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.

Thanks in Advance!!

  • $\begingroup$ It can! And why not suppose the primes are $p,q$ with $p<q$? What do we now know about $|G|$? $\endgroup$ – 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 $p=q$

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