Exercise 3.4.5 (Subgroups and quotients groups of a solvable group are solvable)

Proof. Let $G$ be a solvable group. By definition there is a chain of subgroups

such that $G_{i+1}∕G_i$ is abelian for $i\in [[0,s[[$.

(a)

Let $H$ be any subgroup of $G$. Set $H_i=G_i\cap H$, and note that $$\begin{array}{llll}\hfill H_0& =G_0\cap H=\{1\},& \hfill & \\ \hfill H_s& =G_s\cap H=H,& \hfill & \end{array}$$

and $H_i\le H_{i+1}$ for all $i\in [[0,s[[$.

Consider for every $i$ the restriction ${\lambda }_i$ of the natural projection ${\pi }_i:G_{i+1}\to G_{i+1}∕G_i$ to $H_{i+1}\le G_{i+1}$:

$${\lambda }_i\{\begin{array}{ccc}\hfill H_{i+1}\hfill & \hfill \to \hfill & G_{i+1}∕G_i\hfill \\ \hfill h\hfill & \hfill \mapsto \hfill & h{G}_i.\hfill \end{array}$$

Then ${\lambda }_i$ is an homomorphism. Moreover for all $h\in G_{i+1}$,

$$\begin{array}{llllll}\hfill h\in \ker <!--\; FUNCTION\; APPLICATION\; -->({\lambda }_i)& ⟺h\in H_{i+1}\text{ and }h{G}_i=G_i& \hfill & & \hfill & \\ \hfill & ⟺h\in H_{i+1}\text{ and }h\in G_i& \hfill & & \hfill & \\ \hfill & ⟺h\in H_{i+1}\cap G_i=G_{i+1}\cap H\cap G_i=H\cap G_i& \hfill (\text{since }{G}_i\subseteq G_{i+1})& & \hfill & & \hfill \\ \hfill & ⟺h\in H_i,& \hfill & & \hfill & \end{array}$$

so

$$\ker <!--\; FUNCTION\; APPLICATION\; -->({\lambda }_i)=H_i.$$

In particular $H_i⊴H_{i+1}$.

The First Isomorphism Theorem shows that

$$H_{i+1}∕H_i\simeq {\lambda }_i(H_{i+1})=\mathrm{im}({\lambda }_i)\le G_{i+1}∕G_i.$$

Since the subgroup $\mathrm{im}({\lambda }_i)$ of the abelian group $G_{i+1}∕G_i$ is abelian, $H_{i+1}∕H_i$ is isomorphic to an abelian group, so is itself abelian.

In conclusion, the chain

$$\{1\}=H_0⊴H_1⊴H_2⊴\cdots ⊴H_s=H$$

is such that $H_{i+1}∕H_i$ is abelian for $i\in [[0,s[[$, so $H$ is solvable.

(b)

Suppose now that $H⊴G$. We want to prove that $G∕H$ is solvable.

Consider the natural projection $\pi :G\mapsto G∕H$. We know that $\pi$ is a surjective homomorphism. In particular, every element $\bar{x}\in G∕H$ is of the form $\bar{x}=\pi (x)=\mathit{𝑥𝐻}$ for some $x\in G$.

Put $K_i=\pi (G_i)$ for $i\in [[0,s[[$ (*). Then $K_i⊴K_{i+1}$. Indeed, if $\bar{x}=\pi (x)\in K_{i+1}$ and $\bar{y}=\pi (y)\in K_i$, then for all $i\in [[0,s[[$,

$$\bar{x}\bar{y}{\bar{x}}^{-1}=\pi (x)\pi (y)\pi {(x)}^{-1}=\pi (\mathit{𝑥𝑦}{x}^{-1})\in \pi (G_i)=K_i,$$

because $G_i⊴G_{i+1}$, so $\mathit{𝑥𝑦}{x}^{-1}\in G_i$.

Note that $K_1=\pi (G_1)=\pi (\{1\})=\{\bar{1}\}$ and $K_s=\pi (G_s)=\pi (G)=G∕H$. So we obtain the chain

$$\{\bar{1}\}=K_1⊴K_2⊴\cdots ⊴K_{s-1}⊴K_s=G∕H.$$

It remains to show that $K_{i+1}∕K_i$ is abelian ( $i\in [[0,s[[$).

For this purpose, consider the map

$$\phi \{\begin{array}{ccc}\hfill G_{i+1}\hfill & \hfill \to \hfill & K_{i+1}∕K_i\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & \bar{x}{K}_i.\hfill \end{array}$$

Then $\phi ={\pi }_i\circ \tilde{\pi }$, where $\tilde{\pi }:G_{i+1}\to \pi (G_{i+1})=K_{i+1}$ is the restriction of the natural projection $\pi :G\to G∕H$ to $G_{i+1}$ and $K_{i+1}=\pi (G_{i+1})$, and ${\pi }_i$ is the natural projection ${\pi }_i:K_{i+1}\to K_{i+1}∕K_i$.

Then $\phi$ is a surjective homomorphism, because ${\pi }_i$ and $\tilde{\pi }$ are both surjective homomorphisms.

Moreover $G_i\subseteq \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$: If $x\in G_i$, then $\bar{x}\in K_i$, thus $\phi (x)=\bar{x}{K}_i=K_i$, so $x\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$.

Hence there exists a surjective homomorphism $\bar{\phi }:G_{i+1}∕G_i\to K_{i+1}∕K_i$ such that $\bar{\phi }\circ {\pi }_i=\phi$ (cf. p. 100).

Since $G_{i+1}∕G_i$ is abelian, $K_{i+1}∕K_i=\bar{\phi }(G_{i+1}∕G_i)$ is also abelian for every $i\in [[0,s[[$, therefore $G∕H$ is solvable. □