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 $G$ 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.
Best Answer
Well, you can obtain a contradiction to the simplicity of a finite group $G$ of order $120$ by showing that a Sylow $2$-subgroup $S$ of $G$ can't be a maximal subgroup of $G,$ for example (I won't give the details, but they require somewhat more background than the theorem you want to avoid). Hence $G$ has a subgroup of index $3$ or $5$, but then you are using the embedding in a symmetric group to obtain a contradiction in any case. Or you can do a complicated fusion and transfer analysis with the prime $2,$ but there is a perfect group of order $120$, so that is not straightforward either (the perfect group of order $120$ has a center of order $2$).