Prove if $\left|G\right|=231$ then $Z(G)$ contains a Sylow $11-$subgroup $231=3\cdot7\cdot11.$ 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.
 A: 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 $S\subset Z(G)$.
A: 2nd Sylow's theorem gives you that you have a unique $S_7$, $7$-Sylow, a unique $S_{11}$, $11$-Sylow and either $1$ or $7$ $3$-Sylows.
Take one $3$-Sylow $S$, we know that $G$ acts transitively by conjugation on the set of $3$-Sylow. Remark that for this action the stabilizer of $S$ is $N_G(S)$ (the normalizer of $S$). Using first class formula and the transitivity :
$$|N_G(S)|=231\text{ or } 33 $$
In either case $N_G(S)$ is divisible by $11$ so it contains one $11$-Sylow which cannot but be the unique $11$-Sylow of $G$.
In conclusion $S_{11}$ normalizes both $S$ and $S_7$. Now I claim that this implies that for all $g\in S_{11}$ $gs=sg$ for all $s\in S\cup S_7$ (look at $gsg^{-1}s^{-1}$).
In other words any element of $S_{11}$ commutes with any element of $S$ or $S_7$. Since $S_{11}$ is cyclic of order $11$ any element of $S_{11}$ commutes with any element of $S_{11}$.
Finally any element of $S_{11}$ commutes with $ \langle S,S_7,S_{11} \rangle$ which is equal to $G$ by cardinality argument (its order is divisble by $3$, $7$ and $11$...).
