# Prove that a group G is not soluble if it has a non-trivial perfect subgroup.

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.

• What does "the subnormal series of $H$ with abelian factors" mean? You are trying to prove that there is no such series, so the wording is bad. – Derek Holt Jan 13 '15 at 12:41

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.
• $[H,H]$ is the commutator subgroup of $H$. Your proof is ok, but you use two theorems which you don't need to use in my answer..In other words your killing a fly with a canon. What is your definitin for soluble? – Ofir Schnabel Jan 13 '15 at 11:51
Your proof is almost correct. Since every subgroup of $G$ is solvable, your perfect subgroup $H$ has to be solvable. However, there exists no non-trivial perfect solvable group, because a group $H$ is solvable if and only if the derived series ends with the trivial group. This can never happen for a non-trivial perfect group $H$, because all derived groups $H^{(i+1)}=[H^{(i)},H^{(i)}]$ are equal to $H$ by definition.
P.S. The notation is $H'=H^{(1)}=[H,H]$ for the commutator subgroup (or derived subgroup), and $H^{(2)}=[[H,H],[H,H]]$ etc.