Number of conjugates of a subgroup If $G$ is a simple non-abelian group and $H$ is a subgroup with $[G:H]=7$ then what is the number of conjugates of $H$ in $G$?
So far I found that the order of $H$ cannot be a prime number using Sylow theorems.

The number of conjugates of $H$ is $[G:N_G(H)]$, where $N_G(H)$ is the normalizer of $H$ in $G$. Note that $H$ is normal iff $G=N_G(H)$. Since $H \subseteq N_G(H)$, and $H$ cannot be normal, $H=N_G(H)$, and $H$ has exactly 7 conjugates: $7=[G:H]=[G:N_G(H)] \cdot [N_G(H):H]$. 
