1
$\begingroup$

How do you show that there are no simple groups of order $2^n\times 5$ for $n\geq 4$, without using the theorem that a finite group of order $p^nq^m$, where p, q are primes and $m,n\geq 1$ is not simple.

I have a hint to 'use the coset action determined by the Sylow 2-subgroup', but I'm not sure what this means.

1 Answers 1

3

Hint: There are either 5 or 1 Sylow 2-subgroups (why?). Assuming there are 5 and we let $G$ act by conjugation on the Sylow 2-subgroups we get a non-trivial homomorphism from $G$ to $S_5$. What can we say about its kernel?

  • 0
    Sorry I'm still not quite understanding. I know if the kernel is nontrivial then G can't be simple because the kernel is a normal subgroup. And if there is only 1 Sylow 2-subgroup then it is normal and so G isn't simple. But the group action confuses me, how do I know it has nontrivial kernel?2012-05-17
  • 0
    @09867 What's the order of G and what's the order of $S_5$.2012-05-17
  • 0
    $|S_5|=120$, $|G|=2^n.5$. So if $n\geq 4$ $|G|\geq 80$. So.. this homomorphism can't exist because 80 doesn't divide 120?2012-05-17
  • 0
    @09867 Well, no. But it does mean that the homomorphism can't be injective and since it must be non-trivial, what do we deduce?2012-05-17
  • 0
    or the kernel has to be nontrivial?2012-05-17
  • 0
    Since it must be nontrivial, it is a proper normal subgroup of G or the whole of G, but it can't be all of G because the homomorphism is nontrivial.2012-05-17
  • 0
    @09867 Yes, that's exactly it.2012-05-17