My attempt:
Proceed by contradiction, assume that G is soluble. Then every subgroup and quotient group of G is soluble, so in particular, the non-trivial perfect subgroup, which we call H, is soluble.
But since H is perfect, the smallest normal subgroup (the derived subgroup H') of H where the quotient group H/H' is abelian, is H, so the subnormal series of H with abelian factors won't terminate, since we will have that H contains normal subgroup H' contains normal subgroup H'….. where none of the H'=H is = {e} since H is non-trivial.
Is this argument ok? I'm unsure if using the fact that the subnormal series doesn't terminate is strong enough for a contradiction.
Best Answer
If $H$ is a subgroup of $G$, then $[H,H]$ is a subgroup of $[G,G]$. Hence, for a perfect subgroup $H$ of $G$, $[H,H]=H$ is a subgroup of $[G^{(i)},G^{(i)}]$ and therefore $G$ is not soluble.