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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.

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.

A group action on a nonempty set $X$ is simply transitive when exactly one group element sends any chosen point to any other:Equivalently it is both a transitive group action and a free group action; this is also a regular group action. Transitivity supplies a transport, and freeness makes it unique. Conversely, putting $y=x$ in the uniqueness condition shows that every stabilizer subgroup is $\{e\}$, since the identity element already fixes $x$. In a transitive group action, checking that just one stabilizer subgroup is trivial suffices: all the others are conjugate subgroups.

The cyclic group $C_3$ in the diagram under transitive group action is simply transitive. More generally, the left regular action of any group $G$ on itself is simply transitive: the equation $gx=y$ has the unique solution $g=y{x}^{-1}$, using the inverse element of $x$. This shows why the property is useful: every position has a unique label relative to a chosen starting position.

Choose a point $x_0\in X$. Its orbit mapis a bijection and an equivariant map from the left regular action: ${\varphi }_{x_0}(hg)=h\cdot {\varphi }_{x_0}(g)$. This identifies $X$ with a copy of $G$, but $X$ does not come with a preferred identity element. If $x_1=a\cdot x_0$, then ${\varphi }_{x_1}(g)={\varphi }_{x_0}(ga)$: changing the chosen point composes this identification with right multiplication by $a$. An old label $g$ therefore becomes the new label $g{a}^{-1}$, which also specifies the side of multiplication for a noncommutative group.

The transport from $x$ to $y$ remains independent of the chosen point. Writing $x=b\cdot x_0$ and $y=c\cdot x_0$, it is $c{b}^{-1}$. With $x_1=a\cdot x_0$, the new labels are $b{a}^{-1}$ and $c{a}^{-1}$, andFor a finite group, simple transitivity implies $|X|=|G|$. Conversely, a transitive group action with $|X|=|G|<\infty$ is simply transitive by the orbit-stabilizer theorem. Equality of cardinalities alone is insufficient without transitivity: the cyclic group $C_2$ can fix both points of a two-point set, giving two group orbits and two possible self-transports at each point.

A group action on a nonempty set $X$ is transitive when every two points lie in the same orbit of a group action, equivalently when for all $x,y\in X$ there is a group element $g$ with $g\cdot x=y$. There is exactly one group orbit: every position is reachable from every other. Requiring $X\ne \varnothing$ avoids calling the group action on the empty set transitive merely because the quantified statement has no points to test.^[ref]

Here a left group action of a group $G$ means a function $G\times X\to X$, written $(g,x)\mapsto g\cdot x$, satisfyingwhere $e$ is the identity element. Products move the point by $h$ first, then by $g$. Transitivity says that an orbit map $G\to X$, $g\mapsto g\cdot x$, is surjective for every $x\in X$; it suffices to check one chosen point.^[ref][ref]

For example, the symmetric group $S_3$ acts on $X=\{1,2,3\}$ by $\sigma \cdot i=\sigma (i)$. This is a transitive group action, but it is not a free group action: the stabilizer subgroup of $1$ is $\{e,(23)\}$. Both the transposition $(12)$ and the cyclic permutation $(123)$ send $1$ to $2$. In contrast, the cyclic group $C_3=\mathbb{Z}/3\mathbb{Z}$ acts on itself by $k\cdot x=k+x(\mathrm{mod}3)$, using modular arithmetic. The group element sending $x$ to $y$ is uniquely $k=y-x(\mathrm{mod}3)$. These two group actions reach the same number of points, but only the second has unique transport between them.

The stabilizer subgroup $G_x=\{h\in G:h\cdot x=x\}$ measures the ambiguity of transport. If $g_0\cdot x=y$, thenThe first set is a left coset, and the second identity says that stabilizer subgroups in one group orbit are conjugate subgroups. Briefly, $g\cdot x=g_0\cdot x$ exactly when $g_0^{-1}g\in G_x$. Thus, for a finite group in a transitive group action, the orbit-stabilizer theorem gives $|G|=|X||G_x|$, and there are exactly $|G_x|$ transports for each pair of points. The example counts are $6=3\cdot 2$ for $S_3$ and $3=3\cdot 1$ for $C_3$. This separates the number of reachable positions from the number of ways to reach each one.^[ref]