Free $G$-Set, $G$ being a group 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 $X$?
 A: 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 $g' (g, x) = (g'g, x)$.
A: 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.
A: 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 $x$.
