Let $G$ be a group such that $\vert G\vert=231$. I have to show that the unique Sylow 11-subgroup of $G$ is contained in the center of $G$
I proceed as follows:
Since the number of Sylow 11-subgroup of $G$ is 1, therefore, it is normal in $G$. Similarly, the Sylow 7-subgroup of $G$ is unique and normal. Let $H$ and $K$ denote the Sylow 11-subgroup and Sylow 7-subgroup of $G$ respectively. If $L$ denotes a Sylow $3$-subgroup of $G$, then $KL$ is a subgroup of $G$ of order 21. If I can show that $KL$ is a normal subgroup of $G$, then all elements of $H$ will commute with the elements of $KL$. For this I plan to show that $\vert N(KL)\vert\neq 21$ and so $231$. How to do that. Please help!
A fact that I frequently found useful for these types of problems was the following:
The normalizer of a subgroup H divided by its centralizer is isomorphic to a subgroup of $Aut(H)$, i.e. $$ N(H)/Z(H) \cong K \subset Aut(H) $$
To show that $S_{11}$ belongs to the center is the same as showing that its centralizer is the whole group. To do this, apply the above isomorphism with $H = S_{11}$ and then argue by cardinality (you already know that H is contained in its own centralizer), and lagrange's theorem. Also you will need two facts about groups of prime order: that they are abelian and what their automorphism groups are.