TITLE: Is there a general result that groups of order $2^n\cdot 3$ are solvable?
QUESTION [7 upvotes]: 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.

REPLY [11 votes]: 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.