[Math] Prove if $\left|G\right|=231$ then $Z(G)$ contains a Sylow $11-$subgroup

Question

$231=3\cdot7\cdot11.$ Then $G$ has Sylow subgroups of orders $3,7,11.$ I can prove that Sylow subgroups of orders $7,11$ are normal.
But I can't prove that $Z(G)$ contains a Sylow $11-$subgroup.
I have no idea how to start it. I assume that this has something to do with normalizer or the class equation. I tried proved that $Z(G)$ contains element of order $11$. But I didn't manage to prove it.

Answer 1

By Sylow's third theorem, $N_{11}\equiv 1\mod 11$ and $N_{11}\mid 21$. Thus there is only one $11$-Sylow subgroup of $G$, therefore this Sylow subgroup is normal. Let's denote it $S$. $S$ is cyclic since it has prime order, and $G/C_G(S)$ is isomorphic to a subgroup of $\operatorname{Aut}(S)$.

However, $\lvert\operatorname{Aut}(S)\rvert=10$, which is coprime to $\lvert G\rvert$, hence to $\lvert G/C_G(S)\rvert$. This proves $\lvert G/C_G(S)\rvert=1$, in other words $C_G(S)=G$, or $S\subset Z(G)$.