Solvability is subgroup-closed

Statement
Suppose $$G$$ is a solvable group and $$H$$ is a subgroup of $$G$$. Then, $$H$$ is also a solvable group, and the derived length of $$H$$ is less than or equal to the derived length of $$G$$.

Similar facts

 * Solvability is quotient-closed
 * Solvability is quasivarietal
 * Solvability is extension-closed

Proof using derived series
We can show that the derived series of $$H$$ descends at least as fast as the series obtained by intersecting $$H$$ with the members of the derived series of $$G$$.

Proof using arbitrary normal series
We can show that intersecting $$H$$ with any normal series for $$G$$ with abelian quotients gives a normal series for $$H$$ with abelian quotients. (We could also use a subnormal series instead of a normal series).

Proof using the quasivarietal nature
The proof follows directly from the fact that solvability is quasivarietal, because quasivarietal implies subgroup-closed.