Action of factor group on a group

Question

Suppose $A$ and $G$ are finite groups and $A$ acts on $G$, written $g^a$ for $g\in G$, $a\in A$. If $N\unlhd A$, does $A/N$ then act on $G$ also? By $g^{[a]} = g^a$? Or do I need to assume something about $N$ first?

Answer 1

I suppose that you are interested in an action of $A$ in which every map $g \in G \mapsto g^a \in G$ is an homomorphism of group (indeed an automorphism), otherwise there's no reason to state that $G$ is a group.

In this case it's known that an action of $A$ on $G$ is the same as an homomorphism $act \colon A \to \text{Aut}(G)$.

Your request is equivalent to the existance of an induced homomorphism $\overline {act} \colon A/N \to \text{Aut}(G)$ such that if $\pi \colon A \to A/N$ is the canonical projection (i.e. $\pi(a)=[a]$ for every $a \in A$) then $\overline {act} \circ \pi=act$.

By basic theorem in group theory this is possible if and only if $N \le \ker act$.