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 the number of Sylow $3$-subgroups of $G$.

If there is only one Sylow $3$-subgroup, then it is a normal subgroup, hence $G$ is not simple.

If there are more than one, consider the action of $G$ on those Sylow $3$-subgroups given by conjugation.
Then consider the induced permutation representation.

Observe the prime factorization $72=2^3\cdot 3^2$.
Let $G$ be a group of order $72$.

Let $n_3$ be the number of Sylow $3$-subgroups in $G$.
By Sylow’s theorem, we know that $n_3$ satisfies
\begin{align*}
&n_3\equiv 1 \pmod{3} \text{ and }\\
&n_3 \text{ divides } 8.
\end{align*}
The first condition gives $n_3$ could be $1, 4, 7, \dots$.
Only $n_3=1, 4$ satisfy the second condition.

Now if $n_3=1$, then there is a unique Sylow $3$-subgroup and it is a normal subgroup of order $9$.
Hence, in this case, the group $G$ is not simple.

It remains to consider the case when $n_3=4$.
So there are four Sylow $3$-subgroups of $G$.
Note that these subgroups are not normal by Sylow’s theorem.

The group $G$ acts on the set of these four Sylow $3$-subgroups by conjugation.
Hence it affords a permutation representation homomorphism
\[f:G\to S_4,\]
where $S_4$ is the symmetric group of degree $4$.

By the first isomorphism theorem, we have
\begin{align*}
G/\ker f < S_4.
\end{align*}
Thus, the order of $G/\ker f$ divides the order of $S_4$.
Since $|S_4|=4!=2^3\cdot 3$, the order $|\ker f|$ must be divisible by $3$ (otherwise $|G/\ker f$|$ does not divide $|S_4|$), hence $\ker f$ is not the trivial group.

We claim that $\ker f \neq G$.
If $\ker f=G$, then it means that the action given by the conjugation by any element $g\in G$ is trivial.

That is, $gPg^{-1}=P$ for any $g\in G$ and for any Sylow $3$-subgroup $P$.
Since those Sylow $3$-subgroups are not normal, this is a contradiction.
Thus, $\ker f \neq G$.

Since a kernel of a homomorphism is a normal subgroup, this yields that $\ker f$ is a nontrivial proper normal subgroup of $G$, hence $G$ is not a simple group.

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 […]

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, […]

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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 […]

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 […]

Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4
Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.
Hint.
Use Sylow's theorem.
(See Sylow’s Theorem (Summary) for a review of Sylow's theorem.)
Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is […]

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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 $P$ […]

Are Groups of Order 100, 200 Simple?
Determine whether a group $G$ of the following order is simple or not.
(a) $|G|=100$.
(b) $|G|=200$.
Hint.
Use Sylow's theorem and determine the number of $5$-Sylow subgroup of the group $G$.
Check out the post Sylow’s Theorem (summary) for a review of Sylow's […]

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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.
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We know that $|G/\ker f|$ divides $|S_4|=2^3\cdot 3$. Note that $|G|=2^3 \cdot 3^2$. So if $\ker f$ does not contain the factor $3$, then $|G/\ker f|$ cannot divide $|S_4|=2^3\cdot 3$.

Hi, i didn’t understand why do you say: “the order |kerf| is divisible by 3” ?

Dear cb,

I added more explanations.

We know that $|G/\ker f|$ divides $|S_4|=2^3\cdot 3$. Note that $|G|=2^3 \cdot 3^2$. So if $\ker f$ does not contain the factor $3$, then $|G/\ker f|$ cannot divide $|S_4|=2^3\cdot 3$.