Prove that any simple group $G$ of order $60$ is isomorphic to $A_5$.
I am studying abstract algebra from Artin and studying these assignment from a masters level math course.
It gives hint.
Use the result :
if $H < G$, then $[ G : H]\geq 5$; if $[G : H] = 5$ then $G$ is isomorphic to $A_5$. The assumption that $G$ has no subgroup of index $5$ leads to contradiction.
I am sorry to say but I am not able to follow the hint. Why should $[G:H] \geq 5$ and why $[G:H] = 5$ then $G$ must be isomorphic to $A_5$? and why assumption $G$ has no subgroup of index $5$ leads to contradiction?
Can you please shed some light on this? Any other method is also welcome.