2
$\begingroup$

My problem is as seen in the title:

For positive integer $n>1$, prove that a simple group with order $\geq n!$ cannot have subgroup of index $n$.

Could anyone give me some hints on how to approach this?

$\endgroup$
2
$\begingroup$

If $[G:H]=n$ then $G$ acts on $n$ cosets of $H$. Hence there is a homomorphism $G$ into the symmetric group $S_n$. Since $|S_n|=n!$ then either this homomorphism has the non-trivial kernel or $G=S_n$. But in the last case $G$ also is not simple.

$\endgroup$
6
  • $\begingroup$ Thank you. But actually I prefer hints more than full proof. Like what theorems or techniques should be considered. (Before I asked this question, I've considered group action as homomorphism, but I forgot the fact that kernel is normal subgroup. If anyone would remind me of that, I would be able to solve the problem.) Anyway, I understand that giving hints is harder than giving full proof; I should have mentioned what results I had thus far. $\endgroup$ – 4ae1e1 Oct 20 '13 at 16:19
  • $\begingroup$ Oh, I am sorry! :-) As a compensation I propose (if you want) to prove that $G$ does not contain a subgroup of index $n+1$. $\endgroup$ – Boris Novikov Oct 20 '13 at 16:32
  • $\begingroup$ Uh please don't feel sorry about that. By the way I'm a bit confused about the $n+1$ problem you proposed. What about alternating group $A_{n+1}\ (n \geq 4)$? It is simple, and has an index $n+1$ subgroup $A_n$ (fix $n+1$). Its order is $(n+1)!/2 > n!$. Anything wrong here? $\endgroup$ – 4ae1e1 Oct 20 '13 at 18:06
  • $\begingroup$ Yes, you are right and I mistaked. What one can prove instead of my question is: if $G$ contains a subgroup of index $n+1$ then $G$ is embedded into $A_{n+1}$. Sorry onсe more. $\endgroup$ – Boris Novikov Oct 20 '13 at 19:49
  • $\begingroup$ That's interesting. I'll think about it when I have free time. Thanks! $\endgroup$ – 4ae1e1 Oct 21 '13 at 3:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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