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

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.