LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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Proper subgroup of a finite p group is properly normalized local

Statement

Let $P$ be a finite $p$ -group and let $S < P$ be a proper subgroup. Then $S < N_{P} (S)$ : the normalizer of $S$ in $P$ strictly contains $S$ .

Facts & Assumptions

Given: A prime $p$ , a finite $p$ -group $P$ , and a proper subgroup $S < P$ ; the assertion is proved for all finite $p$ -groups of order $< ∣ P ∣$ (induction hypothesis).

[F1]

$S < N_{P} (S)$ means that $N_{P} (S)$ is a subgroup of $P$ containing $S$ properly, i.e. that there is $x \in P$ with $x S x^{- 1} = S$ and $x \notin S$ (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup, Subgroup).

[F2]

Every subgroup of $P$ is a finite $p$ -group, so $∣ S ∣$ is a power of $p$ ; if $P \neq 1$ then $∣ P ∣ = p^{r}$ with $r \geq 1$ ; $S = P$ if and only if $∣ S ∣ = ∣ P ∣$ (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 $n \in N$ , Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[F3]

If $P \neq 1$ then $Z (P) \neq 1$ ; $Z (P) ⊴ P$ ; and $Z (P)$ consists of the elements commuting with every element of $P$ , so $z S z^{- 1} = S$ for every $z \in Z (P)$ (Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$ , The center $Z (G)$ of a group, The center of a group is a normal subgroup, Normal subgroup: invariance under conjugation, Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

[F4]

For $Z ⊴ P$ the quotient $P / Z$ is a finite group with $∣ P / Z ∣ = [P : Z] = ∣ P ∣ / ∣ Z ∣$ , and $P / Z$ is a $p$ -group, of order $< ∣ P ∣$ whenever $Z \neq 1$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , 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 $G$ , A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[F5]

For $Z ⊴ P$ and $S \leq P$ the image $S Z / Z$ is a subgroup of $P / Z$ , and if $Z \leq S$ then $S Z / Z = S / Z$ with $∣ S / Z ∣ = ∣ S ∣ / ∣ Z ∣$ ; moreover conjugation $x \mapsto u x u^{- 1}$ is an automorphism, so $∣ x S x^{- 1} ∣ = ∣ S ∣$ and $x S x^{- 1} Z$ is the image of $x S x^{- 1}$ in $P / Z$ (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup, Monoid homomorphism and group homomorphism, If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ , If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ , Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

[F6]

Strong induction on the natural number $∣ P ∣$ : if, for every $N$ , truth for all finite $p$ -groups of order $< N$ implies truth for all of order $N$ , then the statement holds for all finite $p$ -groups (Strong (complete) induction).

Proof

technique · direct

1.1

The case $P = 1$ is vacuous, since it has no proper subgroup. Assume the assertion known for every finite $p$ -group of order $< ∣ P ∣$ .

F2F6given

1.2

If $S = 1$ then every $x \in P$ satisfies $x S x^{- 1} = S$ , so $N_{P} (S) = P$ ; if also $S < P$ then $P \neq 1$ and $N_{P} (S) = P > S$ , which is the claim.

F1F3given

1.3

Suppose $Z (P) ⊈ S$ : choose $z \in Z (P)$ with $z \notin S$ . By [F3] $z$ normalizes $S$ , so $z \in N_{P} (S) ∖ S$ and $S < N_{P} (S)$ .

F1F3choose

1.4

It remains to treat the case $Z \leq S$ with $Z : = Z (P)$ , under $S \neq 1$ and $P \neq 1$ . Then $P / Z$ is a finite $p$ -group of order $∣ P ∣ / ∣ Z ∣ < ∣ P ∣$ by [F3] and [F4], and $T : = S / Z$ is a subgroup of it by [F5]. If $T = P / Z$ then $S = P$ by [F5], contrary to hypothesis, so $T < P / Z$ .

F2F3F4F5given

2.1

The induction hypothesis of step 1.1 applies to the finite $p$ -group $P / Z$ and its proper subgroup $T$ : there is a coset $x Z \in N_{P / Z} (T)$ with $x Z \notin T$ . By the definition of the normalizer this means $(x Z) T (x Z)^{- 1} = T$ in $P / Z$ .

F1F4F5step 1.4assume-hyp

3.1

Translating back: $(x S x^{- 1}) Z / Z = S Z / Z$ . Since $Z \leq S$ , this says $x S x^{- 1} Z = S$ , so every element of $x S x^{- 1}$ lies in $S$ ; thus $x S x^{- 1} \subseteq S$ , and since conjugation is injective with $∣ x S x^{- 1} ∣ = ∣ S ∣$ , actually $x S x^{- 1} = S$ . Hence $x \in N_{P} (S)$ by [F1].

F5step 2.1algebra

3.2

Moreover $x \notin S$ : otherwise $x Z \in S / Z = T$ , contrary to the choice in step 2.1.

F5step 2.1

4.1

So in the case $Z (P) \leq S$ there is $x \in N_{P} (S) ∖ S$ as well, and together with steps 1.2 and 1.3 this proves $S < N_{P} (S)$ in every case, completing the induction. ∎

F1step 1.2step 1.3step 3.1step 3.2

Depends on

Used by

Local sylow conjugacy ascent for fusion Lemma

Dependency tree · two levels

56 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)