subgroup of order $11$ lies inside $Z(G).$ I need help to solve this problem:
Let $G$ be group of order $231.$ we need to show that the subgroup of order $11$ lies inside $Z(G).$
 A: Here is an outline for a different solution, you can try to fill in the details.
By Sylow's theorem, the group $G$ has a normal subgroup $P$ of order $11$ and also a normal subgroup $H$ of order $7$. Because every group of order $33$ is cyclic, $G/H$ is abelian and $G' \leq H$. Therefore $P$ is a normal subgroup and $P \cap G' = \{1\}$, which implies that $P \leq Z(G)$.
This is same approach sometimes works for similar exercises (that is, exercises like "Prove that subgroup $X$ is central"). However, it does not always work since $P \leq Z(G)$ does not imply $P \cap G' = \{1\}$. In general I would recommend using the normalizer/centralizer theorem as in Tobias answer.
A: Here is an alternative way to do it:
Show that group has a unique subgroup of order $11$ (to make the question make sense). This follows directly from the Sylow theorems.
Let $H$ be this subgroup (which is normal). $G/C_G(H)$ is isomorphic to a subgroup of $\rm{Aut}(H)$ by the normalizer/centralizer theorem. But the automorphism group of $H$ has order $10$, which is coprime to $|G|$, so the only possibility is that $C_G(H) = G$, which means that $H$ is central in $G$.
Edit: To see that $G/C_G(H)$ is isomorphic to a subgroup of $\rm{Aut}(H)$, we define a map from $G$ to $\rm{Aut}(H)$ by sending each $g\in G$ to conjugation by $g$ (that is, the map from $H$ to itself given by $h\mapsto ghg^{-1}$). Then we check that this map is a homomorphism and has kernel equal to $C_G(H)$.
A: Yet another approach:
Firstly, since $[G:H] \equiv [N_GH:H] \equiv -1 \pmod{11}$, we conclude that $H$ is normal in $G$. Then, define an action of $G$ on $H$ by conjugation. Again by the class-equation, we have
$|H|=|K| +\Sigma |O_i|$, where K is the set of fixed points and $O_i$ are orbits. But $|H|$ is prime, so either $|K|=0$, or $|K|=11$. Since the identity is always fixed, we conclude that $K=H$, i.e. $H$ lies in the center of $G$.
