An orthogonal representation is a real group representation preserving a positive-definite inner product, equivalently a group homomorphism into an orthogonal group. Every finite-dimensional continuous real representation of a compact group has this form after averaging an inner product over Haar measure, by orthogonalization of a compact-group representation.

A projective representation is a group homomorphism into the projective linear group of a vector space. Choosing invertible linear representatives makes multiplication hold up to nonzero scalar factors. A projective unitary representation chooses unitary operators as representatives, whose scalar factors have modulus one.

A Tietze transformation changes a group presentation without changing the presented group, by adding or removing a consequence of the relators, or by adding or eliminating a generator together with a defining equation expressing it in the others. For example a generator $t$ defined by $t=xs{x}^{-1}$ can be eliminated by substitution in every remaining relator.

For a fixed finitely generated group with specified generators, its word problem asks for an algorithm deciding whether an input word represents the identity. For a finitely presented group, positive answers can be enumerated from the relators, but there need not be an algorithm deciding both positive and negative answers. An effective reduction from a nonrecursive halting set proves unsolvability.

A recursive presentation has an effectively enumerable alphabet of generators and a recursively enumerable set of relators. Usually the alphabet is finite; when it is countably infinite this is specified explicitly. Enumerating finite products of conjugates of relators and their inverses enumerates all null words. Conversely the enumerable null words can themselves serve as the relators. Recursive presentability does not imply a soluble word problem for a group.

A group homomorphism between abelian groups, conventionally written additively. Its kernel and image are subgroups, and the additive formula implies preservation of zero, negation and every integer multiple. Maps on Mordell-Weil groups induced by isogenies of elliptic curves and by good reduction are examples.

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]

For $G=⟨g_i\mid R⟩$, put $B=G\ast F(a,b)$, $u_0=a$, $v_0=b$, $u_i=b^{-i}a{b}^i$ and $v_i=g_i{a}^{-i}b{a}^i$ for $i\ge 1$. The $u_i$ freely generate by free conjugate bases in a rank-two free group. The $v_i$ do too: projection killing $G$ maps them to a free conjugate basis, so no reduced word in them is trivial. The HNN extension adjoining $t^{-1}{u}_{i}t=v_i$ embeds $G$ by Britton's lemma. Since $b=t^{-1}at$ and $g_i=t^{-1}{u}_{i}t(a^{-i}b{a}^i{)}^{-1}$, its generators reduce to $a,t$. For finitely many $g_i$ and $R$ this is a finite group presentation; for effective countably many generators and enumerable $R$ it is a two-generator recursive presentation of a group.

For a topological group module, use continuous maps with pointwise addition and the inhomogeneous group-cohomology differential. The kernel modulo the image defines continuous group cohomology as an abelian group. With general topological coefficients the quotient topology need not be Hausdorff; the cohomology here is the algebraic quotient of continuous cocycles by continuous coboundaries. With trivial discrete coefficients in a prime field, its first cohomology is the space of continuous group homomorphisms to that field's additive group.

For a finite group $G$, Tate cohomology joins group cohomology and group homology through the norm map $N_G={\sum }_{g\in G}g$. For $r\ge 1$ it equals ordinary group cohomology; for $r\le -2$ it equals $H_{-r-1}(G,M)$. The middle terms are ${\hat{H}}^0(G,M)=M^G/N_{G}M$ and ${\hat{H}}^{-1}(G,M)=\ker (N_G)/I_{G}M$, where $I_{G}M$ is generated by $gm-m$. For the trivial module $\mathbb{Z}$, ${\hat{H}}^{-2}(G,\mathbb{Z})=G^{\mathrm{ab}}$. With multiplicative coefficients the norm is the product of the translates.

Embed a module $M$ over the group ring $\mathbb{Z}[G]$ in a module $I$ with vanishing positive group cohomology, and write $M^{\prime }=I/M$. The connecting map in the long exact group cohomology sequence of $0\to M\to I\to M^{\prime }\to 0$ is an isomorphism $H^r(G,M^{\prime })\to H^{r+1}(G,M)$ for $r\ge 1$, since the groups involving $I$ on both sides vanish. For a finite group and a module $I$ with vanishing Tate cohomology of a finite group in every degree, the same argument shifts every Tate degree. Repeating this construction transports computations to a convenient resolution.

For a prime $p$, a classifying Eilenberg–MacLane space for $C_p$ is the infinite lens space $S^{\infty }/C_p$, using scalar multiplication by a primitive $p$th root on the complex-coordinate sphere. Its contractible universal cover is $S^{\infty }$. The standard two-periodic free resolution has differentials $g-1$ and $1+g+\cdots +g^{p-1}$, both zero after applying cochains with trivial ${\mathbb{F}}_p$ coefficients. Thus every cohomology degree has dimension one. If $p=2$, the ring is ${\mathbb{F}}_2[t]$, $|t|=1$. If $p$ is odd, it is $\Lambda (a)\otimes {\mathbb{F}}_p[b]$, with $|a|=1$, $|b|=2$: graded commutativity gives $a^2=0$ and the periodicity class $b$ induces the degree-two isomorphisms. The Künneth theorem gives two copies of these generators for $C_p\times C_p$; the dimension in degree $d$ is $d+1$ for every $p$.

A Poincare duality group of dimension $n$ has a finite-length resolution of the trivial group ring module $\mathbb{Z}$ by finitely generated projective modules, and $H^k(G,\mathbb{Z}G)$ vanishes except in degree $n$, where it is infinite cyclic as an abelian group. The top module $D$ can have an orientation-sign action. Reversing the dual projective resolution gives $H^k(G,V)\cong {\mathrm{Tor}}_{n-k}^{\mathbb{Z}G}(D,V)$, equivalently $H_{n-k}(G,D\otimes V)$ with the appropriate diagonal action. Untwisted duality requires trivial orientation action. In positive dimension $H^0(G,\mathbb{Z}G)=0$, since a finitely supported invariant element cannot exist for an infinite group.

The integral cohomological dimension is the least length of a projective resolution of the trivial group ring module $\mathbb{Z}$, equivalently the largest degree in which group cohomology with some module can be nonzero. A group containing torsion has infinite integral cohomological dimension. For ${\mathbb{Z}}^r$ the dimension is $r$, realized by its torus classifying space and detected by its top cohomology.

For a connected CW complex $X$ with contractible universal cover, let $\Gamma ={\pi }_1(X)$ and $R=\mathbb{Z}\Gamma$. Lift each cell once and transport its orientation by deck transformations. Its translates form an $R$-basis of the cellular chains of the cover. The augmentation sends every vertex to one, and contractibility makes the augmented chain complex an exact free resolution of the trivial module $\mathbb{Z}$. An $R$-linear cochain with trivial coefficient action is constant on each orbit of lifts, identifying ${\mathrm{Hom}}_R(C_{\ast }(\tilde{X}),\mathbb{Z})$ with the cellular cohomology complex of $X$. Consequently $H^i(X;\mathbb{Z})={\mathrm{Ext}}_R^i(\mathbb{Z},\mathbb{Z})$. The same argument with singular simplex lifts works for spaces with a contractible universal cover.

For a group action of a group $G$ on a set $X$, the orbit of a group action of $x\in X$ is the subset of all positions reachable from $x$:Two points belong to the same group orbit exactly when one can be moved to the other by a group element. This is an equivalence relation: the identity element gives $x\sim x$, the inverse element of a transformation reverses a move, and a product combines two moves. Hence the group orbits form a partition of a set of $X$ into equivalence classes. A transitive group action on a nonempty set has just one group orbit. See Ackerman's algebra notes, lemma 0.5.1.3, for the orbit partition.

For the square and the eight-element dihedral group $D_4$ defined under group action, direct enumeration gives the following separate group orbits:

These counts depend on the chosen group action, not just the group. An orbit of a group action describes all configurations equivalent under the permitted transformations. For ordered pairs under the symmetric group, equality of the two entries is preserved, separating the two group orbits in equality-pattern orbits of ordered pairs; the orbit-stabilizer theorem computes their sizes when the symmetric group is $S_3$.

For nontrivial centerless $T$, the displayed right group 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.

The convention is right-handed: writing $t^{(a,b)}=a^{-1}tb$, one has $(t^{(a,b)}{)}^{(c,d)}=t^{(ac,bd)}$. With left group action notation, the corresponding formula is $(a,b)\cdot t=at{b}^{-1}$; it has the same diagonal stabilizer subgroup.

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.

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 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.