Solvable Group

Definition

For any $a,b\in G$, the commutator of $a$ and $b$ is defined by $[a,b]=a^{-1}{b}^{-1}ab$. The group generated by all commutators of $G$ is called the derived group and denoted as $G^{\prime }=[G,G]$.

Definition

The following construction can be iterated:

• $G^{(0)}:=G$, and
• $G^{(n)}:=[G^{(n-1)},G^{(n-1)}]$ for $n\in {\mathbb{N}}_{+}$.

The group $G^{(2)},G^{(3)}\dots$ are called the second derived subgroup, third derived subgroup, and so forth, and the descending normal series

$$\cdots ⊲G^{(2)}⊲G^{(1)}⊲G^{(0)}=G$$

is called the derived series.

A group is called a perfect group, if its derived group equals to itself. For a finite group $G$, the derived series terminates in a perfect group, denoted as $G^{(\infty )}$. If $G^{(\infty )}=\{1\}$, we say $G$ is solvable, otherwise we say $G$ is nonsolvable.

Proposition

Let $G$ be a finite group and let $N⊲G$ be a normal subgroup. Then $G/H$ is abelian if and only if $G^{\prime }⩽H$.

\begin{proof} Define the canonical map as $\pi :G\to G/H,g\mapsto \overline{g}$. If $G/H$ is abelian, then $\pi (ab{a}^{-1}{b}^{-1})=\overline{a}\overline{b}{\overline{a}}^{-1}{\overline{b}}^{-1}=\overline{e}$ and so $ab{a}^{-1}{b}^{-1}\in \ker \pi =H$. It yields that $G^{\prime }⩽H$. Conversely, if $ab{a}^{-1}{b}^{-1}\in H$ for any $a,b\in G$, then $\overline{ab{a}^{-1}{b}^{-1}}=\overline{e}$ and $\overline{a}\overline{b}=\overline{b}\overline{a}$. Therefore, we have $G/H$ is abelian. \end{proof}

Proposition

A group is solvable if and only if all its composition factors are abelian.

\begin{proof} Let $G$ be a group and let $1=H_0⊲\cdots ⊲H_n=G$ be a composition series of $G$. We prove the proposition by induction. Assume that the statement is true when the order of group is less than $|G|$. Since $G/H_{n-1}\cong {\mathbb{Z}}_p$ for some prime $p$, there is $H_{n-1}⩾G^{\prime }$ by ^n7ko5p. By induction hypothesis $H_{n-1}$ is solvable and so for $G^{\prime }$. Therefore, the group $G$ is a solvable group. \end{proof}

Proposition

Let $N⊲G$ be a normal subgroup of $G$. Then $(G/N{)}^{\prime }=G^{\prime }N/N$.

\begin{proof} See here. \end{proof}