Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite
1

Let D be a finite simple group, such that 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!

share|improve this question

1 Answer

up vote 7 down vote accepted

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 $p=q.$

share|improve this answer
+1 Very nice and simple. – DonAntonio Dec 14 '12 at 23:47
@DonAntonio: Thank you! – Landscape Dec 15 '12 at 3:49
you mean i have to use the fact that every group of order n is isomorphic to a subgroup of Sn?? – d13 Dec 15 '12 at 5:03
1  
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 $S_q$. – Landscape Dec 15 '12 at 5:13
thankyou now i understand your comment. thanx a lot for ur help. – d13 Dec 16 '12 at 14:13
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.