If $G$ is a group of order $231$, show that every element of order $11$ belongs to its center? By Sylow's theorems, $G$ has a unique $11$-sylow subgroup of order $11$ and then cyclic, say this $11$-sylow subgroup called $N$.
And $N$ contains every element of order $11$ in $G$.
Then how can I complete the proof since I am stuck here.
 A: Let $n_7$ be the number of Sylow 7-subgroup. We have $n_7 \equiv 1$ (mod 7) and $n_7$ divides $33$. So there is a unique Sylow 7-subgroup, named $L$. 
In the same process, there is a unique Sylow 11-subgroup, named $N$. They are both normal.
Let $P$ be a Sylow 3-subgroup. Because $L$ is normal, then $PL$ is a subgroup of order 21 of $G$. 
Consider its normalizer $N_G(PL)$. One has $PL \leq N_G(PL) \leq G$. But the order of $N_G(PL)$ cannot be 21. So $N_G(PL) = G$. It means that $PL \trianglelefteq G$.
So, $N$ and $PL$ are normal subgroup, then $NPL$ is a subgroup of order 231, or $NPL = G$. Moreover, $N$ commute with $PL$ because if $n\in N$, $p \in PL$, then by normality of $N$ and $PL$, $npn^{-1}p^{-1} \in N \cap PL = \{e\}$. So $N$ commute with $G$ since every $g\in G$ can be written as $n'p'$.
A: Let $N=\left<a\right>$ and let $g\in G$. Then $g^{-1}ag=a^r$ for some $r$.
This way $G$ maps to $\text{Aut}(N)$ by conjugation. As $\text{Aut}(N)$
has order $10$ and $G$ has order $231$, which is coprime to $10$, this
map is trivial, so $g^{-1}ag=a$ for all $g\in G$.
