Group of order 396 isn't simple

Question

Prove that group of order $396=11\cdot2^2\cdot3^2$ is not simple. $n_{11}$ is $1$ or $12$, so I assumed $n_{11}=12$ and tried to look at the action of the group on $Syl_{11}\left(G\right)$ by conjugation. Thanks for help

Answer 1

So the action of $G$ on $Syl_{11}(G)$ gives a homomorphism from $G$ into $S_{12}$. Since $G$ is simple, this must be an injective map into $A_{12}$. If $P \in Syl_{11}(G)$, then $[G:N_G(P)] = 12$, so $|N_G(P)| = 33$. Any group of order 33 is cyclic (since both Sylow subgroups are normal). This means that $A_{12}$ must have an element of order 33, which it cannot since that would have to be a product of two disjoint cycles of orders 11 and 3 respectively.