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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.

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up vote 3 down vote accepted

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?

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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? – 09867 May 17 '12 at 12:14
@09867 What's the order of G and what's the order of $S_5$. – JSchlather May 17 '12 at 12:20
$|S_5|=120$, $|G|=2^n.5$. So if $n\geq 4$ $|G|\geq 80$. So.. this homomorphism can't exist because 80 doesn't divide 120? – 09867 May 17 '12 at 12:25
@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? – JSchlather May 17 '12 at 12:28
or the kernel has to be nontrivial? – 09867 May 17 '12 at 12:28

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