A group of order $120$ cannot be simple 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.
 A: 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 $2$).
A: Here is a good one.
Of course $n_5=6, n_3=10 \text{ or } 40.$ Assuming simplicity of G, we can embed G into $S_6$
$n_3$ cannot be 40 because otherwise it contains all the 3-cycles and hence is larger than $A_6$, contradiction.
Hence $n_3 = 10$
Goal: Find an element in G of order 6. Such element must be an odd permutation by studying its cycle structure. Hence we are done.
Observe that $N_G(P_3)$ have order 12, $P_3$ is Sylow-3 group.
Of course $N_G(P_3)/C_G(P_3)$ embed into $Aut(P_3)$.
Hence $C_G(P_3)$ has even order, and hence has an element of order 2. Such element multiplies and element in $P_3$ gives our desired order 6 element. We are done.
