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.

• Just curious: Are you refering to some specific approaches when asking the question? Thanks. – awllower Aug 6 '12 at 3:00
• Well, you could classify all groups of order 120. Why do you want to avoid the theorem? – Kevin Carlson Aug 6 '12 at 4:54
• It does seem to me that this probably is the easiest way to prove that there is no simple group of order $120$. There are other ways to do it, of course, but the ones I can think of are more complicated than that. – Geoff Robinson Aug 6 '12 at 7:02
• You may want to clarify in your question why $A_6$ can have no subgroups of index $3$. – Eric Auld Dec 17 '14 at 2:00
• @EricAuld: this can be done by using the theorem again: $A_6$ is simple, so if it had a subgroup of index 3, then $A_6$ would embed into $A_3$. – Ravi Fernando Oct 23 '16 at 4:47

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 $2$).
• @Geoff May I ask for the reason that a Sylow 2-subgroup $S$ of $G$ can't be a maximal subgroup of $G$? Thanks in advance. – Steve Jacob Mar 18 at 16:41
• @SteveJacob : Suppose that $G$ is simple of order $120$ and a Sylow $2$-subgroup $S$ of $G$ is maximal.By Frobenus's normal p-complement theorem, $S$ has a subgroup $V \neq 1$ such that $N_{G}(V)/C_{G}(V)$ is not a $2$-group. Then $|V| >2$ and clearly $|V| \neq 8$, for otherwise $V = S \not \lhd G$, so $N_{G}(V) = S,$ a contradictio. Similarly, if $V$ has order $4,$ then $V \lhd S$ since maximal subgroups of $2$-groups are normal. Again we obtain $N_{G}(V) = S,$ a contradiction. – Geoff Robinson Mar 18 at 17:03