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 $N$.

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 $N$ 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 $G$.
Hint.
It follows from […]

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, […]

The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$
Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]

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 […]

Subgroup of Finite Index Contains a Normal Subgroup of Finite Index
Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.
Proof.
The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
Hence […]

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 $P$ […]

Any Subgroup of Index 2 in a Finite Group is Normal
Show that any subgroup of index $2$ in a group is a normal subgroup.
Hint.
Left (right) cosets partition the group into disjoint sets.
Consider both left and right cosets.
Proof.
Let $H$ be a subgroup of index $2$ in a group $G$.
Let $e \in G$ be the identity […]

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 […]

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