= Free group action
{title2=$gx=x\Longrightarrow g=1$}
A <group action> is free if the only element of the <group> fixing any point is the <identity element>. Thus every <stabilizer subgroup> is trivial. This is a property of an action of any <group>, not an assertion that the acting group is a <free group>.
Explicitly, for a left <group action> of a <group> $G$ on a <set> $X$ with <identity element> $e$, freeness means
$$
g\cdot x=x\Longrightarrow g=e\qquad(g\in G,\ x\in X),\qquad G_x=\{e\}\quad(x\in X).
$$
Equivalently, for any $x,y\in X$ there is at most one $g\in G$ with $g\cdot x=y$. Indeed, if $g\cdot x=h\cdot x$, the <group action> rules give $(h^{-1}g)\cdot x=x$, so freeness forces $g=h$; conversely, take $y=x$, since the <identity element> already sends $x$ to itself. Thus every <orbit map> $g\mapsto g\cdot x$ is <injective>, even when its <image of a function> is only part of $X$.
For a <permutation> example, let the <cyclic group> $C_2=\{e,a\}$ act on the <set> $X=\{1,2,3,4\}$ by $a=(12)(34)$. The nonidentity element exchanges both pairs and fixes no point, so every <stabilizer subgroup> is $\{e\}$. Its two <group orbits> are $\{1,2\}$ and $\{3,4\}$. This is a <free group action> but not a <transitive group action>: there is no <group> element sending $1$ to $3$. Each restriction to one <group orbit> is a <simply transitive group action>.
\Image[/free-actions-two-orbits-and-fixed-points.png]
{title=A <free group action> can have several <group orbits>}
{id=free-actions-two-orbits-and-fixed-points}
{description=Two <group actions> of the same <cyclic group> $C_2=\{e,a\}$ on four points. On the left, $a=(12)(34)$ exchanges two pairs. On the right, both <group> elements fix every point. Green outlines show <group orbits>; gray loops show the <identity element>, and blue arrows show $a$.}
The <identity element> fixes every point in every <group action>; freeness excludes fixed points only for nonidentity elements. This also differs from a <fixed point of a group action>, which must be fixed by every <group> element. In the left diagram there are no such <fixed points of a group action>. In the right diagram, the same abstract <cyclic group> $C_2$ acts by $e\cdot x=a\cdot x=x$: all four points are <fixed points of a group action>, every <stabilizer subgroup> is $C_2$, and each <group orbit> is a <singleton set>. Consequently this second <group action> is neither free nor transitive. A <group action> with no <fixed points of a group action> can still fail freeness: the natural <group action> of the <symmetric group> $S_3$ has no point fixed by all of $S_3$, but $(23)$ fixes $1$.
For a <finite group>, the <orbit-stabilizer theorem> makes each <group orbit> of a <free group action> have <cardinality> $|G|$. If $X$ is also finite, $|X|$ is therefore a multiple of the <group order>, with one copy of $|G|$ for each <group orbit>. The <trivial group> acts freely on every <set>, and acts transitively exactly on a <singleton set>. On the <empty set> freeness is vacuous; our convention under <transitive group action> requires a nonempty <set> for transitivity.
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