Composition length of semidirect product is sum of composition lengths

For external semidirect product
Suppose $$N$$ and $$H$$ are groups, and $$\rho:H \to \operatorname{Aut}(N)$$ gives an action of $$H$$ on $$N$$ by automorphisms. Suppose the composition length of $$N$$ is $$a$$ and the composition length of $$H$$ is $$b$$. Then the composition length of the external semidirect product $$N \rtimes H$$ is $$a + b$$.

For internal semidirect product
Suppose $$G$$ is a group, $$N$$ is a normal subgroup and $$H$$ is a complement to $$N$$ in $$G$$. In other words, $$G$$ is an internal semidirect product of $$N$$ and $$H$$.

Suppose the composition length of $$N$$ is $$a$$ and the composition length of $$H$$ is $$b$$. Then the composition length of the external semidirect product $$N \rtimes H$$ is $$a + b$$.

Related facts

 * Composition length of direct product is sum of composition lengths
 * Composition length of extension group is sum of composition lengths