# Subgroup of Solvable Group is Solvable/Proof 2

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## Theorem

Let $G$ be a solvable group.

Let $H$ be a subgroup of $G$.

Then $H$ is solvable.

## Proof

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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:

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

for some finite $j$.

Now it is trivial that:

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- $\map D H \le \map D G$

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.

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$\blacksquare$

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