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Let $H$ be a simple group of order $60$. I am trying to see why it is embedded in $A_6$.

$H$ must have $6$ Sylow-$5$ subgroups and $H$ acting by conjugation on these subgroups gives an embedding of $H$ into $S_6$. But this is as far as i have gone.

Please give me only a hint.

Manos
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2 Answers2

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Identify $H$ with its embedded copy in $S_6$ and consider $H \cap A_6$.

Chris Eagle
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  • If $H\not\subseteq A_6$, then $S_6=HA_6$. By the $2$nd isomorphism theorem, $H \cap A_6$ has index $2$ in $H$, a contradiction since $H$ can't have proper subgroup of index less than $5$. – 19021605 Sep 01 '25 at 07:17
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Edit. This answer mistakenly read the question as "prove that there exists some simple group of order $60$ which embeds into $S_6$", and not "prove that every simple group of order $60$ embeds into $S_6$."

Hint: there exists a canonical embedding of $S_5$ in $S_6$. That is, $S_5\leq S_6$. Why does this embedding imply that $A_5\leq A_6$?

user1729
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  • How does this help solve the original question? – 19021605 Sep 01 '25 at 07:15
  • @19021605 Why do you think it doesn't help solve the original question? – user1729 Sep 01 '25 at 08:20
  • If the canonical embedding you meant was mapping $\tau$ to $\tau'$ where $\tau'=\tau$ restricted to ${1,2,3,4,5}$ and $\tau'(6)=6$, then the map is clearly injective by just looking at the images of the first $5$ numbers under each $\tau'$ and maps $A_5$ to a copy of $A_5$ in $S_6$ by applying the sign function to each $\tau'$. If $\sigma$ is OP's map, I can't tell what $\sigma(\tau)$ looks like to apply the sign function. – 19021605 Sep 01 '25 at 08:29
  • @19021605 Is your point "the question is actually asking if every simple group order 60 embeds into $A_6$, not just some"? I can see how the question could be interpreted like this, but the OP accepted the other answer which answers the same question as mine (maybe we were all confused - it was a while ago!). Also, there is only one simple group of order $60$, up to isomorphism. – user1729 Sep 01 '25 at 08:40
  • Yes, if we accept that there is only one simple group of order $60$, up to isomorphism, then the proof works. But statement in the original question can be used to prove that fact (for example, the proof of D&F Abstract Algebra Proposition 4.5.23 and this question: https://math.stackexchange.com/questions/3370045/exercise-about-simple-group-of-order-60?noredirect=1&lq=1). The other answer doesn't require the use of that fact. – 19021605 Sep 01 '25 at 08:46
  • @19021605 Ah, thanks - it was just me who was confused then! Thanks for pointing this out. I'll delete this answer in a minute. – user1729 Sep 01 '25 at 09:07
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    You don't have to. If you edit the "mistake" into your answer, future visitors can learn why this argument may not work. Though if you found a proof that there is only one simple group of order $60$ without using that fact, then the proof works. – 19021605 Sep 01 '25 at 09:14