TITLE: About the proof that a simple group of order 60 is isomorphic to A5
QUESTION [7 upvotes]: I am stuck proving that a simple group $G$ of order $60$ is isomorphic to $A_5$.
In particular:

I have shown $|Syl_5(G)|=6$ and $|Syl_3(G)|=10$, so there must be $6\cdot(5-1)=24$ elements of order $5$ and $10\cdot(3-1)=20$ elements of order $3$.
I could further show $|Syl_2(G)|\notin\{1,3\}$.
Now, where I get stuck: I need to show $|Syl_2(G)|\neq15$ and therefore $|Syl_2(G)|=5$.
I know I have to use the fact that there are already $20+24=44$ elements of order coprime to $2$. That leaves only $60-44-1=15$ elements to have order 2 or 4, which somehow has to lead to a contradiction to $|Syl_2(G)|=15$. But the 2-Sylow-groups have order $2^2$ and can thus intersect non-trivially. So how can I argue from here?

This can probably be done in a very elementary way, but I just don't see it... so thanks for any help!

REPLY [2 votes]: I'll give another proof based on counting which I found easy to follow:
It is $60=2^2\cdot3\cdot5$. Let $H\leq G$ with $[G:H]=n\not=1$. Then since $G$ is simple we have $\rho:G\hookrightarrow S_n\Rightarrow |G| | n!\Rightarrow 2^2\cdot3\cdot5 |n!\Rightarrow n\ge5 $.

If $n=5$ then $[S_5:\rho(G)]=2 \Rightarrow \rho(G)\lhd S_5\Rightarrow \rho(G)=A_5 \Rightarrow G\cong A_5$
We suppose now that every subgroup of $G$ has index $>5$ so $n_p=[G:N_G(P)]>2$ for $p=2,3,5$. We have that:

$n_5|2^2\cdot3$ and $n_5\equiv 1\pmod{5}\Rightarrow n_5=6$
$n_2|3\cdot 5$ and $n_2\equiv 1\pmod{2}\Rightarrow n_2=15$
Let $P,Q$ be two distinct Sylow $2$-subgroups of $G$ with $P\cap Q\not=1$ and let $N=N_G(P\cap Q)$. Since $P,Q$ are abelian subgroups we have $P,Q\subseteq N$. It is also $P,Q\subsetneq N$ since $P\not=Q$. So we have that
$4||N|,4<|N|,|N|| 2^2\cdot 3\cdot 5\Rightarrow |N|=2^2\cdot 3$ or $2^2\cdot 5$ or $2^2\cdot 3\cdot 5$

If $|N|=2^2\cdot 3\cdot 5=|G|$ then $1\not= P\cap Q\lhd N=G$ contradiction since $G$ is simple
If $|N|=2^2\cdot 5$ then $[G:N]=3$ contradiction
If $|N|=2^2\cdot 3$ then $[G:N]=5$ contradiction

So it follows that $P\cap Q=1$ for every pair of Sylow $2$- subgroups of $G$. Hence $|G|>n_5(5-1)+n_2(4-1)=69$ contradiction.
Hence, only possible case $G\cong A_5$
EDIT: As David A. Craven pointed, in the case where we have two distinct Sylow $2-$subgroups $P,Q$ with $P\cap Q\not=1$ then $2||P\cap Q|\Rightarrow \exists x\in P\cap Q$ with $o(x)=2$. Then $P,Q \subset C_G(x)\Rightarrow |C_G(x)|>4\Rightarrow  12\leq |C_G(x)|\Rightarrow [G:C_G(x)]\leq 5\checkmark$