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Write the translation group action as $(A,B,C)\mapsto (e^{i{\theta }_1}A,e^{i{\theta }_2}B,e^{-i({\theta }_1+{\theta }_2)}C)$. Modulo lattice translations, the stabilizer subgroup of $(h,h,h)$, with nonzero real $h$, is the sixfold rotation group. Its fixed-point subspace of a group action is the real line $(a,a,a)$. The stabilizer subgroup of $(r,0,0)$, with nonzero real $r$, consists of the circle ${\theta }_1=0$ together with a half-turn; its fixed-point subspace of a group action is the real line $(a,0,0)$. The equivariant branching lemma therefore supplies generic branches of both axial types.

The full symmetry group of a cube acts as all signed permutation matrices. The stabilizer subgroups of $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$ have orders eight, four and six, respectively, and their fixed-point subspaces of a group action are the lines through those vectors. The three group orbits have sizes six, twelve and eight. They therefore give distinct axial types for the equivariant branching lemma. More generally the fixed-space dimension of a vector stabilizer subgroup equals the number of distinct nonzero absolute coordinate values, so these are precisely its one-dimensional types.

// Destination: dynamical-systems.bigb

For an orthogonal group gauge transformation $R\in O(3)$, a scalar vector transforms as $\varphi \mapsto R\varphi$. The gauge connection belongs to the Adjoint representation, identified with axial vectors: the displayed cross-product identity makes its homogeneous transformation $A\mapsto (\det R)RA$. The matrix connection also has the usual inhomogeneous derivative term. Thus $D\varphi$ transforms as an ordinary vector, including under improper orthogonal transformations. Treating both fields as ordinary vectors would fail because a cross product changes sign under an orientation reversal. A nonzero vector vacuum has stabilizer subgroup $O(2)$; its connected part is $SO(2)$.

An opposite-vertex line in the full octahedral symmetry group is an unoriented coordinate axis. Its stabilizer subgroup has the block form $\mathrm{diag}(A,\epsilon )$, where $A$ is a signed two-dimensional permutation matrix and $\epsilon =\pm 1$. The eight possibilities for $A$ form the dihedral group of a square, of order eight; the independent axial sign gives a cyclic group of order two. Thus the stabilizer is their direct product, of order sixteen. It is larger than the stabilizer of an individual vertex, because exchanging the two endpoints still fixes the line.

For vertices $(\pm 1,\pm 1,\pm 1)$, the symmetry group of a cube consists of signed permutation matrices. A matrix preserving the unoriented line through $(1,1,1)$ must have all three signs equal, so it has the form $\epsilon P$, where $P$ is a permutation matrix and $\epsilon \in \{1,-1\}$. Central inversion commutes with every $P$ and is not itself a permutation matrix. Consequently $(P,\epsilon )\mapsto \epsilon P$ identifies this twelve-element stabilizer subgroup with the direct product of groups $S_3\times C_2$. A body diagonal is different from a face-normal coordinate axis, whose line stabilizer has order sixteen.

Every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. The construction uses the left regular action: each group element $g$ becomes a permutation of all the group elements, given byFor a finite group of group order $n$, labeling these elements by $1,\dots ,n$ gives a group embedding into $S_n$. The same construction works for an infinite group, with the symmetric group of all bijections of its underlying set.^[ref]

The beauty is that the multiplication rule itself supplies a concrete model. Whatever the group elements originally represent, each one rearranges the entire group by multiplication. A product becomes function composition, and no distinctions between group elements are lost. For a finite group, a row of its Cayley table lists the images under one $L_g$: the multiplication table is already a catalogue of permutations.

Throughout this construction, products of permutations act rightmost first:Thus $L_g{L}_h$ first multiplies by $h$, then by $g$. This is a left group action, not a license to reverse the multiplication order in a nonabelian group. Ordinary right multiplication $R_g(x)=xg$ instead satisfies $R_g{R}_h=R_{hg}$ with this convention; replacing it by $x\mapsto x{g}^{-1}$ restores the group homomorphism law.^[ref]

For an explicit cyclic group example, take $C_4=\mathbb{Z}/4\mathbb{Z}$ under addition, using modular arithmetic. Its left regular action consists of $L_k(x)=k+x(\mathrm{mod}4)$. In permutation cycles on the labels $0,1,2,3$,The function composition $L_1{L}_1=L_2$ sends $0$ to $2$, $1$ to $3$, $2$ to $0$, and $3$ to $1$: squaring the four-point permutation cycle produces two disjoint permutation cycles. Also $L_1{L}_3=L_0$. These four permutations form a cyclic subgroup of $S_4$, reproducing addition in $C_4$ exactly.

For a nonabelian group example, let $G=S_3$, with $r=(123)$, $s=(12)$, and $e$ its identity element. The six group elements are $e,r,r^2,s,rs,r^{2}s$, where $r^3=s^2=e$ and $sr=r^{2}s$. Now the labels being permuted are themselves permutations of three points. Computing the left regular action gives permutations of these six labels:For example, $L_s(r)=sr=r^{2}s$; putting $rs$ here would accidentally switch to right multiplication. Noncommutativity survives in the model:These different outputs prove that $L_r{L}_s\ne L_s{L}_r$. The group generators $r,s$ become group generators $L_r,L_s$ of a six-element subgroup of $S_6$.

Why does a group action here produce a group embedding? Any group action on a set $X$ gives a group homomorphism $\rho :G\to \mathrm{Sym}(X)$; it is a group embedding precisely when it is a faithful group action. The group kernel consists of the group elements fixing every point of $X$. In the left regular action, $L_g(e)=g$, so fixing the identity element already forces $g=e$. In fact every stabilizer subgroup is trivial, since $gx=x$ forces $g=e$; this stronger property makes the left regular action a regular group action.^[ref]

Compare two group actions of the same cyclic group $C_4$. Its left regular action on four labels distinguishes all four group elements. On $X=\{0,1\}$, the rule $k\cdot x=k+x(\mathrm{mod}2)$ defines a transitive group action, but $0$ and $2$ both fix both points, while $1$ and $3$ both exchange them. Its group kernel is $\{0,2\}$, so it is not a faithful group action. Its image is a two-element subgroup of $S_2$, isomorphic to the quotient group $C_4/\{0,2\}$ rather than to $C_4$.

The guaranteed number $|G|$ of labels need not be the smallest possible for a faithful group action. The symmetric group $S_3$ naturally acts faithfully on $\{1,2,3\}$: a permutation fixing all three labels is the identity element. Its left regular action uses six labels instead. Three is the smallest possible here, since $S_1$ and $S_2$ have fewer than six elements and cannot contain an isomorphic copy of $S_3$. Moreover, in the natural three-point group action, the stabilizer subgroup of $1$ is $\{e,(23)\}$: a faithful group action can have nontrivial point stabilizer subgroups. Cayley's construction supplies a universal model, while a smaller model can reveal particular symmetries.

The proof is short. Each $L_g$ is a bijection with inverse $L_{g^{-1}}$. Associativity givesso $\lambda$ is a group homomorphism. If $L_g=L_h$, evaluating both at the identity element gives $g=L_g(e)=L_h(e)=h$. Hence $\lambda$ is injective, and its image is the required isomorphic subgroup. No finiteness assumption entered the argument.

The historical connection is unusually close to the notation. The opening paragraph of Arthur Cayley's 1854 paper already treats permutations as one example of operations on a collection of quantities and explicitly performs the rightmost operation first in a product. It also allows noncommutativity while insisting on associativity. The modern left regular action shows how that viewpoint reaches every abstract group, by choosing the group itself as the set to be acted on.^[ref]

The centralizer of an element $x$ of a group $G$ consists of all elements that commute with $x$:It is the stabilizer subgroup of $x$ under the conjugation action, hence a subgroup of $G$. In conjugacy classes of the symmetric group on three letters, the three representative computations areFor the transposition $(12)$, a commuting permutation must preserve the exchanged pair $\{1,2\}$ and fix $3$; only $e$ and $(12)$ do so. For $r=(123)$, a commuting permutation is determined by where it sends $1$: the equation $g(r(i))=r(g(i))$ forces the images of $2$ and $3$, leaving exactly $e,r,r^2$. For $e$, every permutation commutes. These centralizer sizes $6,2,3$ give conjugacy class sizes $1,3,2$.

The centralizer is useful because it measures how much freedom remains after fixing one operation. Its index of a subgroup counts the distinct results of conjugation, and the intersection of all element centralizers is the center of a group. A centralizer need not be an abelian group: $C_G(e)=G$ even when $G$ is a nonabelian group. In an abelian group, all element centralizers equal $G$. These observations also explain why the surrounding group matters: the centralizer of $(123)$ inside the alternating group $A_3$ is the whole $A_3$, whereas its centralizer inside $S_3$ is a proper subgroup. See McKernan, the centralizer definition preceding lemma 23.6.

For an arbitrary subset $A\subseteq G$, the centralizer of a subset fixes every element of $A$ individually under the conjugation action:This is a pointwise stabilizer of the elements of $A$, not of the vertices of a polygon in a geometric realization. In particular, $r=(123)$ belongs to $C_{S_3}(r)$ although its rotation moves every vertex of a polygon of the equilateral triangle. If $A$ generates a subgroup $H$, commuting with every element of $A$ implies commuting with every product of those elements and their inverse elements; hence $C_G(A)=C_G(H)$. For the triangle rotation subgroup $H=⟨r⟩$, this gives $C_{S_3}(H)=C_{S_3}(r)=H$.

To see the commutation test directly on the equilateral triangle, compare a rotation $r$ with the reflection $s=(12)$. Their permutations satisfyThe two orders send vertex of a polygon $1$ to different places: $rs(1)=3$, but $sr(1)=1$. Thus $s\notin C_{S_3}(r)$. By contrast $r$ and $r^2$ commute, with $r{r}^2=r^{2}r=e$. The reflection reverses the chosen rotation while preserving the subgroup of all triangle rotations, so $s\in N_{S_3}(H)$. The distinction is visible before any counting: commuting preserves an operation itself; membership in the normalizer permits exchanging it with another operation in the same subgroup.

Both constructions are stabilizer subgroups for the conjugation action. The centralizer $C_G(x)$ fixes an individual element $x$; the normalizer $N_G(H)$ preserves a subgroup $H$ as a whole, while possibly moving its elements. The centralizer of a subset $C_G(H)$ fixes every member of $H$ individually, soFor a cyclic subgroup $H=⟨r⟩$, commuting with the group generator $r$ is equivalent to commuting with all its powers, giving $C_G(H)=C_G(r)$.

In the symmetric group $S_3$, let $r=(123)$ and $H=⟨r⟩=\{e,(123),(132)\}$. ThenEvery transposition sends $r$ to $r^{-1}$ under conjugation. In particular, for $t=(12)$,Thus $t$ belongs to the normalizer of $H$ but not to the centralizer of $r$. Preserving the collection of powers allows the two possible directions of its group generator to be exchanged.

This distinction lets one ask separately which operations leave a chosen operation unchanged and which operations preserve the subgroup it generates. The normalizer also supplies a group homomorphism to the automorphism group of $H$, by $g\mapsto (h\mapsto gh{g}^{-1})$; its group kernel is $C_G(H)$. In this example the group homomorphism has two possible outputs, fixing $r$ or replacing it by $r^{-1}$. The quotient group $N_{S_3}(H)/C_{S_3}(H)$ is therefore a cyclic group of group order two. Milne develops the normalizer as a stabilizer subgroup in Group Theory (v4.01) by James S. Milne.

A rigid Skyrmion orientation is quantized on a cover of its Skyrmion collective-coordinate orbit. A lifted combined rotation in the static field's stabilizer subgroup identifies the same classical field and imposes ${\hat{D}}^J(R){\hat{D}}^I(A)\Psi ={\chi }_{\mathrm{FR}}\Psi$, where the sign is the chosen Finkelstein-Rubinstein constraints character of that configuration-space loop. Choosing the nontrivial character models unit baryons as fermions. A $2\pi$ spatial rotation or isorotation of charge $B$ then has sign $(-1{)}^B$: deforming the field to separated unit lumps adds the unit-lump loop classes in the fundamental group ${\mathbb{Z}}_2$, while the labelled orbital paths can be contracted in three dimensions. The resulting Finkelstein-Rubinstein constraints sign is the product of the unit-lump signs. Hence spin and isospin are half-integer for odd $B$ and integer for even $B$. Further stabilizer constraints restrict which pairs $(J,I)$ and which body-fixed states occur. The energy operator comes from the inertia tensors on the orbit; symmetry selects allowed states but does not by itself determine their energies. A bosonic choice of the trivial topological character is mathematically possible but does not model a fermionic nucleon.

Let a group action of $G$ on a set $X$ be given, and let a function $c:X\to C$ attach a colour label to each point. The colour-preserving subgroup of $G$ isIt is a subgroup: the identity map preserves all labels, compositions of two preserving maps preserve them, and replacing $x$ by $g^{-1}x$ shows that inverse maps preserve them too. Extra object data restrict the allowable transformations; the labels are now part of the object rather than bookkeeping for a drawing.

For a geometric square, take $G=D_4$, of group order eight. If all four vertices have the same colour, all eight maps remain. If two adjacent vertices are red and the others blue, only the identity and one reflection remain. Putting the two red vertices opposite one another instead leaves four maps: the identity, a half-turn rotation, and the two diagonal reflections. Giving all four vertices distinct colours leaves only the identity.

The count alone does not specify the pattern. Both two-red examples have the same number of red and blue vertices, but their stabilizer subgroups differ. For a single red vertex, the orbit-stabilizer theorem gives a group orbit of size four and a stabilizer subgroup of group order two, agreeing with the existing group action illustration. Such counts identify equivalent placements without treating each placement as a new geometric case.

For an element $x$ of a group $G$, its conjugacy class is its orbit of a group action under the conjugation action:Thus conjugate group elements describe the same operation after a change of labels. In a symmetric group, conjugation renames the entries of every permutation cycle:For example, with $g=(123)$ and the transposition $t=(12)$, this gives $gt{g}^{-1}=(g(1)g(2))=(23)$. Products use function composition of permutations, with the rightmost factor applied first. The complete six-element example is conjugacy classes of the symmetric group on three letters.

The conjugacy classes form a partition of a set of $G$, since conjugacy is an equivalence relation. In a symmetric group, two permutations lie in the same conjugacy class precisely when they have the same cycle type: renaming preserves the lengths of permutation cycles, and matching equal-length permutation cycles constructs a renaming in the other direction. This compresses many labelled permutations into a few structural types. See McKernan, Conjugation in Sn, lemma 6.2 and proposition 6.6.

The stabilizer subgroup of $x$ under the conjugation action is its centralizer $C_G(x)$. Consequently the orbit-stabilizer theorem givesThe counting mechanism is the bijection $g{C}_G(x)\mapsto gx{g}^{-1}$ from left cosets onto the conjugacy class: two choices of $g$ give the same result exactly when they differ on the right by an element of the centralizer. A large centralizer means many choices of labels leave the operation unchanged, hence a small conjugacy class. A conjugacy class has size one precisely when its element belongs to the center of a group. The class equation counts these orbits of a group action together.

For a subgroup $H$ of a group $G$ and $g\in G$, the conjugate subgroup is the conjugate subsetIt is a subgroup because conjugation is a group automorphism. Restricting that group automorphism gives a group isomorphism from $H$ to $gH{g}^{-1}$, so conjugate subgroups have the same group order. This describes the same operations after moving their reference configuration. Conjugacy of subgroups is an equivalence relation. ^[ref]

For an equilateral triangle with vertices of a polygon labeled $1,2,3$ counterclockwise, its symmetry group is the symmetric group $S_3$, also the dihedral group of group order six. Let $r=(123)$ be the counterclockwise rotation through $120$ degrees, and let $s=(12)$ be the reflection fixing vertex of a polygon $3$. Products apply the rightmost permutation first. The cyclic subgroup $K=⟨s⟩=\{e,(12)\}$ contains the identity element and one reflection. Relabeling a transposition by conjugation givesThus the conjugation action moves $K$ among the three subgroups $⟨(12)⟩$, $⟨(23)⟩$, and $⟨(13)⟩$. Geometrically, the rotation moves the axis of the reflection to the next axis. Each reflection fixes a different vertex of a polygon, although group isomorphisms relate all three subgroups.

The normalizer tells when this change of reference leaves the subgroup unchanged. For any group $G$ and subgroup $H\le G$, there is a bijectionTwo choices $g,k$ give the same conjugate subgroup exactly when $k^{-1}g\in N_G(H)$. Consequently the index of a subgroup $[G:N_G(H)]$ counts the distinct conjugate subgroups, even for an infinite group. For a finite group the count is $|G|/|N_G(H)|$. These are cosets; they need not form a quotient group, since $N_G(H)$ need not be a normal subgroup of $G$. In the triangle example there are $6/2=3$ conjugate subgroups, while the cyclic subgroup $⟨r⟩$ has only one. ^[ref]

A group acts on the left cosets of a subgroup by left multiplication; its kernel is the core of that subgroup.

Let $H\le G$ be a subgroup. The notation $G/H$ here means the set of left cosets $\{aH:a\in G\}$, even when $H$ is not a normal subgroup. DefineThis gives a group action without needing a quotient group structure on $G/H$.^[ref]

The representative check is essential: if $aH=bH$, then $a=bh$ for some $h\in H$, and $(ga)H=(gbh)H=(gb)H$. Thus equal input cosets give equal output cosets. The identity element fixes every coset, and associativity gives $g_1\cdot (g_2\cdot aH)=(g_1{g}_2)\cdot aH$. Each induced map is a permutation, with inverse induced by $g^{-1}$. This is a transitive group action because $b{a}^{-1}$ sends $aH$ to $bH$.

The stabilizer subgroup of $H$ is $H$ itself. At a general coset, the stabilizer subgroup and the group kernel areConsequently the coset action is a faithful group action exactly when the subgroup core is trivial.^[ref]

Indeed, $g\cdot (aH)=aH$ exactly when $a^{-1}ga\in H$, or $g\in aH{a}^{-1}$. Intersecting these membership conditions over all cosets gives the group kernel. Conjugation permutes the subgroups in this intersection, making it a normal subgroup of $G$ contained in $H$. Conversely, any normal subgroup of $G$ contained in $H$ lies in every $aH{a}^{-1}$. This explains why the subgroup core is the largest such normal subgroup: it is exactly the portion of $H$ invisible at every label, rather than merely at the label $H$.

For a normal subgroup $H$, every $aH{a}^{-1}$ equals $H$, so $\ker {\rho }_H=H$. For $H=\{e\}$, the construction is the regular group action behind Cayley's theorem. Between these extremes, a nontrivial subgroup with trivial subgroup core can give a faithful group action on fewer labels. If the index of a subgroup is finite, $[G:H]=n$, the first isomorphism theorem givesThus a subgroup of small index of a subgroup exposes a small permutation group associated with $G$. The two examples below and in nonfaithful coset action of the symmetric group on three letters show exactly which information can survive.

The full symmetry group of a cube acts transitively and faithfully on its twelve edges. For vertices $(\pm 1,\pm 1,\pm 1)$, its signed permutation matrices act transitively on the edges. The stabilizer subgroup of $\{(1,1,z):-1\le z\le 1\}$ consists of independently swapping $x,y$ and reversing $z$, and is a Klein four-group. The orbit-stabilizer theorem gives order $12\cdot 4=48$. Fixing every edge fixes every vertex, as each vertex is the unique intersection of its three incident edges, so the group action is faithful. The resulting subgroup of $S_{12}$ has index $12!/48$ and is not normal: central inversion has cycle type $2^6$, whose conjugacy class in $S_{12}$ has $12!/(2^{6}6!)=10395>48$ elements.

An elliptic point of a modular curve is the image of a point of the complex upper half-plane with a nontrivial stabilizer subgroup in the effective modular subgroup. For the full modular group the stabilizer subgroup orders are two and three, represented by $i$ and $e^{2\pi i/3}$. A local disk coordinate $u$ upstairs descends as $u^e$ when the stabilizer subgroup order is $e$. The central matrices $\pm I$ are factored out before counting this order.

The general linear group over a finite field acting diagonally on projective triples has five orbits of a group action: all equal, three different positions for the repeated coordinate when exactly two are equal, and all distinct. The corresponding representative stabilizer subgroups are upper triangular, diagonal, and scalar matrices. Their orders are $p(p-1{)}^2$, $(p-1{)}^2$, and $p-1$, respectively; the orbit sizes are $p+1$, $p(p+1)$, and $p(p+1)(p-1)$. Transitivity follows from the action being sharply three-transitive on a projective line.

Take the symmetric group $G=S_3$ and the subgroup $H=⟨(12)⟩=\{e,(12)\}$ of group order two. With the rightmost permutation applied first, its three left cosets areThe subscripts record the image of $3$: every member of $C_i$ sends $3$ to $i$. Because $H$ is the stabilizer subgroup of $3$ in the usual group action, $aH\mapsto a(3)$ is a bijection onto $\{1,2,3\}$. It is an equivariant map, since $(ga)(3)=g(a(3))$. The coset action therefore becomes the familiar group action on three letters. In particular,

The three stabilizer subgroups areTheir intersection is $\{e\}$, so the subgroup core and the group kernel are trivial. This proves that the coset action is a faithful group action even though every point has a nontrivial stabilizer subgroup. The orbit-stabilizer theorem reads $6=3\cdot 2$, while the induced group homomorphism gives six distinct permutations of the three cosets. For example, $(12)$ and $e$ agree at $C_3$, but disagree at $C_1$; inspecting one label would miss their difference.

The subgroup need not be a normal subgroup for the coset action to exist: $(123)H(123{)}^{-1}=\{e,(23)\}\ne H$. Here the group action is informative because the three conjugate stabilizer subgroups have no common nonidentity element. For the normal subgroup $A_3$ instead, nonfaithful coset action of the symmetric group on three letters shows that only the sign of a permutation survives.

A group action is faithful when only the identity acts as the identity on every point.

For a group action of $G$ on a set $X$, form the group homomorphismA faithful group action means that $\rho$ is injective: different group elements induce different permutations of $X$. Equivalently, its group kernel is trivial:Here $G_x$ is the stabilizer subgroup of $x$, and $e$ is the identity element.^[ref]

The test is global. A faithful group action requires each nonidentity group element to move at least one point; a free group action requires it to move every point. In the faithful coset action of the symmetric group on three letters, $(12)$ fixes one coset and exchanges the other two. Thus a nontrivial stabilizer subgroup does not rule out faithfulness. The distinction follows directly from the order of the quantifiers:On a nonempty set, a free group action is therefore a faithful group action. A transitive group action instead asks whether every point can reach every other; it allows several group elements to perform exactly the same rearrangement. The nonfaithful coset action of the symmetric group on three letters is a counterexample to the claim that every transitive group action is a faithful group action.

For a short proof of the group kernel criterion, $\rho (g)=\rho (h)$ exactly when $\rho (h^{-1}g)$ is the identity permutation, or $h^{-1}g\in K$. If $K=\{e\}$, this forces $g=h$; if $K$ contains a nonidentity element, that element and $e$ act identically. Thus a faithful group action turns $G$ into an actual subgroup of a symmetric group, retaining its group operation as function composition. Choosing a suitable coset action can give a much smaller set of labels than the regular group action used in Cayley's theorem.

A group action is free when only the identity fixes any given point, or equivalently every stabilizer subgroup is trivial. The trivial group acts freely on every space. A nontrivial group acting freely must act by a fixed-point-free map for each nonidentity element; the fixed-point property of even-dimensional complex projective space obstructs such actions. For a finite group acting freely on a Hausdorff manifold, the quotient map is a covering map.

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.