Let $G$ be a group of order $$72=2^3\cdot 3^2$$ Without using Burnside's Theorem, how to show that $G$ is solvable?


If we can show that $G$ has at least one non-trivial normal subgroup $N$, then it would be easy to show it is solvable. Indeed, $$1\longrightarrow N\longrightarrow G\longrightarrow G/N\longrightarrow 1$$ would be a short exact sequence with $N$ and $N/G$ of order $2^i\cdot3^j$ for some $i,j\in\{0,1,2\}$ and it is not too hard to show that such groups are always solvable. However, I can't find a way to show that $G$ is not simple.

Added: If $G$ is not simple, then Sylow's Theorem implies that there are $4$ subgroups of order 9 and 3 or 9 subgroups of order 8. Then, I don't see how to use that to show that $G$ is not simple.


1 Answer 1


If $G$ has 4 Sylow-3 subgroups, $G$ acts on those subgroups via conjugation, inducing a homomorphism $G\to S_4$. Since $|S_4|=24<72=|G|$, this map must have a non-trivial kernel. If the morphism is not the trivial map, you are done. What can you say if the kernel is all of $G$?

  • 3
    $\begingroup$ Great ! Thanks. If the kernel is all of $G$ then each $g\in G$ normalizes each Sylow 3-subgroup. Then they are all normal which can't happen (if $G$ has a normal Sylow 3-subgroup then $n_3=1$). Hence, the kernel of this action is a non-trivial normal subgroup of $G$ and we are done. Right? $\endgroup$
    – Spenser
    Nov 17, 2013 at 2:48
  • 1
    $\begingroup$ That is correct. $\endgroup$ Nov 17, 2013 at 2:49

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