Subgroup of Solvable Group is Solvable/Proof 2

Proof
Firstly we, know that a group is solvable if and only if its derived series:

$\map D G = \sqbrk {G, G} \, \ \map {D^i} G = \sqbrk {\map {D^{i - 1} } G, \map {D^{i - 1} } G}$

becomes trivial after finite iteration.

Meaning:

$\map {D^j} G = \set 1$

for some finite $j$.

Now it is trivial that:


 * $\map D H \le \map D G$

since $H$ is smaller than $G$.

Further since $\map {D^i} H$ is dominated by $\map {D^i} G$, it too has to become trivial after a finite amount of steps.