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 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.

For a review of the Sylow’s theorem, check out the post “Sylow’s Theorem (summary)“.

Proof.

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.

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