Semidirect product acts on normal subgroup

Statement
Suppose $$G$$ is a semidirect product with normal subgroup $$N$$ and a complement $$H$$. In other words:

$$G = N \rtimes H$$.

The elements of $$G$$ are written as $$(a,b)$$, with $$a \in N, b \in H$$.

Then, we have an action of $$G$$ on the underlying set of $$N$$ given as follows:

$$(a,b) \cdot n = a (b.n)$$.

In other words, it is the product of $$a$$ and the element obtained by acting $$b$$ on $$n$$.

This action has the property that the restriction of the action to $$N$$ is the left-regular group action, and the restriction of the action to $$H$$ is simply the induced action of $$H$$ on $$N$$ by automorphisms.

Proof that the identity element acts trivially
This is easy to see:

$$(e,e) \cdot n = e(e.n) = n$$

Proof that the product condition is satisfied
Consider two elements $$(a,b)$$ and $$(c,d)$$, and consider $$n \in N$$:

$$(a,b)(c,d) \cdot n = (a(b.c),bd) \cdot n = a(b.c)((bd).n)$$.

On the other hand:

$$(a,b) \cdot ((c,d) \cdot n) = (a,b) \cdot (c(d.n)) = a(b.(c(d.n))) = a(b.c)(b.(d.n)) = a(b.c)((bd).n)$$.

Thus, the condition for a group action is satisfied.