Every normal subgroup of a solvable group must have an abelian subgroup which is normal in G.

Question

Let G be a solvable group, and let H be a nontrivial normal subgroup of G. Prove that there exists a nontrivial subgroup A of H that is Abelian and normal in G. Actually this problem is from DF. It was answered here by Derived series. But before this problem DF did not introduce Derived series. So It must have another way to prove. Can anyone prove?

Answer 1

Since derived series cannot be used, I assume that an equivalent definition of a solvable subgroup is used: a group $G$ is called solvable if it has a normal series $1 = G_0 < G_1 \dots < G_n = G$ (that is, all subgroups $G_i$ are normal in $G$) such that all factor groups $G_i/G_{i-1}$ are abelian, $i = 1, 2,\dots, n$. Let now $H$ be a nontrivial normal subgroup in $G$. Let $i$ be the maximal possible integer such that $H \cap G_i = 1$. Notice that $i < n$ since $H$ is non-trivial. Then $A= H \cap G_{i+1}$ is a nontrivial subgroup, and, being the intersection of two normal subgroups, it is normal in $G$. Also, the image of $A$ in the abelian factor group $G_{i+1}/G_i$ is isomorphic to $A/(A\cap G_i) \simeq A/1 \simeq A$. Hence $A$ is isomorphic to a subgroup of an abelian group $G_{i+1}/G_i$ and is therefore abelian.

My answer would perhaps be more precise and useful if I knew what specific definition (of several equivalent definitions used) of solvable groups is used in your question.