Free group action ($gx=x⟹g=1$)

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 meansEquivalently, 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.

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.

Let a finite group $G$ have free group actions on finite sets $S,T$ of equal cardinality. There is a bijection $f:S\to T$ satisfying $f(g\cdot s)=g\ast f(s)$ for every $g\in G$, $s\in S$.

By the orbit-stabilizer theorem, every orbit of a group action in either set has cardinality $|G|$. Thus both have $r=|S|/|G|$ orbits of a group action. Choose representatives $s_1,\dots ,s_r$ and $t_1,\dots ,t_r$, and defineEvery point has a unique displayed representation: its orbit of a group action determines $i$, and the free group action property makes $g$ unique. Thus $f$ is well defined, and $g\ast t_i\mapsto g\cdot s_i$ defines its inverse. Finallyproving the required compatibility. Empty sets are covered by the unique empty bijection. This shows that a finite free group action is determined up to such an isomorphism by its number of orbits of a group action.

If a nontrivial finite group acted freely on ${\mathbb{R}}^n$, a prime-order cyclic subgroup $C_p$ would too. The quotient is an $n$-manifold with contractible universal cover. The singular version of cellular free resolution from a contractible universal cover identifies its cohomology with group cohomology of $C_p$. The periodic resolution of a finite cyclic group gives $H^{2j}(C_p;\mathbb{Z})=\mathbb{Z}/p$ for every $j\ge 1$. But Poincaré duality for noncompact manifolds makes the quotient cohomology zero above dimension $n$. This contradiction excludes the action, for arbitrary homeomorphisms, not only Euclidean isometries.

For every nonidentity element of a free group action on $S^{2r}$, its action map has no fixed point and hence degree minus one by fixed-point-free sphere maps are homotopic to the antipodal map. Degree of homeomorphisms defines a homomorphism into $\{1,-1\}$, whose kernel is trivial, since no nonidentity action map can have degree one. Thus the acting group has order at most two. For $S^0$, freeness on its two points gives the same conclusion directly.

Suppose a finite group $G$ acts freely on $S^n$, $n>1$. The quotient has fundamental group $G$ and higher homotopy groups equal to those of $S^n$. Attach one $(n+1)$-cell killing a generator of ${\pi }_n\cong \mathbb{Z}$, then attach higher cells successively to kill all remaining higher homotopy. This constructs a Eilenberg–MacLane space $K(G,1)$ with only one $(n+1)$-cell. The one-cell bound on next-degree cohomology gives dimension at most one in degree $n+1$. But if $G=C_p\times C_p$, the mod-p cohomology ring of a cyclic group and Künneth theorem give dimension $n+2$ there. This contradiction excludes that group, and also excludes such a subgroup of any group acting freely on a sphere.