# Exercise 26, Chapter 25 of Gallian's Contemporary abstract Algebra

This is problem 26, chapter 25 of Gallian's Contemporary abstract algebra,

Let $G$ be a finite simple group and suppose that it contains subgroups $H$ and $K$ such that $[G:H]$ and $[G:K]$ are prime. Show that $|H|=|K|$.

I am not getting any idea to prove this. If G is simple if $H$ is a subgroup of $G$ such that $[G:H]=m$ , then G is isomorphic to a subgroup of $A_{m}$. Can this help in my problem? Any help will be appreciated.

• It can! And why not suppose the primes are $p,q$ with $p<q$? What do we now know about $|G|$? – ancientmathematician May 6 '17 at 13:59
Let $|G:H|=p$ and $|G:K|=q$ . Assume that $q<p$ . Now there homomorphism $\phi :G \rightarrow S_{G/H}$ ( the symmetric group on the cosets ) .
Since $G$ is simple , the $\ker \phi$ is exactly $\{1\}$ and the homorphism is injective . Now $|\phi(K)| | |S_{G/H}|$ i.e $q|p!$ . But by the choice of $p,q$ we see that $q!$contains no multiple of $p$ . So this is impossible . Contradiction to the fact that $q<p$ .
Do the case $q>p$ similarly and obtain contradiction . So only case possible is $p=q$