Hi,

I am stuck proving that a simple group $\displaystyle G$ of order $\displaystyle 60$ is isomorphic to $\displaystyle A_5$.

In particular:

- I have shown $\displaystyle |Syl_5(G)|=6$ and $\displaystyle |Syl_3(G)|=10$, so there must be $\displaystyle 6\cdot(5-1)=24$ elements of order $\displaystyle 5$ and $\displaystyle 10\cdot(3-1)=20$ elements of order $\displaystyle 3$.

- I could further show $\displaystyle |Syl_2(G)|\notin\{1,3\}$.

- Now, where I get stuck: I need to show $\displaystyle |Syl_2(G)|\neq15$ and therefore $\displaystyle |Syl_2(G)|=5$.

I know I have to use the fact that there are already $\displaystyle 20+24=44$ elements of order coprime to $\displaystyle 2$. That leaves only $\displaystyle 60-44-1=15$ elements to have order 2 or 4, which somehow has to lead to a contradiction to $\displaystyle |Syl_2(G)|=15$. But the 2-Sylow-groups have order $\displaystyle 2^2$ and can thus intersect non-trivially. So how can I argue from here?

Any help would be much appreciated!