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.


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.

  • $\begingroup$ Thanks Steve D, I get if $P$ is a normal Sylow $2$-subgroup, then both $P$ and $G/P$ are solvable, so $G$ is solvable. I'm not familiar with the second result about a nontrivial homomorphism into $S_3$. Do you have a reference where I can find out about this? Thanks. $\endgroup$ – yunone Sep 13 '11 at 19:30
  • 2
    $\begingroup$ @yunone: Let's call the three Sylow 2-subgroups $P_1$, $P_2$, and $P_3$. Then by Sylow's theorems we know that for any $g\in G$, $P_1^g$ is some $P_i$ for $1\le i\le3$. Thus if we consider the set $X=\{P_1,P_2,P_3\}$, $G$ is acting on this (3-element) set. This gives a homomorphism into $S_3$. For example, if there is a $g\in G$ with $P_1^g=P_2$, $P_2^g=P_1$, and $P_3^g=P_3$, then we would map $g$ to the element $(1,2)(3)$ in $S_3$. It is fairly easy to check this is a group homomorphism. This idea is very powerful: given an action of a group $H$ on some $n$-element set, we get a... $\endgroup$ – user641 Sep 13 '11 at 19:34
  • $\begingroup$ homomorphism from $H$ to $S_n$. $\endgroup$ – user641 Sep 13 '11 at 19:35
  • $\begingroup$ Thanks for clearing that up Steve, I really appreciate it. $\endgroup$ – yunone Sep 14 '11 at 2:43

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