I want to show that the group G, where $|G|=60$ and has 20 elements of order 3 is simple.
Here's what I did:
Suppose that G is not simple this implies that $n_3>1, n_2>1, n_5>1$ where these n's denote the number of sylow p-subgroups in G.
Consider $n_5=(1+5k)|12 \Rightarrow n_5=6$ (as we've said it cant be one)
But then $|G:N_G(P)|=6$ ( P denoting the sylow 5 subgroups here).
This implies there is a homomorphism induced by G acting on the left cosets of G/P by left multiplication. $|G/P|>1$ and operation is transitive so the kernel is not the whole group. since g is not simple the map is not injective and also since the groups are finite not surjective. which implies $|G|>|S_n|$ a contradiction .....so G is simple.
Is this correct ?
If not what specifically is wrong with it ?