# Subgroup Containing All $p$-Sylow Subgroups of a Group

## Problem 227

Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$.
Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$.

Then show that $N$ contains all $p$-Sylow subgroups of $G$.

Contents

## Hint.

We give two proof.
The first one uses Sylow’s theorem. The second one consider indexes of groups.

To review Sylow’s theorem, read the post Sylow’s Theorem (summary)

## Proof 1.

Since $p$ does not divide the index $|G: N|$, the order of $N$ is of the form
$|N|=p^am, \text{ where } m|n.$ By Sylow’s theorem, the group $N$ has a $p$-Sylow subgroup $P$.

Since the order of $P$ is $p^a$ and $P$ is a subgroup of $G$, it is also a Sylow subgroup of $G$.
Let $P’$ be any $p$-Sylow subgroup of $G$. Then by Sylow’s theorem, two $p$-Sylow subgroups are conjugate.

Thus there exists $g\in G$ such that
$g^{-1}Pg=P’.$ Then since $N$ is normal in $G$, we have
\begin{align*}
P’=g^{-1}Pg< g^{-1}Ng=N \end{align*} and $P'$ is a subgroup of $N$.

## Proof 2.

Let $P$ be any $p$-Sylow subgroup of $G$. Seeking a contradiction, assume that $P \not \subset N$. Thus there exists $x \in P \setminus N$.
Let $H=\langle x \rangle$ be a subgroup of $G$ generated by $x$.
Since the order of $x$ is a power of $p$, the order of $H$ is a power of $p$.

Since $N \triangleleft G$, by the second isomorphism theorem, we have
$|HN :N|=| H: H \cap N| \tag{*}.$ Also, the chain of subgroups $N < HN < G$ implies $|G: N|=|G: HN| |HN: N|.$ Combining this with (*), we obtain \begin{align*} |G: N|=|G: HN|| H: H \cap N|. \end{align*} Since $H\not \subset N$, $| H: H \cap N|$ is a positive power of $p$. However, this contradicts that $p$ does not divide $|G: N|$. Hence $P$ must be contained in $N$.

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1. 02/06/2017

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##### If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup

Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$. Let...

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