Exercise 3.4.11 (Abelian subgroup of a solvable group)

Proof. We proceed as in Exercise 3.4.8 (iv).

Let $H$ be a nontrivial normal subgroup of the solvable group $G$. Consider the set $𝒮$ of all non trivial subgroups $L$ of $H$ with $L⊴G$:

Then $𝒮\ne \varnothing$, since $H\in 𝒮$, and $𝒮$ is finite, so $𝒮$ has a minimal element $A$, satisfying

We will show that $A$ is abelian.

Since $A$ is a subgroup of the solvable group $G$, $A$ is also solvable (see Exercise 5), so its composition factors are cyclic of prime order $p$ (Exercise 8). Therefore the penultimate element $N$ of a composition series of $A$ satisfies $N⊴A$ and $A∕N$ is cyclic of prime order $p$. A fortiori $A∕N$ is abelian, so

for all $x,y\in A$.

This gives $\mathit{𝑥𝑦𝑁}=\mathit{𝑦𝑥𝑁}$ and $x^{-1}{y}^{-1}\mathit{𝑥𝑦𝑁}=N$, so

Let $g$ be any element of $G$. Then ${\gamma }_g:G\to G$ defined by ${\gamma }_g(x)=\mathit{𝑔𝑥}{g}^{-1}$ is an automorphism of $G$. Since $N⊴A$, then ${\gamma }_g(N)⊴{\gamma }_g(A)=A$, so

Moreover $|\mathit{𝑔𝑁}{g}^{-1}|=|N|$, thus $|A∕\mathit{𝑔𝑁}{g}^{-1}|=|A∕N|=p$, so $\mathit{𝑔𝑁}{g}^{-1}$ is a normal subgroup of $A$ of index $p$ in $A$. The same argument as above shows that $x^{-1}{y}^{-1}\mathit{𝑥𝑦}\in \mathit{𝑔𝑁}{g}^{-1}$:

Therefore if $x,y\in A$,

Put $K=\underset{g\in G}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}\mathit{𝑔𝑁}{g}^{-1}$. Then $K⊴G$, and $K\subseteq N⊊A$. The minimality of $A$ among the nontrivial normal subgroups of $G$ shows that $K=\{1\}$, so

Hence $x^{-1}{y}^{-1}\mathit{𝑥𝑦}=1$, so $\mathit{𝑥𝑦}=\mathit{𝑦𝑥}$ for all elements $x,y\in A$: $A$ is abelian.

If $H$ is a nontrivial normal subgroup of the solvable group $G$ then there is a nontrivial subgroup $A$ of $H$ with $A⊴G$ and $A$ abelian. □