If $|G| = 1920$ and $|H| = 80$ are groups, $f:G \rightarrow H$ epimorphism then $G$ is solvable.

Question

I have problems proving the following proposition

Let $G,H$ be groups with $|G| = 1920$ and $|H| = 80$. If $f:G \rightarrow H$ is an epimorphism then $G$ is a solvable group.

I've done a lot of research and learnt a couple of things I didn't know but I have no idea on how to solve it. Any ideas?

Answer 1

First: since $\;80=2^4\cdot5\;$ , any group of this order has either a unique Sylow $\;5\,-$ subgroup or a unique Sylow $\;2\,-$ subgroup (why? Count elements), so either way it has a normal Sylow subgroup, and either it has a quotient of order $\;2^4\;$ , which is a $\;2\,-$ group and this solvable, or a quotient of order $\;5\;$ which is abelian and trivially solvable, so the whole group will be solvable.

Also, the kernel of $\;\phi:G\to H\;$ has order $\;24\;$, and also this group is solvable (can you prove it? No need of Burnside's Theorem), so again $\;G\;$ is a solvable extension by a solvable group and we're done.

Answer 2

$G$ is an extension of $H$ by a normal subgroup of $G, N,$ that has order 24. Both $|H|$ and $|N|$ have two distinct prime factors, so both groups are solvable, and an extension of a solvable group by a solvable group is solvable.