Group action ($G\times X⟶X$)

A left group action of a group $G$ on a set $X$ is a function $(g,x)\mapsto g\cdot x$ satisfyingwhere $e$ is the identity element of $G$. The second rule means that $h$ moves the point first and $g$ moves it next. These rules let group multiplication describe consistent rearrangements of a set; see Ackerman's algebra notes, definition 0.4.1.1.

For each $g$, the function $x\mapsto g\cdot x$ is a bijection: its inverse is $x\mapsto g^{-1}\cdot x$, using the inverse element of $g$. Equivalently, a group action is a group homomorphism $\rho :G\to \mathrm{Sym}(X)$ into the symmetric group of all permutations of $X$, with $\rho (g)(x)=g\cdot x$ and $\rho (gh)=\rho (g)\circ \rho (h)$. See MIT's group-action notes, definition 23.1.

For a concrete group action, put the vertices of a polygon of a square at $A=(1,1)$, $B=(-1,1)$, $C=(-1,-1)$ and $D=(1,-1)$. Let $r(x,y)=(-y,x)$ be a counterclockwise quarter-turn rotation and let $s(x,y)=(y,x)$ be a reflection. Their eight permutations of the vertices of a polygon form the dihedral groupHere $D_4$ means the symmetry group of a four-sided polygon, with group order eight; the alternative convention calls this same group $D_8$. Products apply the rightmost operation first. For example, $s\cdot B=D$ and $(rs)\cdot B=r\cdot D=A$. The same group also has separate group actions on the sides of a polygon, the diagonals of a polygon, and the singleton set containing the centroid of the square. Specifying $X$ matters: changing what is moved changes the orbits of a group action and the stabilizer subgroups.

A group action need not be a faithful group action. For example, the cyclic group $C_4=\mathbb{Z}/4\mathbb{Z}$ acts on $X=\{0,1\}$ by $k\cdot x=x+k(\mathrm{mod}2)$, using modular arithmetic. Only the parity of $k$ matters: $0,2$ fix both points and $1,3$ exchange them. The associated group homomorphism $C_4\to \mathrm{Sym}(X)$ has group kernel $\{0,2\}$, of group order two. Either point has a group orbit of size two and a stabilizer subgroup of group order two, so the orbit-stabilizer theorem gives $4=2\cdot 2$.

The useful questions are which positions can be exchanged, and which transformations survive after imposing a constraint such as keeping one marked point fixed. An orbit of a group action answers the first question; a stabilizer subgroup answers the second. This turns a list of transformations into a count of distinct configurations, and explains why several transformations can produce the same configuration.

For a finite group $G$, transitive finite group actions are classified up to equivalence by conjugacy classes of subgroups: the class of $H$ gives the coset action on $G/H$. Every finite action decomposes uniquely as a disjoint union of its orbits, so its equivalence type is a multiset of these transitive types. Consequently the coefficient of $t^n$ in the displayed product, with one factor for every subgroup conjugacy class, counts actions on $n$ points up to relabeling. Each factor expands as a geometric series recording any multiplicity of that orbit type. For both $C_6$ and $S_3$ there is exactly one transitive type of each size $1,2,3,6$; their coefficient at degree six is $4+2+1+1=8$.

Suppose a binary operation on the underlying set $\mathbb{Z}$ makes it a group, and $\rho (n,x)=x^n$ is a group action on another group $G$. Every acting element must induce a bijection, with inverse function given by its acting inverse element. The integer $0$ instead induces the constant function $x\mapsto e$, where $e$ is the identity element of $G$, so $G$ must be the trivial group. Conversely, on the trivial group, ordinary addition on $\mathbb{Z}$ supplies such a group action.

A group action on a set is n-transitive when it can send any ordered tuple of $n$ distinct points to any other such tuple. Here ordered means a sequence of entries, not increasing in a total order. For order-preserving actions the corresponding notion is order n-transitivity.

For a group of Euclidean isometries, every element preserves the displayed union of images of a set, because left multiplication permutes its group indices. Its full setwise isometry group can be larger even when $T$ itself has no nonidentity symmetry. Take $G$ to be the integer translations of the plane and $T=\{(0,0),(1,0),(0,2)\}$. By rigidity of a scalene triangle, $T$ has trivial stabilizer subgroup. Yet $S={\mathbb{Z}}^2$, whose quarter-turn about the origin is a symmetry outside the translation group.

In the right multiplication action, the orbit of the coset $H$ under $K$ is $\{Hk:k\in K\}$. It is the whole right coset space exactly when every $g$ has form $hk$, proving the displayed criterion. The factorization need not be unique; for finite groups $|HK|=|H||K|/|H\cap K|$.

A circle action is a group action by $S^1$. Hamiltonian circle actions are especially useful in symplectic reduction.

A symmetry group collects bijections of an object to itself that preserve a specified structure in both directions, with function composition as the operation. For a geometric object one may require Euclidean distances to be preserved, giving an isometry group; for a graph, one preserves adjacency, giving a graph automorphism group. For an unstructured finite set, every permutation is allowed. The structure to be preserved is part of the definition.

For a square with its usual geometric structure, the symmetry group consists of four rotations and four reflections. These form the dihedral group of group order eight. The existing group action example shows how its eight elements move the vertices of a polygon; changing the object to its edges or a marked configuration changes the group action being studied.^[ref]

Specifying the preserved data makes it possible to test a proposed symmetry exactly. After the preserved data are declared, function composition, the identity map, and inverse maps automatically preserve them too. Symmetry transformations therefore form a group, and questions about equivalent configurations become questions about group orbits, stabilizer subgroups, and group homomorphisms. This supplies reusable mathematics for comparing many different kinds of objects.

A symmetry transformation is a bijection preserving the data that define the object's symmetry group. Its individual group elements are symmetry transformations; the group operation combines them by function composition.

Take the four-point setthe vertices of a polygon of a rectangle whose side lengths are four and two. Three choices of structure give three different symmetry groups on these same points.

As an unstructured set, $R$ has all $4!=24$ permutations as symmetries, forming the symmetric group $S_4$. If Euclidean distance is part of the data, the allowed maps are only the identity, a half-turn rotation, and the reflections across the two coordinate axes. They form a Klein four-group of group order four. To see that nothing is missing, fix the image of one vertex of a polygon: its three other distances, $2$, $4$, and $\sqrt{20}$, are distinct, so they uniquely determine the images of the remaining vertices.

Preserving affine structure permits more. The affine mapcycles all four points of $R$. It sends a horizontal side of length four to a vertical side of length two, so it is not an isometry. Yet it preserves the rectangle's vertex set and all affine combinations among those points.

Let $A=\mathrm{diag}(2,1)$ and let $r(x,y)=(-y,x)$ be the square's quarter-turn rotation. ThenConjugating every square symmetry by $A$ gives eight affine maps preserving $R$. This group isomorphism makes the affine symmetry group a dihedral group, although some of its maps do not preserve the original Euclidean distances. There are no further affine symmetries: an affine map preserves the common midpoint of the diagonals, so it must preserve opposite pairs. The image of one vertex has four choices and the image of an adjacent vertex has two; those choices determine the map through three noncollinear points.

ThusAn arbitrary permutation need not be an affine map. For example, exchanging the two top vertices while fixing the two bottom vertices sends the common diagonal midpoint to two different points, which no single affine map can do. The phrase "the symmetry of these points" becomes unambiguous only after stating which structure matters.

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.

Suppose a permutation group $G$ acts on an underlying atom set $A$ and on a collection of objects built over it. A finite subset $S\subseteq A$ supports an object $x$ when every permutation fixing all members of $S$ fixes $x$. Pairing supported objects uses the union of their supports. A definable subset or functional image inherits a support from its domain and parameters. In a hereditarily finite-supported Quine-atom model, every member is recursively required to have such a support; symmetry then rules out a choice function on the pairs of atoms.

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.

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.

For a group action, a fundamental domain selects one representative from each orbit. Geometric fundamental domains are often taken with their boundary included, allowing boundary overlaps while their interiors remain disjoint under the action. For translations of the real line by the integers, $[0,1)$ is an exact representative set and $[0,1]$ is its closed geometric version. A lattice similarly gives a half-open fundamental parallelepiped.

A group action is $t$-transitive if it acts transitively on ordered $t$-tuples of distinct points. In degree $n\ge t$ this implies $n(n-1)\cdots (n-t+1)\mid |G|$. It is sharply t-transitive when each such ordered tuple has trivial stabilizer subgroup.

Transitivity on ordered triples of distinct points. It implies a two-transitive group action, but is stronger: in projective dimension at least two, collinear and noncollinear triples distinguish two orbits, even though the projective action is two-transitive.

For nontrivial centerless $T$, this action of $T\times T$ is faithful and transitive. Its identity stabilizer is the diagonal subgroup. Subgroups above that diagonal are $\{(a,b):a^{-1}b\in N\}$ with $N⊴T$: multiply by a diagonal element to isolate the second coordinate, then use diagonal conjugation. Thus the action is primitive exactly when $T$ is nonabelian simple.

For a group acting on a ring by ring automorphisms, the elements fixed by every group element form a subring containing $1$. For finite $G$ acting on a commutative ring, each $r$ satisfies the monic orbit equation ${\prod }_{g\in G}(T-g(r))$, whose coefficients are fixed. Thus $R$ is integral over its invariant subring. This uses an orbit product, not division by the group order.

A fraction of invariant elements is fixed by every automorphism. If that fraction is integral over $R^G$, its monic equation also makes it integral over $R$. A normal domain therefore places it in $R$, and fixedness places it in $R^G$. This proves integral closedness of the invariant subring in its own fraction field. The proof works for any group and in every characteristic; no averaging or finiteness of the group is needed for normality.

A transitive group action is primitive if the only blocks are singletons and the whole set; a block has every translate either equal to it or disjoint from it. Orbits of a normal subgroup form blocks. Hence a nontrivial normal subgroup in a faithful primitive action is transitive. Two-transitivity implies primitivity. This normal-subgroup observation is a hypothesis engine for Iwasawa's simplicity lemma.

If a primitive subgroup of $S_n$ contains a transposition, make a graph whose edges are the supports of its conjugate transpositions. Connected components form blocks for the group. An edge exists, so primitivity forces the graph to be connected. Its edge transpositions generate all of $S_n$, proving that the original group is $S_n$.

Every nontrivial normal subgroup of a faithful primitive group of prime degree is transitive and contains a Sylow $p$-subgroup $P$. Its normalizer in $A_p$ has order $pq$. For a proper normal subgroup $N$, the Frattini argument gives $G=N{N}_G(P)$; unless $G=C_p$, its $P$-normalizer is the full order-$pq$ normalizer, while $N_N(P)=P$. The Burnside transfer theorem gives a characteristic normal $p$-complement in $N$. This complement is normal in $G$ and cannot be transitive because its order is prime to $p$, so it is trivial. Thus $N=P$ and $|G|=pq$.

The theorem classifies finite primitive permutation groups by their minimal normal subgroups and the action of their socle. The modern types are affine (HA), almost simple (AS), simple diagonal (SD), compound diagonal (CD), product action (PA), holomorph simple (HS), holomorph compound (HC), and twisted wreath (TW). A coarser version describes primitive maximal subgroups of symmetric groups by affine, almost-simple, diagonal and full product-action overgroups. Intransitive and imprimitive maxima are respectively unequal-part subset stabilizers and block-system stabilizers. Primitivity alone is not maximality in the symmetric group; parity can put an otherwise natural candidate entirely inside the alternating group.

For a nonabelian simple group $T$, act on cosets of its full diagonal in $T^r$. A suitable factor-permuting extension is primitive and has point stabilizer intersecting the socle in the diagonal. The full normalizer has order $|T{|}^r|\mathrm{Out}(T)|r!$. For $T=A_5,r=2$, it acts on the sixty elements of $A_5$ by left/right multiplication, automorphisms and inversion, and has order $14400$. Odd conjugation in $S_5$ induces an odd permutation of these sixty points, while inversion is even; the full group is a primitive maximal subgroup of $S_{60}$.

An elementary abelian regular minimal normal subgroup acts as translations of a vector space. Its point stabilizer acts faithfully and irreducibly on that space; invariant subspaces would be blocks. The full affine normalizer is $AG{L}_d(p)$. In degree sixteen it lies inside $A_{16}$: nonzero translations swap eight pairs, and elementary linear transvections swap four pairs. Thus it is not a maximal subgroup of $S_{16}$.

The group of maps $x\mapsto Ax+b$ of a finite-dimensional vector space, with $A$ invertible. Identifying the underlying points with a regular elementary abelian group identifies its normalizer in the symmetric group with this semidirect product.

Conjugate the three-cycle throughout the group. Connected components of the support hypergraph form an invariant block system. Primitivity makes it connected, and connected triple supports generate an alternating group gives the conclusion.

The symmetric group acts on multivariate polynomials by replacing each variable $x_i$ by $x_{\sigma (i)}$. With $(\sigma \tau )(i)=\sigma (\tau (i))$, substitution gives $\sigma \cdot (\tau \cdot f)=(\sigma \tau )\cdot f$. Thus it is a left group action. A polynomial encoding a combinatorial partition has a group orbit encoding all relabelings and a stabiliser subgroup encoding its label symmetries.

A fixed point of a group action is an element fixed by every group element. The set $A^G=\{a:g\cdot a=a\text{ for all }g\in G\}$ is the categorical limit of the action viewed as a functor from the one-object category associated with $G$ to the Category of sets.

The left regular action of a group on its own underlying set is left multiplication. It is faithful, because an element fixing the identity is itself the identity. For a finite group it gives the permutation representation used in Cayley theorem.

A block system for a group action is a partition of the acted-on set such that every group element permutes its parts. For a transitive action, a nontrivial block system has more than one part and more than one element per part. The full symmetric group preserves no such partition: a transposition exchanging one point in each of two parts breaks it.

For a transitive finite permutation group, suppose each point stabilizer has orbits of a common size $r$ on the other points. If a nontrivial block system exists, all two-point stabilizers are trivial. Indeed for block size $b$, $r\mid n-1$ gives $\gcd (r,b)=1$. An orbit of a block not containing $\alpha$ has size at most $br$ and divisible by $br$, so its b point-orbits are disjoint. A stabilizer of $\alpha$ and a point of that block fixes the whole block. Interchanging the two points also fixes the block of $\alpha$. Equal two-point-stabilizer orders then show that each two-point stabilizer fixes every point, and faithfulness makes it trivial.

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]

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 transitive group action is regular if every point stabilizer is trivial. Equivalently each ordered pair of points is connected by exactly one group element. Identifying one point with the identity identifies the action with left multiplication on the group. A regular normal subgroup inside a larger primitive group can supply translations or nonabelian left multiplication; the whole primitive group usually has a nontrivial stabilizer acting by automorphisms of this regular subgroup.

A derangement is an element fixing none of the points of a group action. Every finite transitive action on at least two points has a derangement. Count pairs $(g,x)$ with $gx=x$: transitivity gives ${\sum }_g|\mathrm{Fix}(g)|={\sum }_x|G_x|=|G|$. If every element fixed a point, the identity's extra fixed points would make this sum strictly larger than $|G|$. This is the Burnside lemma counting argument specialized to existence of a derangement.

Inclusion-exclusion applied to the events that a permutation fixes a specified point gives the displayed formula. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ choices of $k$ fixed points and $(n-k)!$ permutations fixing them. The probability that a uniformly chosen permutation is a derangement tends to $e^{-1}$.

// Target: analysis.bigb

The number of point-stabilizer orbits, equivalently the number of orbits on ordered pairs. On $k$-spaces of an $n$-space with $k\le n/2$, $G{L}_n(q)$ has rank $k+1$, indexed by intersection dimension.

A regular permutation subgroup acts transitively with trivial point stabilizers. Each ordered pair of points is related by exactly one subgroup element. A finite regular subgroup on $n$ points has order $n$. Normality and characteristicity are additional properties; they do not follow from regularity alone.

Identify a regular orbit with the group using a chosen base point. A permutation commuting with all left translations satisfies $c(x)=xc(1)$. Thus the centralizer consists of right translations. Defining $R_g(x)=x{g}^{-1}$ makes $R_g{R}_h=R_{gh}$ and gives an isomorphism to $G$, rather than an unmentioned reversal of multiplication.

An action is sharply $t$-transitive if each ordered $t$-tuple of distinct points is carried to any other by exactly one group element. For finite degree $n\ge t$, its order is $n(n-1)\cdots (n-t+1)$ and each ordered-tuple stabilizer is trivial. Sharp one-transitivity is a regular action. Finite sharp two-transitivity gives a regular kernel of a finite sharply two-transitive group.

Exactly one element sends each ordered triple of distinct points to each other such triple. On a projective line, representatives of the first two target points form a basis; scaling them so their sum represents the third point gives the unique projective transformation.

A group action is sharply two-transitive when exactly one element sends each ordered distinct pair to each other ordered distinct pair. The affine group over a finite field provides an example: solve $ax+b=u$ and $ay+b=v$ uniquely, giving $a=(v-u)/(y-x)$.

In a finite sharply two-transitive group, the identity together with all fixed-point-free elements forms a regular normal subgroup. One proof constructs virtual characters from induced point-stabilizer characters and the augmentation character, proves their norms are one, and realizes this set as an intersection of representation kernels. Conjugation by the point stabilizer is transitive on the kernel's nonidentity elements. This forces the kernel to be elementary abelian, so it is a unique Sylow subgroup and is characteristic. Closure of the fixed-point-free set is a theorem, not something supplied merely by counting its size.

Adjoin a new point $\omega$ fixed by a transitive group $G$. A one-point extension is a transitive group on the enlarged set whose stabilizer of $\omega$ is exactly $G$, acting on the remaining points in the prescribed way. Its order is $(n+1)|G|$ in the finite degree-$n$ case. The double-coset criterion for a one-point extension gives a concrete generator test.

If $x$ interchanges the new point and $\alpha$, put $H=G_{\alpha }$. Then $⟨G,x⟩$ is a one-point extension exactly when $x^2\in H$, $xH{x}^{-1}=H$, and $xgx\in GxG$ for every $g\notin H$. These conditions make $G\cup GxG$ multiplication-closed, hence a group in the finite setting, with new-point stabilizer $G$. Necessity comes from the two double cosets in the extended two-transitive action.

A transitive subgroup $G\le S_p$ for prime $p$ that contains a transposition is $S_p$. Form a graph on the $p$ letters with edges given by all conjugates in $G$ of that transposition. The transitive action permutes connected components, so their common size divides $p$. An edge rules out size one, hence the graph is connected. Its edge transpositions generate $S_p$: successive swaps along a path generate the swap of its endpoint with any chosen start. All these edge transpositions belong to $G$.

A group action on a set is two-transitive when it acts transitively on ordered pairs of distinct points. Equivalently, a point stabilizer acts transitively on the remaining points.

This is the $k=2$ case of a multiply transitive group action.

A point stabilizer in a doubly transitive group acts transitively by conjugation on the nonidentity elements of a regular normal subgroup $N$. All those elements have the same order; Cauchy's theorem makes this a prime p and excludes other primes from $|N|$. A finite p-group has a nontrivial characteristic center. Transitivity then forces all of $N$ into its center, so $N$ is elementary abelian. Choosing a vector-space basis identifies the original permutation group with a subgroup of the corresponding general affine group.

There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$ partitions of six letters into two unordered triples. A stabilizer is $S_3\wr S_2$, also the normalizer of the Sylow subgroup generated by one 3-cycle in each triple. Relative to one partition, all other partitions have intersection pattern $1+2$ up to interchange, giving one orbit of the stabilizer and hence a two-transitive action. A natural $S_5$ fixing one letter is transitive on these partitions, giving $S_6=S_5(S_3\wr S_2)$.

A minimal normal subgroup of a finite faithful two-transitive group action is either regular elementary abelian or primitive nonabelian simple. The imprimitive case reduces to two-point stabilizers being trivial, then counting derangements gives a characteristic regular Sylow subgroup. In the primitive nonregular case, each simple direct factor is transitive. Two commuting transitive factors are regular and are mutual full centralizers. Thus at most two factors occur and $|N|=n^2$. But the common subdegree of $N_{\alpha }$ then divides both $n$ and $n-1$, forcing $N_{\alpha }=1$, a contradiction. Only one factor remains.

Translations are regular on $V$, and the zero stabilizer is $H$. Two-transitivity is equivalent to $H$ being transitive on nonzero vectors. Full affine linear groups and groups with special linear or symplectic linear parts give examples.

For a group $G$ acting on a set $X$ and a point $x\in X$, the orbit map is $G\to X$, $g\mapsto gx$. Its image is the orbit of a group action of $x$, and its fibres are cosets of the stabilizer subgroup of $x$.

For $G$-sets $X$ and $Y$, a map $f:X\to Y$ is equivariant when $f(gx)=gf(x)$ for every $g\in G$ and $x\in X$.

Translating an input translates its output under a map rather than necessarily leaving the output unchanged. Convolution has this property under consistent domain and boundary conventions.

A group acts on itself and on its subgroups by conjugation: $g\cdot x=gx{g}^{-1}$ and $g\cdot H=gH{g}^{-1}$.

For a normal subgroup $K$ of $G$, each $c_h:k\mapsto h^{-1}kh$ is a group automorphism of $K$. Its inverse is $c_{h^{-1}}$, but the dependence on $h$ is a group antihomomorphism under ordinary function composition, because $(gh{)}^{-1}k(gh)=h^{-1}(g^{-1}kg)h$. The assignment $h\mapsto (k\mapsto hk{h}^{-1})$ instead is a group homomorphism to the automorphism group. Equivalently the first convention is a right group action, while the second is a left group action.

For a proper subgroup $B$ of a finite group $G$, the number $m$ of its conjugates is $[G:N_G(B)]\le [G:B]$, because its normalizer contains $B$. All conjugates contain the identity, so their union has size at most $1+m(|B|-1)\le |G|-[G:B]+1<|G|$. Thus some element is not conjugate to any element of $B$. Finiteness is essential to this cardinal counting argument.

Conjugation by $g$ is the group automorphism $x\mapsto gx{g}^{-1}$. Its orbits are the conjugacy classes.

For a subset $S$ of a group and an element $g$, its conjugate subset isConjugation is a bijection and a group automorphism, so it preserves cardinality and every property expressed using the group operation.

Two group elements $x$ and $y$ are conjugate when $y=gx{g}^{-1}$ for some group element $g$, equivalently when they belong to the same conjugacy class. Conjugate elements have the same order of a group element.

Two permutations in the same symmetric group are conjugate if $\tau =\pi \sigma {\pi }^{-1}$ for some permutation $\pi$. Conjugation relabels every permutation cycle, so it preserves cycle type. Conversely, equal cycle types permit a relabelling between corresponding cycles, establishing conjugacy.

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 finite group $G$, write $k(G)$ for its number of conjugacy classes. Exactly the elements of the center of a group $Z(G)$ have singleton conjugacy classes; every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, $k(G)-|Z(G)|\le (|G|-|Z(G)|)/2$. If $G$ is a nonabelian group, the cyclic quotient by the center criterion and the fact that every group of group order at most three is a cyclic group give $[G:Z(G)]\ge 4$, hence $k(G)\le 5|G|/8$. The quaternion group attains this bound, with five conjugacy classes among eight elements.

A conjugacy class is real when it is stable under inversion of conjugacy classes. Equivalently, one, and therefore every, member is conjugate to its inverse. Members need not themselves be involutions.

For a group $G$, inversion sends the conjugacy class of $g$ to that of $g^{-1}$, because $(xg{x}^{-1}{)}^{-1}=x{g}^{-1}{x}^{-1}$. It therefore defines an involution on the collection of conjugacy classes, even in a noncommutative group where inversion is not a group homomorphism.

The symmetric group $S_3$ has exactly six permutations, written using permutation cycles:Here $e$ is the identity element; $(12)$, $(13)$ and $(23)$ are transpositions; and $(123)$ and $(132)$ are permutation cycles of length three. With $t=(12)$ and $r=(123)$, rightmost-first function composition gives $r^2=(132)$ and $r^{-1}=r^2$. The conjugacy classes and representative centralizers are

An explicit check of the conjugation $gt{g}^{-1}$ with $g=r=(123)$ sends $1\mapsto 3\mapsto 3\mapsto 1$, $2\mapsto 1\mapsto 2\mapsto 3$, and $3\mapsto 2\mapsto 1\mapsto 2$, where each chain applies $g^{-1}$, then $t$, then $g$. Thus $1$ is fixed and $2,3$ are exchanged: $gt{g}^{-1}=(23)$. Relabelling also produces $(13)$, so the three transpositions share one conjugacy class. A transposition reverses the length-three permutation cycle, since $tr{t}^{-1}=(213)=(132)=r^{-1}$, putting both length-three permutation cycles in one conjugacy class.

Checking each of the six possible permutations against $gx=xg$ gives the displayed centralizers. Only $e$ has a one-element conjugacy class, so the center of a group here is $Z(S_3)=\{e\}$. The class equation is $6=1+3+2$. The second diagram in centralizer and normalizer uses the same $r$ and $t$ to show how preserving a cyclic subgroup differs from fixing its group generator.