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Sylow subgroups of a normal subgroup are intersections with Sylow subgroups

Statement

Let $G$ be a finite group, $p$ a prime, $K ⊴ G$ a normal subgroup and $P \in {Syl}_{p} (G)$ a Sylow $p$ -subgroup (Sylow $p$ -subgroups of a finite group). Then $K \cap P$ is a Sylow $p$ -subgroup of $K$ . If in addition $[G : K]$ is a power of $p$ , then $K P = G$ .

Facts & Assumptions

Given: A finite group $G$ , a prime $p$ , a normal subgroup $K ⊴ G$ , and a Sylow $p$ -subgroup $P \leq G$ .

[F1]

Write $∣ G ∣ = p^{a} m$ and $∣ K ∣ = p^{c} m_{K}$ with $p ∤ m$ , $p ∤ m_{K}$ ; the $p$ -adic valuations give $c \leq a$ , and $P$ has order $p^{a}$ while a Sylow $p$ -subgroup of $K$ has order $p^{c}$ (Sylow $p$ -subgroups of a finite group, The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ , Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[F2]

If $S \leq G$ is a $p$ -subgroup, then $S \leq Q$ for some Sylow $p$ -subgroup $Q$ of $G$ ; any two Sylow $p$ -subgroups of $G$ are conjugate, $Q = g P g^{- 1}$ for some $g \in G$ ; and $G$ has a Sylow $p$ -subgroup (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).

[F3]

$K$ is normal: $g K g^{- 1} = K$ for every $g \in G$ , and conjugation $x \mapsto g x g^{- 1}$ is an automorphism, so $∣ g S g^{- 1} ∣ = ∣ S ∣$ for every subgroup $S$ (Normal subgroup: invariance under conjugation, Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

[F4]

A subgroup of a finite $p$ -group is a finite $p$ -group, so its order is a power of $p$ (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 $n \in N$ ).

[F5]

If $H \leq G$ then $∣ H ∣$ divides $∣ G ∣$ , and $K P$ is a subgroup with $∣ K P ∣ = ∣ K ∣ ∣ P ∣ / ∣ K \cap P ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ ).

Proof

technique · direct

1.1

By [F2] applied inside the finite group $K$ , there is a Sylow $p$ -subgroup $S$ of $K$ , of order $p^{c}$ by [F1]; $S$ is a $p$ -subgroup of $G$ , so by [F2] there is a Sylow $Q$ of $G$ with $S \leq Q$ , and $Q = g P g^{- 1}$ for some $g \in G$ .

F1F2

1.2

On the other hand $K \cap P \leq P$ , so $K \cap P$ is a finite $p$ -group by [F4]; its order divides $∣ K ∣$ by [F5], hence is a power of $p$ dividing $p^{c} m_{K}$ with $p ∤ m_{K}$ by [F1], and therefore $∣ K \cap P ∣$ divides $p^{c}$ .

F1F4F5

2.1

Then $S^{g^{- 1}} = g^{- 1} S g$ is a subgroup of $K$ , because $S \leq K$ and $K ⊴ G$ , and it has order $∣ S ∣ = p^{c}$ by [F3]; also $S^{g^{- 1}} \leq g^{- 1} Q g = P$ . Hence $S^{g^{- 1}} \leq K \cap P$ , and $∣ K \cap P ∣ \geq p^{c}$ .

F3step 1.1

3.1

Combining steps 2.1 and 1.2, $∣ K \cap P ∣ = p^{c}$ , the order of a Sylow $p$ -subgroup of $K$ ; hence $K \cap P \in {Syl}_{p} (K)$ , the first assertion.

F1step 2.1step 1.2

4.1

Suppose now that $[G : K] = p^{r}$ for some $r \geq 0$ . Then $∣ K ∣ = ∣ G ∣ / p^{r} = p^{a - r} m$ by [F1] and [F5], so the $p$ -part of $∣ K ∣$ is $p^{a - r}$ ; by step 3.1, $∣ K \cap P ∣ = p^{a - r}$ .

F1F5step 3.1algebra

5.1

Since $K P$ is a subgroup of $G$ by [F5], its order $∣ K ∣ ∣ P ∣ / ∣ K \cap P ∣ = p^{a - r} m \cdot p^{a} / p^{a - r} = p^{a} m = ∣ G ∣$ by [F5] and step 4.1; a subgroup of $G$ with as many elements as $G$ is $G$ itself, so $K P = G$ . ∎

F5step 4.1algebra

Depends on

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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)