Normal subgroup acting on a set

Question

I am trying to solve the following problem:

Let $G$ be a group acting on a set $X$ and let $S \lhd G$. Determine the necessary and sufficient conditions so that there exists an action of $G/S$ on $X$ such that $\overline{a}.x=a.x$ for all $a \in G$ and $x \in X$.

I am pretty much stuck with this exercise, I would appreciate any hints.

Answer 1

Hint. Think about what it would mean for the given action ($\bar{a}\cdot x = a\cdot x$) to be well-defined. That means that, whenever $\bar{a}=\bar{b}$, for $a,b\in G$, you must have $\bar{a}\cdot x = \bar{b}\cdot x$, for all $x$ in $X$.

Answer 2

Checking conditions of Group action does not say much as it is natural..

$\overline{ab}\cdot x=ab\cdot x=a\cdot(b\cdot x)$

$\bar{a}\cdot(\bar{b}\cdot x)=\bar{a}\cdot(b\cdot x)=a\cdot(b\cdot x)$

All that you need to see is does the map makes sense??

I mean is it well defined??