Exercise 3.4.6 (First part of Jordan-Hölder Theorem)

Proof. Let $G\ne \{1\}$ be a finite group.

(a)

Let $H\ne G$ be a normal subgroup of $G$ of maximal order (since $G\ne \{1\}$ is finite and $\{1\}⊴G$, such a subgroup exists). We show that $G∕H$ is a simple group.

If $G∕H$ were not simple, $G∕H$ would have a normal subgroup $\bar{K}$,

$$\{\bar{1}\}⊊\bar{K}⊊G∕H,$$

where $\bar{1}=H$ is the identity of $G∕H$. Let $\pi =G\to G∕H$, defined by $g\mapsto \mathit{𝑔𝐻}$, be the canonical projection, and let $K={\pi }^{-1}(\bar{K})$.

Since $\{\bar{1}\}⊊\bar{K}⊊G∕H$, then ${\pi }^{-1}(\{\bar{1}\})⊊{\pi }^{-1}(\bar{K})⊊{\pi }^{-1}(G∕H)$ (because $\pi$ is surjective, so $\pi ({\pi }^{-1}(K))=K$), therefore

$$H⊊K⊊G.$$

We show that $K⊴G$. Let $y\in G,x\in K$. Then $\bar{y}=\mathit{𝑦𝐻}\in G∕H$ and $\bar{x}=\mathit{𝑥𝐻}\in \bar{K}$, where $\bar{K}⊴G∕H$, hence $\pi (\mathit{𝑦𝑥}{y}^{-1})=\bar{y}\bar{x}{\bar{y}}^{-1}\in \bar{K}$, so $\mathit{𝑦𝑥}{y}^{-1}\in K={\pi }^{-1}(\bar{K})$.

$H⊊K⊊G$ and $K⊴G$ is in contradiction with the definition of $H$ as normal subgroup of $G$ of maximal order, so such a subgroup $\bar{K}$ of $G∕H$ doesn’t exist.

$G∕H$ is a simple group.

(b)

If $|G|=2$, then $G\simeq Z_2$ is simple, and has a composition series $\{1\}⊴G$.

Reasoning by complete induction, we suppose that every non trivial group of order less that $n$ has a composition series, and let $G$ be a group of order $n>2$.

By part (a), there exists a normal subgroup $H$ of maximal order such that

$$\{1\}\subseteq H⊊G,H⊴G,$$

where $G∕H$ is simple.

If $H=\{1\}$, then $G$ is simple and $\{1\}⊴G$ is a composition series.

Otherwise, since $1<|H|<n$, the induction hypothesis gives a composition series for $H$:

$$\{1\}=G_1⊴G_2⊴\cdots ⊴G_l=H,$$

where $G_{i+1}∕G_i$ is simple ( $2\le i<l$).

Then

$$\{1\}=G_1⊴G_2⊴\cdots ⊴G_l=H⊴G_{l+1}=G$$

is a composition series for $G$, since $G∕H$ is simple by part (a). So the induction is done.

Every non trivial finite group has a composition series.