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Let $p<q$ be distinct prime numbers. Every group of group order $p^2{q}^2$ is Abelian if and only if $p$ does not divide $q^2-1$.

For sufficiency, the Sylow theorems make the number of Sylow p-subgroups one of $1,q,q^2$, congruent to one modulo $p$. The divisibility hypothesis excludes both nontrivial choices, so that subgroup is normal. The number of Sylow subgroups for $q$ is one of $1,p,p^2$, congruent to one modulo $q$. The choice $p$ is impossible because $p<q$. If $q$ divided $p^2-1=(p-1)(p+1)$, primality and $q>p$ would force $q=p+1$, hence $(p,q)=(2,3)$; this pair violates the hypothesis. Thus the other Sylow subgroup is normal too. Their intersection is trivial, and their group commutators lie in that intersection, so the group is their direct product of groups. A group of prime-square order is abelian: its nontrivial center of a group is either the whole group or has cyclic quotient, and a cyclic central quotient also forces commutativity. Both factors are therefore abelian.

For necessity, suppose $p\mid q^2-1$. The multiplicative group of a finite field is cyclic, so ${\mathbb{F}}_{q^2}^{\times }$ contains $\alpha$ of order $p$. Let the cyclic group $C_{p^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by mapping its generator to multiplication by $\alpha$. This is a well-defined nontrivial group action, since ${\alpha }^{p^2}=1$. The resulting semidirect product has order $p^2{q}^2$ and is nonabelian, because the generator does not commute with a nonzero field element. Thus a nonabelian group of this order exists exactly when $p\mid q^2-1$.

Let $D_8$ denote the dihedral group of group order eight acting on a square. Here the subscript specifies group order; the same group is called $D_4$ in the convention that counts vertices of a polygon, as in the group action example. Let $r$ be a counterclockwise quarter-turn rotation and $s$ a reflection in a diagonal. ThenProducts act rightmost first. The group commutator of the generators of a group isIt is a half-turn rotation, so forcing $r$ and $s$ to commute must identify a half-turn with $e$.

More concretely, write each element uniquely as $r^i{s}^j$, with $0\le i<4$ and $j\in \{0,1\}$. The functionis a surjective group homomorphism. Indeed, $srs=r^{-1}$ giveswhere the exponents are reduced modulo four and two. Modulo two, $-k$ and $k$ agree, so $\varphi$ adds the two coordinates separately. Its group kernel is $N=\{e,r^2\}$. Since the target is an abelian group, $D_8^{\prime }\subseteq N$; since $[r,s]=r^2$, also $N\subseteq D_8^{\prime }$. ThusThe four cosets are $\{e,r^2\}$, $\{r,r^3\}$, $\{s,r^{2}s\}$, and $\{rs,r^{3}s\}$. Each coset forgets the difference of a half-turn. The quotient group still distinguishes rotations from reflections, and even from odd numbers of quarter-turns; the two surviving choices are independent. This is the Klein four-group, rather than the cyclic group $C_4$, because each nonidentity element squares to the identity element. For the general dihedral group calculation, see Conrad, Dihedral groups I, theorem A.1; that source uses the vertex-count convention and $xy{x}^{-1}{y}^{-1}$ for individual group commutators.

A nonempty set $A$ equipped with a group action of the additive group of a vector space $V$, such that for every $x,y\in A$ there is a unique $v\in V$ with $x+v=y$. There is no distinguished origin. Choosing one identifies $A$ with $V$; changing it gives a translation.

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.

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 group action on $X$ and a contractible free $G$-space $EG$, the diagonal orbit space $EG{\times }_{G}X$ maps to the classifying space $BG$ with fibre $X$. It defines equivariant cohomology. When $X\to X/G$ is a principal $G$-bundle, the map $EG{\times }_{G}X\to X/G$ has contractible fibre $EG$ and gives a homotopy equivalence for spaces of CW type. Freeness without the bundle hypotheses should not be substituted for this assertion for arbitrary topological groups.

A Cayley graph turns multiplication in a group into travel: the chosen generators of a group are the allowed one-step moves, and a word over an alphabet becomes a route. This lets us study a finitely generated group using graph distances, closed balls, and geodesics, rather than only equations between products.

Let $G$ be a group with a finite generating set of a group $S$ satisfying $S=S^{-1}$ and $1\notin S$. Such a symmetric subset of a group includes the inverse element of every allowed move. The Cayley graph ${\mathrm{Cay}}_S(G)$ has vertex set $G$ and joins $g$ to $gs$ for each $s\in S$. We choose right multiplication for these moves. Reversing the move uses $s^{-1}$, so for graph distance we regard $g\to gs$ and $gs\to g$ as the two orientations of one undirected edge. An involution therefore gives one edge, not two parallel edges. Each edge has length one, regardless of its length in a drawing.^[ref]

A sequence $s_1,\dots ,s_n$ labels the walkIt ends at $h$ exactly when $g^{-1}h=s_1\cdots s_n$. Thus the Cayley graph is a connected graph precisely because $S$ generates $G$, and its graph distance on vertices is the word metric. Left multiplication by $a$ sends the edge from $g$ to $gs$ to the edge from $ag$ to $(ag)s$, preserving its generator label: this is a group action by graph automorphisms and isometries. Viewing groups as symmetries of geometric objects is the motivation of Clay and Margalit's introduction.^[ref]

The useful geometric object includes its allowed moves, not just its drawing. Cayley graphs of the integers show how changing the moves adds shortcuts; word metric balls in the integer lattice show how the same group can acquire visibly different closed balls. Meier's book develops this passage from group operations to graphs as its central motivation.^[ref]

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]

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

The coadjoint action of a Lie group on the dual space of its Lie algebra is $({\mathrm{Ad}}_g^{\ast }\eta )(X)=\eta ({\mathrm{Ad}}_{g^{-1}}X)$. The inverse in this formula makes it a left group action.

An isometric group action on a metric space is cobounded if some orbit is within a fixed distance $R$ of every point. Equivalently, translates of one bounded ball cover the space. Coboundedness does not require finite stabilizers, a proper space, or a finitely generated group. Every action on a point is cobounded.

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 a subgroup $H\le G$ of a group $G$, a left coset is $gH=\{gh:h\in H\}$; a right coset is $Hg=\{hg:h\in H\}$. The name refers to the side on which the representative $g$ multiplies $H$. Think of a coset as a translated copy of a subgroup: it remembers the relative differences allowed by $H$, while moving the whole collection to another location in the group.

Two left cosets of the same subgroup are either equal or disjoint. Thus distinct left cosets form a partition of a set covering $G$; the same statement holds separately for right cosets. The useful membership test isThis also defines an equivalence relation: changing a representative by right multiplication with a member of $H$ leaves its left coset unchanged. These facts hold even for infinite groups. See MIT's coset notes, definitions 3.6 and 3.9.

For an explicit additive example, take the cyclic group $G=\mathbb{Z}/6\mathbb{Z}$, with modular arithmetic, and the subgroup $H=\{0,3\}$. Adding each representative to both members gives exactlyRepresentatives $3,4,5$ repeat these three cosets, respectively. For example, $4+H=\{4,1\}=1+H$, since $4-1=3\in H$. Every residue appears once in the displayed partition of a set. Each coset has two members, so the group order of $H$ is $2$ and its index of a subgroup is $3$. Left and right cosets coincide here because $G$ is an abelian group.

A coset need not be a subgroup. In this example, $\{1,4\}$ lacks the identity element $0$, and $1+1=2$ leaves it. In fact, the only left or right coset of $H$ that is a subgroup is $H$ itself: a subgroup must contain the identity element, and the coset containing that element is $H$. Equal sizes come instead from the bijection $h\mapsto gh$, whose inverse is $x\mapsto g^{-1}x$; for right cosets, use $h\mapsto hg$ and $x\mapsto x{g}^{-1}$.

The nonabelian group $S_3$, the symmetric group on three letters, separates the two sides. Use function composition with the rightmost permutation applied first, and let $H=\{e,(12)\}$. Then $(13)(12)=(123)$ but $(12)(13)=(132)$, giving the differing pairThe complete partition of a set into left cosets consists of $H$, $\{(13),(123)\}$ and $\{(23),(132)\}$; the complete partition of a set into right cosets consists of $H$, $\{(13),(132)\}$ and $\{(23),(123)\}$. Each lists all six permutations exactly once. Both have three parts of size two, although their parts differ.

Left cosets support a group action by $a\cdot (gH)=(ag)H$ for every subgroup, as in the coset action. Multiplication of representatives defines a quotient group operation $(aH)(bH)=(ab)H$ precisely when $H$ is a normal subgroup. Otherwise the answer can depend on the representatives: in this $S_3$ example, $e$ and $(12)$ represent the same coset $H$, yet $e(13)H=(13)H$ and $(12)(13)H=(23)H$ differ. In the additive example, $H$ is a normal subgroup, making $(1+H)+(2+H)=3+H=H$ valid. See MIT's quotient-group notes, theorem 9.1.

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.

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 $4+2+1+1=8$.

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.

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.

The diagonal group action of the symmetric group on ordered pairs has two orbits of a group action when $n\ge 2$: equal pairs, of size $n$, and distinct pairs, of size $n(n-1)$. Equality is preserved by a bijection, and a prescribed bijection on one or two distinct letters extends to a permutation of all letters. At $n=1$ only the equal-pair orbit remains.

The Borel version of equivariant cohomology is the ordinary cohomology of the Borel construction. It retains information about both the space and its group action. For a point it is the cohomology of the classifying space; for a free action with the usual bundle hypotheses it is the cohomology of the orbit space.

An estimator respects the group action if the displayed identity holds. If it takes values in the transformation group itself, normalization $T(y{)}^{-1}y$ is a maximal invariant: equivariance proves invariance, and equality of normalized samples reconstructs the transformation between them. For estimates in a homogeneous parameter space with a nontrivial stabilizer, normalization is defined only modulo that stabilizer; quotienting the remaining action is essential.