$G$ be a finite simple group and $H,K$ be subgroups of prime index ; then is it true that $H,K$ are of same size?

Question

Let $G$ be a finite simple group and $H,K$ be subgroups of prime index ; then is it true that $|H|=|K|$ ?

Answer 1

Let $G$ be simple and assume $G$ has a subgroup of prime index, as stated. Define an action of $G$ on $G/H$ ($|G/H|=p$) and $G/K$ ($|G/K|=q$) as usual where $p$ and $q$ are, as usual, primes.

Then the resultant homomorphism $G\to S_p$ has trivial kernel and thus $G$ is isomorphic to a subgroup of $S_p$. Similarly, $G$ is isomorphic to a subgroup of $S_q$. Thus, $|G|$ divides $q!$ and $p!$.

Since $G$ is isomorphic to a subgroup of $S_p$ and $p$ divides the order of $G$ (NB: $|G|=p|H|$) and since $G$ is isomorphic to a subgroup of $S_q$ as well and $q$ divides the order of $G$ (NB: $|G|=q|K|$), it must be that $q\le p$ if $q$ divides $|G|$ and thus $|S_p|$ and also $p\le q$ if $p$ divides $|G|$ and thus $|S_q|$, whence $p=q$.