Non-Abelian simple group of order $120$

Question

Let's assume that there exists simple non-Abelian group $G$ of order $120$. How can I show that $G$ is isomorphic to some subgroup of $A_6$?

Answer 1

Hint: How many Sylow 5-subgroups will $G$ have? Do you know of a way $G$ acts on them?

Answer 2

A group of order 120 can not be simple. Let's assume that there exists simple non-Abelian group G of order 120. Then we know the number sylow 5-subgroups of $G$ is 6. Hence, the index of $N_{G}(P)$ in $G$ is 6 ($P$ is a sylow 5-subgroup of $G$). Now there exists a monomorphism $\phi$ of $G$ to $S_{6}$. We claim that $Im\phi\leq A_{6}$. Otherwise $Im\phi$ has an odd permutation and so $G$ has a normal subgroup of index 2, a contradiction. Hence, $G\cong Im(\phi)\leq A_{6}$.