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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