# Finite simple group with subgroups of same order

Let $D$ be a finite simple group with $H < D$ and $K < D$. Also $[D:H]=q$ and $[D:K]=p$, where $p$, $q$ are primes. Want to show that $p=q$.

I want to come up with a contradiction with one of the subgroups being normal. I just couldn't do it with the given information. Please help!

Thank you!

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Consider the left multiplication action of $D$ on the left cosets of $H$, which induces a group homomorphism from $D$ to $S_q$, the symmetric group of $q$ symbols. Since $D$ is simple, this homomorphism is injective, and therefore the order of $D$ divides $q!$, the order of $S_q$. On the other hand, since $[D:K]=p$, $p$ divides the order of $D$. As a consequence, $p$ divides $q!$, i.e. $p\le q$. A similar argument shows that $q\le p$, so $p=q.$
No, I mean if $D$ a finite group and $H$ is a subgroup of $D$ with $[D:H]=q$, then there is a group homomorphism $\rho:D\to S_q$ induced by the left multiplication action of $D$ on the left cosets of $H$. Since the action is transitive, $\rho$ is nontrivial, i.e. $\mathrm{Ker}\rho\ne D$ . Moreover, since in this question, $D$ is simple, $\rho$ has to be injective. Therefore, $D$ is isomorphic to a subgroup of $S_q$. – 23rd Dec 15 '12 at 5:13