Take the 2-minute tour ×
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.

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?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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. –  KevinSayHi Oct 20 '13 at 16:19
    
Oh, I am sorry! :-) As a compensation I propose (if you want) to prove that $G$ does not contain a subgroup of index $n+1$. –  Boris Novikov Oct 20 '13 at 16:32
    
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? –  KevinSayHi Oct 20 '13 at 18:06
    
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. –  Boris Novikov Oct 20 '13 at 19:49
    
That's interesting. I'll think about it when I have free time. Thanks! –  KevinSayHi Oct 21 '13 at 3:55
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.