Exercise 4.5.56 (If every proper subgroup of $G$ is abelian, then $G$ is solvable)

Proof.

Every group of order $n\le 2$ is abelian and solvable.

Reasoning by induction, suppose that every group of order less that $n$ ( $n\ge 2$), in which every proper subgroup is abelian, is solvable.

Let $G$ be a group of order $n$, in which every proper subgroup is abelian. If $G$ is abelian, then $G$ is solvable. Suppose now that $G$ is not abelian. By Exercise 53, $G$ is not simple, so there is some non trivial proper subgroup $N$ of $G$. Then $N$ is abelian by hypothesis, so $N$ is solvable.

It remains to prove that $G∕N$ is solvable. Let $\bar{H}$ be a proper subgroup of $G∕N$. Put $H={\pi }^{-1}(\bar{H})$, where $\pi :G\to G∕N$ is the natural projection. Since $\pi$ is surjective, $\pi (H)=\bar{H}$. Then $N\le H\le G$. Moreover $H\ne G$, otherwise $\bar{H}=\pi (H)=G∕N$ is not a proper subgroup. So $H$ is a proper subgroup of $G$, and by hypothesis $H$ is abelian. Then $\bar{H}=\pi (H)$ is abelian: if $\mathit{𝑎𝑁},\mathit{𝑏𝑁}\in \bar{H}$, where $a,b\in H$, then $\mathit{𝑎𝑁}\mathit{𝑏𝑁}=\mathit{𝑎𝑏𝑁}=\mathit{𝑏𝑎𝑁}=\mathit{𝑏𝑁}\mathit{𝑎𝑁}$.

This shows that every proper subgroup of $G∕N$ is abelian, where $|G∕N|<|G|$ since $N$ is not trivial. By the induction hypothesis, $G∕N$ is solvable.

Since $N$ and $G∕N$ are solvable, $G$ is solvable (see Proposition p. 105).

The induction is done: if every proper subgroup of $G$ is abelian, then $G$ is solvable. □