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There are many famous proofs. However, I cannot find a solution using the following method which is also clear. It’s absolutely correct. Has anyone seen this proof before, and do you have a reference? By the way, why few people seem to prefer this method? As for me, it’s very strong a way to solve many problems.

PS: I have also applied this method to the proofs that groups of order $180$, $540$, $1080$ are not simple.

I will apply the well-known Sylow theorem, strong Cayley theorem, $N/C$ theorem, Burnside's transfer theorem and some basic theories in group-theory.

$G$ is a simple group of order $60$.

$|G|=60=2^2\cdot 3\cdot 5.$ So a Sylow $2$-subgroup would have order $4$.

By Sylow theorem, $n_2\mid 3\cdot 5$, $n_2\equiv 1~(\text{mod}~2)$. There are four possibilities: $1$, $3$, $5$, $15$. And we denote one of the Sylow $2$-subgroups by $P_2$.

$1)$ If $n_2=1$, then $P_2$ will be a normal subgroup of $G$, contradicting the simplicity of $G$.

$2)$ If $n_2=3$, [ strong Cayley theorem ] $G$ simple, $\ker\Phi$ is $G$ itself or the trivial subgroup$\{e\}$, furthermore, $\ker \Phi$ must be $\{e\}$ $($if $\ker \Phi=G$, then $G\leqslant N_G(P_2)$, which is ridiculous $)^{[1]}$. Hence, $G=G/\{e\}=G/\ker\Phi\lesssim S_3$. It contradicts the fact that $|G|=60>6=|S_3|.$

$3)$ If $n_2=15$, $|G:N_G(P_2)|=15$, $|N_G(P_2)|=4$. However, $|P_2|=4$, hence abelian, $P_2\trianglelefteq C_G(P_2)\trianglelefteq N_G(P_2)$, $C_G(P_2)=N_G(P_2)$. By Burnside's transfer theorem, there exists $K\triangleleft G$, which contradicts the simplicity of $G$.

Thus, $n_2$ must be $5$, $G=G/\{e\}=G/\ker\Phi\cong \Phi(G)\leqslant S_5$, where $|\Phi(G)|=|G|=60$. Hence $|S_5:\Phi(G)|=2$, $\Phi(G)$ must be $A_5$. That is, $G\cong A_5$.

Note:

[1] $\Phi$ is a map introduced in strong Cayley theorem ; $G$ is simple, hence $\ker\Phi $, as a normal subgroup, must be $G$ itself or the trivial subgroup$\{e\}$. If $\ker \Phi=G $, then according to strong Cayley theorem, $$G=\ker\Phi=\bigcap\limits_{x\in G} x^{-1}N_G{(P_2)}x\leqslant N_G(P_2).$$ Whereas, $|G:N_G{(P_2)}|>1$, it will be ridiculous.

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  • What is the question? Whether this is correct? – Mathematician 42 Feb 15 '18 at 08:32
  • Then the question bears repeating: What is your question? – Alec Feb 15 '18 at 08:38
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    @Benny Not really! This forum is for asking questions, not for general discussion. You could ask "is my proof correct?",or I guess you could just ask "has anyone seen this proof before, and do you have a reference?". – Derek Holt Feb 15 '18 at 08:46
  • @Benny Good. Then do please write that explicitly in your question... – DonAntonio Feb 15 '18 at 09:32
  • MSE has many helful answers for this question, see for example here for a stronger Sylow variant. – Dietrich Burde Feb 15 '18 at 09:35
  • In your (2) you talk of $;\Phi;$ ...what is that? Why its kernel is trivial? Why to write after "trivial" that "furthermore it must be $;\ker\Phi={e};$ ? This is precisely what being trivial means within this context, isn't it? – DonAntonio Feb 15 '18 at 09:52
  • @DonAntonio : $\Phi$ is a map introduced in strong Cayley theorem ; $G$ is simple, hence $\ker\Phi $, as a normal subgroup, must be trivial. If $G=\ker \Phi $, then according to strong Cayley theorem, $G=\ker\Phi=\cap_{x\in G} x^{-1}N_G{(P_2)}x\leqslant N_G(P_2)$. Whereas, $|G:N_G{(P_2)}|=3$, it will be ridiculous. –  Feb 15 '18 at 10:12
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    @Benny Great. Why won't you write down that in your question? "Strong Cayley theorem" is something I never heard. That theorem is a usual, and rather easy, consequence of other facts, though I can understand why the Cayley name there... – DonAntonio Feb 15 '18 at 10:23

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It seems to me that there is a significant overlap with the standard proofs, e.g., see here. "Proof that there is only one simple group of order sixty, isomorphic to the alternating group of degree five": The key idea is to prove that $G$ has a subgroup of index five. After that, we use the fact that $A_5$ is simple to complete the proof. Again, Burnside is used. For another idea to use a stronger version of Sylow see here:

About the proof that a simple group of order 60 is isomorphic to A5

So to your question Has anyone seen this proof before? Yes, up to minor variations, which are always possible. In general, this theorem has been treated so often, that almost everything has been said. I am not sure that it is interesting to come up with yet another variation. On the other hand, it is certainly useful to write this up for oneself, and for further topics in group theory.

Dietrich Burde
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