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 $G$.

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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