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 $G$.

Subgroup Containing All $p$-Sylow Subgroups of a Group
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 […]

Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]

Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]

Sylow Subgroups of a Group of Order 33 is Normal Subgroups
Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.
Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]

A Group of Order $20$ is Solvable
Prove that a group of order $20$ is solvable.
Hint.
Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem.
See the post summary of Sylow’s Theorem to review Sylow's theorem.
Proof.
Let $G$ be a group of order $20$. The […]

Every Group of Order 72 is Not a Simple Group
Prove that every finite group of order $72$ is not a simple group.
Definition.
A group $G$ is said to be simple if the only normal subgroups of $G$ are the trivial group $\{e\}$ or $G$ itself.
Hint.
Let $G$ be a group of order $72$.
Use the Sylow's theorem and determine […]

Every Sylow 11-Subgroup of a Group of Order 231 is Contained in the Center $Z(G)$
Let $G$ be a finite group of order $231=3\cdot 7 \cdot 11$.
Prove that every Sylow $11$-subgroup of $G$ is contained in the center $Z(G)$.
Hint.
Prove that there is a unique Sylow $11$-subgroup of $G$, and consider the action of $G$ on the Sylow $11$-subgroup by […]

If a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itself
Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$.
Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.
Then show that $N_G(H)=H$.
Hint.
Use the conjugate part of the Sylow theorem.
See the second statement of the […]

## 1 Response

[…] If a Sylow subgroup is normal in a normal subgroup, it is a normal subgroup […]