Show $g.S = \{g.x|x\in S\}$ defines an action on the power set

Question

Let $X$ be a finite set and let $\mathscr{P}(X)$ be its power set. A group $G$ acts on $X$. Given $g \in G$ and $S \subseteq X$ show $g.S = \{g.x|x\in S\}$ defines an action on $\mathscr{P}(X)$.

I know this means that I have to show that $(gh).S = g.(h.S)$ but I am unsure as to how actually show this.

Answer 1

You want to show an equality of sets: $(gh) \cdot S = g \cdot (h \cdot S)$, for all $g,h \in G$ and $S \subseteq X$. This is proven by double inclusion:

• Suppose $y \in (gh) \cdot S$, i.e. $y = (gh) \cdot x$ for some $x \in S$. Then since the action of $G$ on $X$ is a group action, $y = g \cdot (h \cdot x)$. Then $$h \cdot x \in h \cdot S \implies g \cdot (h \cdot x) \in g \cdot (h \cdot S).$$

• Conversely suppose $y \in g \cdot (h \cdot S)$. Then $y = g \cdot x'$ with $x' \in h \cdot S$, thus $x' = h \cdot x$ for some $x \in S$. But then $y = g \cdot (h \cdot x) = (gh) \cdot x \implies y \in (gh) \cdot S$.

Finally $(gh) \cdot S = g \cdot (h \cdot S)$.

To prove that this is a group action, you also need to show that $e \cdot S = S$ for all $S$ (where $e$ is the unit of $G$). But if $x \in S$, then $x = e \cdot x \in e \cdot S$, and conversely $e \cdot S \subseteq S$ is easy to show too.