# The order of $H$ is relatively prime to its index $[G:H]$ [closed]

Suppose that a subgroup $H$ of a finite group $G$ satisfies one of the following two conditions:

(i) For any nonidentity element $x$ of $H$ we have $C_{G}(x) \subset H$

(ii) If $K$ is a subgroup of $H$ and $K \neq 1$, then $N_{G}(K) \subset H$

Prove that the order of $H$ is relatively prime to its index $[G:H]$

• For (ii) you could take $K$ to be a Sylow subgroup of $H$. Dec 17 '13 at 9:06
• Did you imply (ii) is equivalent to $K$ is a Sylow subgroup of $H$?
• No, I meant that, to prove that condition (ii) implies the result, you could apply condition (ii) in the case when $K$ is a Sylow $p$-subgroup of $H$, where $p$ is a prime dividing $|H|$, and use the fact that proper subgroups of $p$-groups are strictly contained in their normalizers to deduce that $K$ must be a Sylow $p$-subgroup of $G$, and hence $p$ does not divide $|G:H|$. Dec 17 '13 at 12:38