There is no simple group of order $36$. I tried to do this as an exercise and wanted to ask if my proof is correct or if it is missing something. Thank you so much.
Let $G$ be a group such that $\lvert G \rvert = 36 = 2^2 \cdot 3^2.$ Show that $G$ is not simple.
$\underline{Proof:}$
Assume $G$ is simple.

*

*We start with the $2-$Sylow subgroups of $G$.
Then:


*

*$s_2(G) \vert 9 \implies s_2 \in \{1,3,9\}$


*$s_2 \equiv 1 \mod2 \implies s_2 \in \{1,3,9\}$.
But by assumption, $s_2 \neq 1$, since then the only $2-$Sylow subgroup would be normal in $G$.
Let $S_1,S_2 \in Syl_2(G)$ be two $2-$Sylow subgroups of $G$. Consider then
$S_1 \cap S_2 <S_1.$ Then we have:
$$\lvert S_1 \cap S_2\rvert \vert \lvert S_1\rvert = 4$$
Thus $\lvert S_1 \cap S_2 \rvert \in \{1,2\}$.
Consider now $P_2$, the set of elements of order $2$ or $4$. We have that $(S_1 \cup S_2) - \{1\} \subset P_2$ and thus:
$$\lvert P_2 \rvert \geq \lvert S_1 \rvert + \lvert S_2\rvert - \lvert S_1\cap S_2\rvert -1 \geq 4+4-2-1 =5 $$
So we know there are at least $5$ elements of order 2 or 4.


*We now consider the $3$-Sylow subgroups of $G$.


*

*$s_3 \vert 4 \implies s_3 \in \{1,2,4\}$

*$s_3 \equiv 1 \mod 3 \implies s_3 \in \{1,4\}$
Again by assumption on $G$, $s_3 \neq 1$. So we have that $s_3 = 4$.
Let $H_1 , H_2 \in Syl_3(G)$ be two $3$-Sylow subgroups. We consider again its intersection as a subgroup, then:
$$\lvert H_1 \cap H_2 \rvert \vert \lvert H_1\rvert = 9$$
So we have that $\lvert H_1 \cap H_2\rvert \in \{1,3\}$.

*

*If $\lvert H_1 \cap H_2\rvert = 1$ and consider $P_3$ the set of elements of order $3$ or $9$, we have that:

$$P_3 = \bigsqcup_{H\in Syl_3} (H-\{1\})$$
Which implies that:
$$\lvert P_3 \rvert = s_3 \cdot (9-1) = 4\cdot 8 = 32$$
But then $\lvert P_2\rvert + \lvert P_3\rvert = 37 > \lvert G \rvert$. Which is a contradiction. So $\lvert H_1 \cap H_2\rvert = 3$


*Define now $T:= H_1 \cap H_2$ and consider its normalizer $N:= N_G(T)$. We then know:


*

*$\lvert N\rvert \vert 36 =\lvert G \rvert$

*$H_i$ are abelian for $i =1,2$ since they have cardinality $3^2 = 9$.

*$9= \lvert H_1\rvert \vert \lvert N\rvert $
It follows that $\lvert N \rvert \in \{9,18,36\}$. But $\lvert N \rvert \neq 36$ since then $N = G$ and we would have two normal subgroups in $G$. Contradicting that $G$ is simple.
Notice that $\lvert N \rvert \geq \lvert H_1 \cup H_2\rvert \geq 9+9-3 > 9$
So we have that $\lvert N \rvert = 18$.


*We now consider the $3$-Sylow subgroups of $N$. We have:


*

*$s_3(N)\vert 2 \implies s_3(N) \in \{1,2\}$

*$s_3(N) \equiv 1 \mod 3 \implies s_3(N) = 1$
But this is a contradiction since
$$H_1, H_2 \subset N$$
So $G$ has only one $3$-Sylow subgroup. Which implies that $G$ has to be simple.
$\square$
 A: A simpler approach, let $P \in Syl_3(G)$ and assume $G$ to be simple. Then $|G:P|=4$. $G$ acts by left multiplication on the left cosets of $P$ and the kernel of this action is core$_G(P)=\bigcap_{g \in G}P^g$. This is a normal subgroup of $G$ and since $G$ is simple, it it trivial. This means that $G$ embeds isomorphically into $S_4$. This implies $36 \mid 24$, by Lagrange's Theorem, a contradiction. Hence $G$ cannot be simple.
Note on the solution The following is a generalization of the well-known Cayley imbedding, which can be recovered by taking $H=1$ in the sequel. Let $H \leq G$, with $|G:H|=n$ a positive integer. Put $\Omega=\{g_1H,g_2H, \ldots ,g_nH\}$, the set of left cosets of $H$. $G$ works on $\Omega$ by multiplication from the left: if $g \in G$, then $g(g_iH)=gg_iH$ is again one of the elements of $\Omega$. So each $g \in G$ induces a permutation $\sigma_g$ on $\Omega$. Hence we can define a map $f: G \rightarrow S_n$, by $f(g)=\sigma_g$. It is easy to verify that $f$ is a homomorphism (observe that $\sigma_g \circ \sigma_h=\sigma_{gh}$ for all $g,h \in G$). Now let us compute ker$(f)$. If $x \in $ ker$(f)$, this means that $x$ induces the trivial permutation, so for any left coset $gH$, we must have $xgH=gH$, implying $x \in gHg^{-1}$ for all $g \in G$. This amounts to $x \in \underset{g \in G}\bigcap H^g$ (where $H^g=g^{-1}Hg$; note that when $g$ runs through the entire $G$, $g^{-1}$ does the same, so intersection over $g^{-1}Hg$ is the same as over $gHg^{-1}$). Conversely, if $x \in \underset{g \in G}\bigcap H^g$, then it induces a trivial permutation.
We conclude that ker$(f)=\underset{g \in G}\bigcap H^g$ and by the first isomorphism theorem $G/$ker$(f) \cong $ im$(f)$, being a subgroup of $S_n$.
In general, if $H \leq G$, the core$_G(H)$ is the largest normal subgroup of $G$ contained in $H$. It is easy to show that core$_G(H)=\underset{g \in G}\bigcap H^g$.
