Exercise 4.5.22(If $|G| = 132$ then $G$ is not simple)

Proof. Here $|G|=132=2^2\cdot 3\cdot 11$. By Sylow’s Theorem

Therefore

Assume for the sake of contradiction that $G$ is simple. Then $n_{11}\ne 1$ and $n_3\ne 1$, so $n_{11}=12$ and $n_3=4$ or $n_3=22$. But $n_3=22$ is impossible, otherwise $G$ contains $n_{11}\cdot 10=120$ elements of order $11$, and $n_3\cdot 2=44$ elements of order $3$, which gives $|G|\ge 120+44=164$. So

This gives $120$ elements of order $11$ and $4\cdot 2=8$ elements of order $3$. It remains a set $S$ of $4$ elements, whose orders are neither $11$ nor $3$.

Let $P$ be any Sylow $2$-subgroup. The order of the elements of $P$ are $1,2$ or $4$, so $P\subset S$. Since $|P|=|S|=4$, we obtain $P=S$, so there is a unique $2$-Sylow of $G$, which is normal in $G$, so $G$ is not simple. This is a contradiction, which proves that $G$ is not simple.

If $|G|=132$ then $G$ is not simple. □