Solvable Groups and Derived Subgroups. Why does $D(G_i) \subset G_{i + 1} \implies D_i(G) \subset G_i$.

Question

I'm working through the proof of the following theorem:

$G$ is a solvable group $\Leftrightarrow$ There exists $n$ such that $D_n(G) = {1}$, where $D_i(G)$ is the $i$th derived subgroup of $G$.

Proof: We let $G$ be a solvable group with $G = G_0 \supset G_1 \supset G_2 \cdots \supset G_n = {1}$. As $G_i/G_{i+1}$ is abelian, $D(G_i) \subset G_{i + 1}$. It follows inductively that $D_i(G) \subset G_i$. The proof continues...

I don't understand how $D_i(G) \subset G_i$ follows from $D(G_i) \subset G_{i + 1}$. I tried incrementing $D$ upwards, but then I can't find a way of going from $G_i$ to $G$.

I feel like I'm missing something obvious. Could someone please help.

Answer 1

Your inclusion symbols are reversed it should be

$1 = G_n \subset G_{n-1} \subset G_{n-2} \cdots \subset G_0 = G$.

Now $G_0/G_1$ is abelian, so $D(G) \subseteq G_1$. Hence $D_2(G) \subseteq D(G_1) \subseteq G_2$. Can you finish?