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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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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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    @09867 Yes, that's exactly it.2012-05-17