Let's assume that there exists simple non-Abelian group $G$ of order $120$. How can I show that $G$ is isomorphic to some subgroup of $A_6$?


Hint: How many Sylow 5-subgroups will $G$ have? Do you know of a way $G$ acts on them?

  • $\begingroup$ I know that G acts on them by conjugation, but that's all :( $\endgroup$ – Martin Jan 17 '13 at 11:08
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    $\begingroup$ Good! So if $X$ is the set of Sylow 5-subgroups, then conjugation action gives you a group homomorphism from $G$ to $S_n$, where $n=|X|$. Can you work with that? $\endgroup$ – Jyrki Lahtonen Jan 17 '13 at 11:11
  • $\begingroup$ Yes, I think I can. Thanks a lot :) $\endgroup$ – Martin Jan 17 '13 at 11:13
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    $\begingroup$ Great! If you can complete the argument, I would encourage you to type your solution as an answer to this question. Then the rest of us can comment on it and upvote it. You get more feedback that way. You can also accept that answer. It is not clear that a hint like this should be the accepted answer, for the readers who join the party later may benefit more from a full solution. $\endgroup$ – Jyrki Lahtonen Jan 17 '13 at 11:16
  • $\begingroup$ No problem :) When I get complete solution, it will be posted $\endgroup$ – Martin Jan 17 '13 at 11:21

A group of order 120 can not be simple. Let's assume that there exists simple non-abelian group $G$ of order 120. Then we know the number Sylow 5-subgroups of $G$ is 6. Hence, the index of $N_{G}(P)$ in $G$ is 6 ($P$ is a Sylow 5-subgroup of $G$). Now there exists a monomorphism $\phi$ of $G$ to $S_{6}$. We claim that $\operatorname{Im\phi}\leq A_{6}$. Otherwise $\operatorname{Im\phi}$ has an odd permutation and so $G$ has a normal subgroup of index 2, a contradiction. Hence, $G\cong \operatorname{Im(\phi)}\leq A_{6}$.

  • $\begingroup$ @ Jyrki Lahtonen: Ok. I edited it. $\endgroup$ – maryam Jan 17 '13 at 13:18

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