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For the past day or so I've been trying to solve an exercise in Lang showing all groups of order less than $60$ are solvable.

Excluding the case of order $56$, most cases are taken care of by other theorems. The ones that are giving me trouble are groups of order $2^n\cdot 3$, namely, $12, 24, 48$. I can solve these individually by counting arguments, but I'd rather knock them all out in one go with a more general result.

I've been searching for a proof that groups of order $2^n\cdot 3$ are solvable, but haven't found one. Does anyone have a clean proof of this fact? Thank you.

P.S. I'd rather not use Burnside's Theorem, since I think that's a little overkill for the spirit of this problem.

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Let $G$ be a group of order $2^n\cdot 3$. If $G$ has a normal Sylow 2-group, you are done (can you see why?).

Otherwise, $G$ has 3 Sylow 2-subgroups, and that gives a non-trivial homomorphism of $G$ to $S_3$ via the conjugation action on these Sylow 2-subgroups. The kernel is a 2-group, and the image is solvable, so again, you are done.

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    Thanks for clearing that up Steve, I really appreciate it.2011-09-14