Show that a group of order 66 has a normal subgroup of order 33.

Question

This question is somewhat similar to: A group of order $66$ has an element of order $33$.

However, I do not understand how I would show that the subgroup of order 33 is normal. So far I have that there is a unique Sylow 11-subgroup, but I don't know how to continue.

Any hints are appreciated.

Answer 1

Let $P$ be the Sylow $11$-subgroup you've shown exists. Since you've shown $P$ is unique, we have that $P$ is normal. Now if we let $Q$ be a Sylow $3$-subgroup, then $PQ=QP$ by normality of $P$, and thus $PQ$ is a subgroup of $G$. But $PQ$ has order $33$ (hence index $2$) so $PQ$ is normal in $G$.

Answer 2

We know by the linked answer that there exists an element of order $33$. Let this element be $x$. Hence, $\langle x\rangle$ has order $33$. We know that all subgroups with index 2 in a group are normal.