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I have searched the site for some information on this problem, but the results aren't illuminating enough.

Let $G$ be a simple group of order $60$. Since it is simple, then there must be $6$ Sylow $5-$subgroups in $G$. Conjugation by elements of $G$ give rise to a permutation representation $\varphi:G\rightarrow S_6$ since all Sylow $5-$subgroups are conjugate. Since $G$ is simple, then the $\operatorname{ker}\varphi=\{1\}$. Then by the first isomorphism theorem for groups $\varphi(G)\cong G\le S_6$.

Since $G$ is simple, then $\varphi(G)$ is simple, so that $G\cong\varphi(G)\le A_6$, the largest simple subgroup of $S_6$.

Is this sufficient? Or is there a better way to argue this?

Laars Helenius
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  • This seems fine. What part seems lacking? – Nick Mar 01 '16 at 03:26
  • It just seems like the last observation was so trivial that I might have been overlooking something. I am preparing for a qualifier, so I just want to make sure I have this right. – Laars Helenius Mar 01 '16 at 03:28
  • You would have to prove first that every simple subgroup of $S_6$ of order $60$ is contained in a copy of $A_6$ in order for your argument to work. We could brute force this by looking at the lattice of $S_6$ but that seems too computational. – 19021605 Aug 21 '25 at 13:47
  • Similar: https://math.stackexchange.com/questions/75411/why-is-a-simple-group-of-order-60-embedded-in-a-6?rq=1 – 19021605 Sep 01 '25 at 07:20

2 Answers2

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I have hard time understanding why $G\le A_6$ just because $A_6$ is the largest simple subgroup of $S_6$.

I'd argue differently, by looking at the signature homomorphism: $$G\to S_6\to C_2.$$ That must be trivial, otherwise $G$ has a subgroup of index $2$. Thus $G$ is a subgroup of $A_6$.

Quang Hoang
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  • In general: https://math.stackexchange.com/questions/2830721/a-6-does-not-have-a-subgroup-of-order-120 – 19021605 Aug 21 '25 at 14:19
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If $\varphi(G)\not\leq A_6$, then $S_6=\varphi(G)A_6$. By the $2$nd isomorphism theorem, $\varphi(G)\cap A_6$ has index $2$ in $\varphi(G)$, a contradiction since $\varphi(G)$ can't have any proper subgroup of index less than $5$.

19021605
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