nth Derived Subgroup Criterion for a Group to be Solvable

# nth Derived Subgroup Criterion for a Group to be Solvable

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., $G^{(n)} = \{ e \}$. |

**Proof:**$\Rightarrow$ Suppose that $G$ is a solvable group. Let:

\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 $G$.

- In particular, since $G_n/G_{n-1} = G/G_{n-1}$ is abelian, we have that:

\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:

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

- Continuing on in this manner and we get that:

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

- Thus $G^{(n)} = \{ e \}$.

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

\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. $\blacksquare$