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Any simple group of order 60 is isomorphic to $A_5$

Let $G$ be a simple group of order $60$

.(Assumption) $G$ has a subgroup of order 12 say $H$ .

Then by Extended Cayley's Theorem $\exists f:G\rightarrow S_5$ such that $ker f\subset H$. Since $G$ is simple $ker f=\{e\}$ ;hence $G$ is isomorphic to a subgroup $T$ of $S_5$

I think if I can show that $T=A_5$ I am done .But how to show that?any help

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  • $\begingroup$ What is the Extended Cayley's Theorem? I haven't heard of it and it doesn't show up on Google. $\endgroup$ – Qudit Mar 12 '15 at 19:22
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Since index of $G$ is 2 AND G is contained in $A_{5}$ (by G is simple) , $G$ is $A_{5}$.

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  • $\begingroup$ why is G in $A_5$ $\endgroup$ – Learnmore Mar 13 '15 at 3:07
  • $\begingroup$ Because $G$ is simple. Think about it! $\endgroup$ – Derek Holt Mar 13 '15 at 9:23
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Use the fact that $A_n$ is the only normal subgroup in $S_n$ of index 2: For $n = 2$ this is obvious, thus consider $n \geq 3$. Let $N \subseteq S_n$ be a normal subgroup of index 2. Then for any 3-cycle $\sigma \in S_n$ we have $\sigma^2N = N$ and thus $\sigma^2 \in N$. Since ord($\sigma$) = 3 we have $<\sigma^2> = <\sigma>$ and it follows that $\sigma \in N$. Since $A_n$ is generated by all 3-cycles we have $A_n \subseteq N$ and for cardinality reasons $A_n = N$.

Now you have already shown that $G$ is a subgroup of $S_5$ of index 2. This implies that $G$ is also normal in $S_5$ (Since for any $\sigma \in S_5 \setminus G$ we have $S_5 = G \overset{\cdot}{\cup} \sigma G$ and thus $\sigma G \sigma^{-1} \subseteq G$). Hence we obtain $G \cong A_5$.

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