I find that there are either $1$ or $3$ Sylow-$2$ subgroups, and $1, 4$ or $16$ Sylow-$3$ subgroups. I need one of them to be $1$, so that it is normal, so I can mod out by it and have $2$ $p$-groups $N$, $G/N$, which are thus both solvable which implies that $G$ is solvable.


Hint: If $G$ has $3$ Sylow $2$-subgroups, then $G$ acts non-trivially on them by conjugation, therefore there is a non-trivial mapping $f:G\to S_3$. Then $|G/\ker f|=2,3,6$. This action is transitive, so $|G/\ker f|=3,6$ and $|\ker f|=16,8$ and both these groups are solvable.

  • $\begingroup$ $f$ is surjective group hom. so how is $G/Kerf = 2,3$? Isn't $G/Kerf$ supposed to be isomorphic to $S_{3}$ which implies its order to be $6$? $\endgroup$ – masszz Nov 1 '13 at 8:16
  • $\begingroup$ I added the answer. $\endgroup$ – Boris Novikov Nov 1 '13 at 8:25
  • $\begingroup$ Why can you not get $|G/\rm{ker}(f)| = 6$? $\endgroup$ – Tobias Kildetoft Nov 1 '13 at 8:29
  • $\begingroup$ @Tobias Kildetoft Oh, thank you! $\endgroup$ – Boris Novikov Nov 1 '13 at 8:41
  • $\begingroup$ @sergey You are right, I corrected. $\endgroup$ – Boris Novikov Nov 1 '13 at 8:42

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