Solvable Groups

# Solvable Groups

 Definition: Let $G$ be a group and let $1$ denote the identity in $G$. Then $G$ is said to be Solvable Group if there is a subnormal series $\{ 1 \} = G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_k = G$ such that every factor $G_{i+1}/G_i$ for $i \in \{ 0, 1, ..., k-1 \}$ is an abelian group.
 Theorem 1: Let $G$ be a finite group. Then $G$ is a solvable group if and only if the factors of any composition series of $G$ are cyclic and of prime order.

Recall from The Jordan-Hölder Theorem page that if $G$ is a finite group then $G$ has a composition series, and that the Jordan-Hölder theorem states that any two composition series in a group are equivalent. Hence, if $G$ is a finite group, the factors of any two composition series are necessarily the same.

• Proof: $\Rightarrow$ Suppose that $G$ is a solvable group. Then there exists a subnormal series:
(1)
\begin{align} \quad \{ 1 \} = G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_k = G \end{align}
• where the factors $G_{i+1}/G_i$ are abelian groups for each $i \in \{ 0, 1, ..., k-1 \}$. But since $G$ is a finite group, $G$ also has a composition series. By the Schreier refinement theorem, these two subnormal series have equivalent refinements. So any factors of the form $G_{i+1}/G_i$ are simple and abelian.
• The subnormal series above can be refined to a composition series where for each $i \in \{ 0, 1, ..., k-1 \}$ we may need to insert subgroups $H$ and $K$ of $G$ in the form:
(2)
\begin{align} \quad ... \trianglelefteq G_i \trianglelefteq H \trianglelefteq K \trianglelefteq G_{i+1} \trianglelefteq ... \end{align}
• where $H / G_i$, $K / H$, and $G_{i+1} / K$ are simple groups. Since the factor $G_{i+1}/G_i$ is abelian, so are $H / G_i$ and $K / G_i$. Hence $(K/G_i) / (H/G_i)$ is abelian. So $K/H$ is abelian. Therefore $H/K$ is a simple abelian group. But every simple abelian group is isomorphic to a cyclic group $\mathbb{Z}_p$ where $p$ is prime.
• So all of the factors of any composition series of $G$ are cyclic and of prime order.
• $\Leftarrow$ Suppose that the factors of any composition series of $G$ are cyclic and of prime order. Consider any composition series $\{ 1 \} = G_0 \triangleleft G_1 \triangleleft ... \triangleleft G_k = G$ in $G$. Then this series is also a subnormal series. And we have that $G_{i+1} / G_i$ is cyclic and of prime order for all $i \in \{ 0, 1, ..., k-1 \}$. But since the factors are cyclic, they are abelian. So $G$ has a subnormal series whose factors are abelian groups. Hence $G$ is a solvable group. $\blacksquare$