Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.

It follows from Sylow’s theorem that if $Q_1$ and $Q_2$ are both $p$-Sylow subgroups of a group $H$, then they are conjugate.
Namely, there exists $h\in H$ such that $h^{-1}Q_1h=Q_2$.

To prove the problem, let $g\in G$ be any element and try to show that both $P$ and $g^{-1}Pg$ are $p$-Sylow subgroups of $N$.
Then use the fact above with $Q_1=P$, $Q_2=g^{-1}Pg$, and $H=N$.

We use the following notations: $A < B$ means that $A$ is a subgroup of a group $B$, and $A \triangleleft B$ denotes that $A$ is a normal subgroup of $B$.

Proof.

For any $g \in G$, since $P < N$ and $N \triangleleft G$, we have
\begin{align*}
g^{-1}Pg < g^{-1}Ng=N.
\end{align*}
Thus $g^{-1}Pg$ is a $p$-Sylow subgroup in $N$. In general, any two $p$-Sylow subgroups in a group are conjugate by Sylow's theorem.
Since $P$ and $g^{-1}Pg$ are both $p$-Sylow subgroups in $N$, there exists $n \in N$ such that
\[n^{-1}Pn=g^{-1}Pg.\]
Since $n\in N$ and $P$ is normal in $N$, we have $n^{-1}Pn=P$.
Hence we obtain
\[P=g^{-1}Pg.\]
Since $g\in G$ is arbitrary, this implies that $P$ is a normal subgroup in $G$.

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