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

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.

• What is the question? Whether this is correct? – Mathematician 42 Feb 15 '18 at 8:32
• Sorry, I forgot that point. It's absolutely correct. – Math Feb 15 '18 at 8:37
• Then the question bears repeating: What is your question? – Alec Feb 15 '18 at 8:38
• @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 8:46
• @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

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: