Let \(G\) be a group with order 120. We show that \(G\) is not simple.
Note that \(|G| = 2^3\cdot3\cdot5\).
Let us assume that \(G\) is simple and arrive at a contradiction.
By Sylow Theorem (3), we see that \(n_5 \mid 24\) and \(n_5 \equiv 1 \mod 5\). Looking at the divisors of \(24\), we see that \(n_5 = 6.\) (\(n_5 \neq 1\) by our assumption of simplicity.)
Let \(G\) act on Syl\(_5(g)\) by conjugation.
This gives a natural homomorphism \(m : G \to S_6.\)
By transitivity of the action, we know that ker \(m \neq G.\) However, \(G\) is simple by assumption and kernel is a normal subgroup. Thus, ker \(m = (1).\)
This gives us that \(G\) is isomorphic to a subgroup \(H\) of \(S_6.\) (To be precise, \(H = m(G).\))
In fact, more is true. As \(\vert H\vert > 2\) and \(H\) is simple, we also have that \(H \le A_6.\) (See the proof of 112 for the similar result and its proof.)
Now, consider the set of cosets \(A_6/H.\) (This will not be a group but that is okay.)
Note that \(|H| = 120\) and \(A_6 = 6!/2 = 360\) and hence, \(|A_6/H| = 3.\)
Let \(A_6\) act on \(A_6/H\) by left multiplication. Once again, this is a transitive action and thus the kernel of the corresponding homomorphism \(m' : A_6 \to S_3\) is a proper subgroup of \(A_6.\)
As \(A_6\) is simple, it forces that ker \(m' = (1).\) However, this is not possible as \(|A_6| = 120 < 6 = |S_3|.\) Thus, we have arrived a contradiction and we are done!
Note: We have used the fact that \(A_n\) is simple for \(n \ge 5.\)
The last contradiction was using the fact that ker \(m' = (1)\) would imply that \(m'\) is injective.