induction on $n = ∣ Q ∣$ the statement: for every finite $p$ -group $P$ and every nontrivial $Q ⊴ P$ with $∣ Q ∣ = n$ , one has $[Q, P] < Q$ and $[Q, P] ⊴ Q$ . Let such $P$ and $Q$ be given. By [F1] the subgroup $Z_{0} : = Q \cap Z (P)$ is nontrivial; it is normal in $P$ as the intersection of the normal subgroups $Q$ and $Z (P)$ (the center is normal), and $[Z_{0}, P] = {1}$ because every element of $Z_{0}$ commutes with every element of $P$ .

F1F2given

2.1

If $Z_{0} = Q$ , that is $Q \leq Z (P)$ , then every generator $[q, u]$ of $[Q, P]$ equals $1$ , so $[Q, P] = {1} < Q$ since $Q$ is nontrivial; also $[Q, P] = {1} ⊴ Q$ .

F2F5step 1.1

2.2

Otherwise $Z_{0} < Q$ , so $\overset{ˉ}{Q} : = Q / Z_{0}$ is a nontrivial normal subgroup of the finite $p$ -group $\overset{ˉ}{P} : = P / Z_{0}$ by [F4]; since $∣ \overset{ˉ}{Q} ∣ = ∣ Q ∣ / ∣ Z_{0} ∣ < ∣ Q ∣ = n$ , the induction hypothesis of step 1.1 applies to $\overset{ˉ}{P}$ and $\overset{ˉ}{Q}$ and gives $[\overset{ˉ}{Q}, \overset{ˉ}{P}] < \overset{ˉ}{Q}$ .

F3F4F5step 1.1

3.1

Let $π : P \to \overset{ˉ}{P}$ be the quotient map. By [F3], $π (Q) = \overset{ˉ}{Q}$ , $π (P) = \overset{ˉ}{P}$ , and $π ([Q, P])$ is generated by the elements $π ([q, u]) = [π (q), π (u)]$ for $q \in Q$ , $u \in P$ , that is $π ([Q, P]) = [\overset{ˉ}{Q}, \overset{ˉ}{P}]$ ; on the other hand $π ([Q, P]) = [Q, P] Z_{0} / Z_{0}$ .

F2F3step 2.2

4.1

The inclusion $[\overset{ˉ}{Q}, \overset{ˉ}{P}] < \overset{ˉ}{Q}$ of step 2.2 therefore says $[Q, P] Z_{0} / Z_{0} < Q / Z_{0}$ , which means $[Q, P] Z_{0} < Q$ ; as $[Q, P] \leq [Q, P] Z_{0}$ , we get $[Q, P] < Q$ .

F4step 2.2step 3.1

5.1

Finally $[Q, P] ⊴ Q$ : for $q \in Q$ and $a \in Q$ , $b \in P$ one has $q [a, b] q^{- 1} = [q a q^{- 1}, q b q^{- 1}] = [a^{q}, b^{q}]$ , and $a^{q} \in Q$ , $b^{q} \in P$ because $q \in Q \leq P$ and $P^{q} = P$ ; hence conjugation by $q$ permutes the generators of $[Q, P]$ , so $[Q, P]^{q} = [Q, P]$ for every $q \in Q$ , which is normality of $[Q, P]$ in $Q$ . This completes the induction and the proof. ∎

F2F3step 1.1step 4.1

Depends on

Used by

Fusion control forces trivial Sylow intersection with the p residual Lemma

Dependency tree · two levels

45 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Paul Flavell, An Introduction to Transfer and Fusion in Finite Groups, §§2–5 (standard reference, not scraped)
Hans Kurzweil and Bernd Stellmacher, The Theory of Finite Groups, §§7.1–7.2 (standard reference, not scraped)