Exercise 8.1.2

Proof. Let $G$ solvable, and $H$ a normal subgroup of $G$. We must prove that $G∕H$ is solvable.

There exist subgroups $G_i$ of $G$ such that

where $G_i◃G_{i-1}$, and $G_{i-1}∕G_i$ is of prime order.

(a)

Let $\pi :G\to G∕H,g\mapsto \mathit{gH}$ the canonical projection, and ${\tilde{G}}_i=\pi (G_i)$.

$\text{∙}$ As $G_0=G$, ${\tilde{G}}_0=\pi (G)=G∕H$ since $\pi$ is surjective.

$\text{∙}$ As $G_n=\{e\},{\tilde{G}}_n=\pi (\{e\})=\{\mathit{eH}\}=\{\bar{e}\}$, where we write $\bar{e}$ the identity of $G∕H$.

$\text{∙}$ We know that $G_i◃G_{i-1}$. Then we show that ${\tilde{G}}_i◃{\tilde{G}}_{i-1}$.

Let any $\bar{x}\in {\tilde{G}}_{i-1}$, and $\bar{y}\in {\tilde{G}}_i$, with $\bar{x}=\mathit{xH},\bar{y}=\mathit{yH},x\in G_{i-1},y\in G_i$.

$\bar{x}\bar{y}{\bar{x}}^{-1}=\mathit{xHyH}{x}^{-1}H=\mathit{xy}{x}^{-1}H=\mathit{zH}$, where $z=\mathit{xy}{x}^{-1}\in G_i$, since $G_i◃G_{i-1}$.

Therefore $\bar{x}\bar{y}{\bar{x}}^{-1}=\pi (z)\in \pi (G_i)={\tilde{G}}_i$. So ${\tilde{G}}_i◃{\tilde{G}}_{i-1}$.

$\text{∙}$ Let $\phi$ be the mapping

$$\phi :\{\begin{array}{ccc}\hfill G_{i-1}\hfill & \hfill \to \hfill & \hfill {\tilde{G}}_{i-1}∕{\tilde{G}}_i\hfill \\ \hfill g\hfill & \hfill \mapsto \hfill & \hfill \pi (g){\tilde{G}}_i\hfill \end{array}$$

(if $g\in G_{i-1},\pi (g)\in {\tilde{G}}_{i-1}$, thus $\pi (g)\widetilde{G_i}\in {\tilde{G}}_{i-1}∕{\tilde{G}}_i$).

If $g\in G_i,\pi (g)=\mathit{gH}\in {\tilde{G}}_i$, thus $\phi (g)=\pi (g){\tilde{G}}_i=\widetilde{G_i}$, which is the identity of ${\tilde{G}}_{i-1}∕{\tilde{G}}_i$, so $g\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$. This proves

$$G_i\subset \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi ).$$

As $G_i\subset \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, if two elements $g,g^{\prime }$ of $G_{i-1}$ are congruent modulo $G_i$, i.e. $g{G}_i=g^{\prime }{G}_i$, then $g^{-1}{g}^{\prime }\in G_i\subset \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, so $g,g^{\prime }$ have the same image by $\phi$. Consequently $\phi (g)$ depends only of the class $g{G}_i$ of $g$ in $G_{i-1}∕G_i$.

The mapping $\bar{\phi }:G_{i-1}∕G_i\to {\tilde{G}}_{i-1}∕{\tilde{G}}_i$ defined by $g{G}_i\mapsto \pi (g){\tilde{G}}_i$ is thus well defined, and is a group homomorphism.

Let $y=v{\tilde{G}}_i$ be any element of ${\tilde{G}}_{i-1}∕{\tilde{G}}_i$, where $v\in {\tilde{G}}_{i-1}$, so $v=\pi (g)$ for some $g\in G_{i-1}$. Therefore $y=\pi (g){\tilde{G}}_i=\phi (g)=\bar{\phi }(g{G}_i)$, so $\bar{\phi }$ is surjective.

(b)

Let $\varphi :M_1\to M_2$ be a surjective group homomorphism, and suppose that $|M_1|=p$ is prime.

Then $M_2\simeq M_1∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\varphi )$. The order of the subgroup $\ker <!--\; FUNCTION\; APPLICATION\; -->(\varphi )$ of $M_1$ divides $|M_1|=p$, so $|\ker <!--\; FUNCTION\; APPLICATION\; -->(\varphi )|=1$ or $p$, thus $|M_2|=|M_1|∕|\ker <!--\; FUNCTION\; APPLICATION\; -->(\varphi )|=p$ or $1$.

(c)

By hypothesis $|G_{i-1}∕G_i|=p$ is prime. By part (b) applied to the surjective group homomorphism $\varphi =\phi :G_{i-1}∕G_i\to {\tilde{G}}_{i-1}∕{\tilde{G}}_i$, we know that $|{\tilde{G}}_{i-1}∕{\tilde{G}}_i|=1$ or $p$.

Therefore ${\tilde{G}}_{i-1}∕{\tilde{G}}_i$ is trivial, or of prime order.

To conclude:

$$\{\bar{e}\}={\tilde{G}}_n\subset \cdots \subset {\tilde{G}}_i\subset {\tilde{G}}_{i-1}\subset \cdots \subset {\tilde{G}}_0=G∕H,$$

with ${\tilde{G}}_i◃{\tilde{G}}_{i-1},{\tilde{G}}_i∕{\tilde{G}}_{i-1}$ trivial or of prime order. By discarding duplicates, we obtain a composition series which proves that $G∕H$ is solvable.