We know that:
Theorem: If a simple group $G$ has a proper subgroup $H$ such that $[G:H]=n$ then $G\hookrightarrow A_n$.
This fact can help us to prove that any group of $G$ of order $120$ is not simple. In fact, since $n_5(G)=6$ then $[G:N_G(P)]=6$ where $P\in Syl_5(G)$ and so $A_6$ has a subgroup of order $120$ which is impossible. My question is:
Can we prove that $G$ of order $120$ is not simple without employing the theorem? Thanks.