Free $G$-Set, $G$ being a group

Question

How is defined the free $G$-Set on a set $X$, i.e. how the action $\mu$ looks like, $\mu:G\times X\to X$ to define a free $G$-Set on $X$?

Answer 1

Just like a free group on $X$ is not $X$, but is made from $X$, a free $G$-set on $X$ is not $X$ itself, but another set.

First, think about a free $G$-set generated by a single element $x$. Free means imposing no additional relations, so this is just the set of formal products $\{gx, g \in G\}$.

Now, a free $G$-set on a set $X$ is just the disjoint union of such one-element-generated free $G$-sets: $\{gx, g \in G, x \in X\} \cong G \times X$. Here, a $G$-action is defined as $g' (g, x) = (g'g, x)$.

Answer 2

Sometimes the free $G$-set on $X$ is defined as the set $G \times X$, with $$ \mu : G \times (G \times X) \to G \times X $$ being defined by $$ g \mapsto ( (h, x) \mapsto (h g, x)). $$ (My actions are right actions.)

Here the elements of $X$ (to be identified with those of $\{1 \} \times X$) generate $G \times X$, in the sense that they are a complete set of representatives of the orbits.

And each stabilizer is clearly trivial.

Answer 3

If $G_x = \{1\}$ for all $x \in X$ , the action $\mu$ is free. In this case, X is called a free G-set. Here $G_x = \{ g \in G \mid gx = x\}$ is the isotropy group, or stabiliser of $x$.