nth Derived Subgroup Criterion for a Group to be Solvable

Table of Contents

Theorem 1: Let

$G$

be a group. Then

$G$

is solvable if and only if there exists an

n \in \mathbb{N}

such that the

n^{\mathrm{th}}

derived subgroup is the trivial subgroup, i.e.,

G^{(n)} = \{ e \}

(1)

\begin{align} \quad \{ e \} = G_0 \leq G_1 \leq ... \leq G_n = G \end{align}

Be a successive chain of subgroups such that $$G_i$$ is a normal subgroup of $G_{i+1}$ for all $0 \leq i \leq n - 1$ and such that $G_{i+1}/G_i$ is abelian for all $0 \leq i \leq n - 1$ .

Recall from the Criterion for a Quotient Group to be Abelian page that $$G/H$$ is abelian if and only if $$H$$ contains the commutator subgroup (derived subgroup) of $$G$$ .

(2)

\begin{align} \quad G^{(1)} \subseteq G_{n-1} \end{align}

And so $G^{(2)} \subseteq G_{n-1}^{(1)}$ . Now since $G_{n-1}/G_{n-2}$ is abelian we have that $G_{n-1}^{(1)} \subseteq G_{n-2}$ . So:

(3)

\begin{align} \quad G^{(2)} \subseteq G_{n-1}^{(1)} \subseteq G_{n-2} \end{align}

(4)

\begin{align} \quad G^{(n)} \subseteq G_{n-1}^{(n-1)} \subseteq ... \subseteq G_1^{(1)} \subseteq G_0 = \{ e \} \end{align}

$\Leftarrow$ Suppose that $G^{(n)} = \{ e \}$ . Consider the chain of subgroups:

(5)

\begin{align} \quad \{ e \} = G^{(n)} \leq G^{(n-1)} \leq ... \leq G^{(1)} \leq G^{(0)} = G \quad (*) \end{align}

From one of the results on the Basic Theorems Regarding the Derived Subgroup of a Group page we see that $G^{(1)}$ is a normal subgroup of $$G$$ . Similarly, $G^{(2)}$ is a normal subgroup of $G^{(1)}$ , and so forth. So for each $1 \leq i \leq n$ we have that $G^{(i)}$ is a normal subgroup of $G^{(i-1)}$ .

So each quotient $G^{(i)} / G^{(i-1)}$ is well-defined. From the theorem referenced above we have that $G^{(i)} / G^{(i-1)}$ is abelian if and only if $G^{(i)'} \subseteq G^{(i - 1)}$ , or equivalently, if and only if $G^{(i - 1)} \subseteq G^{(i - 1)}$ , which is trivially true and of course, holds for all $1 \leq i \leq n$ .

So the chain $$(*)$$ is a chain of successive subgroups of $$G$$ satisfying the properties of the definition of $$G$$ being a solvable group. $\blacksquare$

Mathonline