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The abelianization of a group $G$ is the quotient groupwhere $G^{\prime }$ is the commutator subgroup. Its elements are cosets, and its operation is $(g{G}^{\prime })(h{G}^{\prime })=gh{G}^{\prime }$. The result is an abelian group: the two products $gh$ and $hg$ differ by a group commutator, which disappears in the quotient group.

For the symmetric group $S_3$, the sign of a permutation realizes $S_3^{\mathrm{ab}}\cong C_2$, as shown in commutator subgroup of the symmetric group on three letters. For the dihedral group of a square, abelianization of the square symmetry group gives $C_2\times C_2$: two independent yes-or-no pieces of information survive. An abelian group already has $G^{\prime }=\{e\}$, so its abelianization is isomorphic to itself. At the other extreme, a perfect group has trivial abelianization.

Calling this the largest abelian quotient means that every group homomorphism into any abelian group factors uniquely through $q$. It does not require $G$ to be finite, or compare the sizes of unrelated abelian groups. The universal property of abelianization makes this precise. It lets us solve all such mapping problems in one simpler group: any invariant that respects multiplication and takes values in an abelian group sees only $G^{\mathrm{ab}}$.

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.

An abelian quotient is a quotient group $G/N$ that is an abelian group. It exists only for a normal subgroup $N$, and it is abelian exactly when $N$ contains the commutator subgroup. Thus every abelian quotient factors through the abelianization $G/[G,G]$.

The absolute residue degree is the degree of the residue field over its prime field. For a tower above $p$, degrees multiply: $[k_{\mathfrak{q}}:{\mathbb{F}}_p]=[k_{\mathfrak{q}}:k_{\mathfrak{p}}][k_{\mathfrak{p}}:{\mathbb{F}}_p]$. In a Galois extension over $\mathbb{Q}$, the decomposition group has a cyclic quotient group of this order by its inertia group. This can exclude residue degrees even when the whole extension has that degree: $S_3$ has no cyclic quotient group of order six.

For a subgroup $K$ of a finite abelian group $G$, its annihilator isThese are exactly the linear characters that descend to the quotient group $G/K$. The character group of a finite abelian group has the same size as the group, so $|K^{\perp }|=|G/K|=|G|/|K|$. The annihilator is what abelian hidden-subgroup Fourier sampling measures.

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 center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

In coordinates $h(x,y,z)=\left(\begin{array}{ccc}1& x& y\\ 0& 1& z\\ 0& 0& 1\end{array}\right)$, the Heisenberg group law is $h(x,y,z)h(a,b,c)=h(x+a,y+b+xc,z+c)$. Commutation for every $(a,b,c)$ forces $xc=az$, hence $x=z=0$. The center of a group is therefore the $y$-axis. The surjective group homomorphism $h(x,y,z)\mapsto (x,z)$ has exactly this kernel of a group homomorphism, giving the stated quotient group by the first isomorphism theorem.

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.

The commutator subgroup of a group $G$ is the subgroup generated by all its group commutators:Here and throughout, $[x,y]=x^{-1}{y}^{-1}xy$. A group commutator is the identity element precisely when $x$ and $y$ commute. The generated subgroup includes finite products of group commutators and their inverse elements; an element of $G^{\prime }$ need not itself be a single group commutator. Thus $G^{\prime }=\{e\}$ precisely when $G$ is an abelian group.

The useful idea is to record exactly what must disappear to make multiplication commutative. The identity $xy=yx[x,y]$ shows how a group commutator corrects a change in multiplication order. Setting every such correction equal to $e$ produces the abelianization $G/G^{\prime }$, while preserving every piece of information that a group homomorphism to an abelian group can detect. This connects a local comparison of two operations to a canonical quotient group of the whole group.

The commutator subgroup is a characteristic subgroup, hence a normal subgroup: a group automorphism $\alpha$ sends $[x,y]$ to $[\alpha (x),\alpha (y)]$. For any normal subgroup $N$,Indeed, the group commutator of $xN$ and $yN$ is $[x,y]N$. This makes $G^{\prime }$ the smallest normal subgroup whose quotient group is an abelian group. See Conrad, Subgroup series I, section 4: commutators and abelian quotients.

Let $B(x)=\left(\begin{array}{cc}\cosh x& -\sinh x\\ -\sinh x& \cosh x\end{array}\right)$, $T=\mathrm{diag}(1,-1)$ and $P=\mathrm{diag}(-1,1)$. Every real matrix preserving $\mathrm{diag}(1,-1)$ has a unique form $B(x)S$ with $S\in \{I,T,P,TP\}$. The boost subgroup is normal because $TB(x)T=PB(x)P=B(-x)$. Its four cosets form the Klein four-group, giving the displayed quotient group. In contrast, the four-element subgroup generated by $T,P$ is not normal: $B(x)TB(-x)=B(2x)T$ is not diagonal for $x\ne 0$. This distinguishes the normal identity component from a chosen set of component representatives.

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]

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 any group action with group kernel $K$, the quotient group $G/K$ has a faithful group action on the same set, given byTo check independence of the coset representative, replace $g$ by $gk$ with $k\in K$: then $(gk)\cdot x=g\cdot (k\cdot x)=g\cdot x$. If $gK$ fixes every point, then $g\in K$, so $gK=K$, the identity element of the quotient group. The first isomorphism theorem identifies this quotient group with the image of $\rho$ inside the symmetric group. The group orbits stay the same because the possible point movements stay the same.

This construction removes precisely the indistinguishable operations. In nonfaithful coset action of the symmetric group on three letters, it replaces six group elements by two cosets, each representing one visible operation. In faithful coset action of the symmetric group on three letters, the group kernel is already trivial, so all six distinct operations survive.

Let $\pi :G\to G/N$ be a quotient group map, let $A$ be finite and symmetric, and suppose $P\subseteq \pi (A^m)$ has $|P|\ge \delta |\pi (A)|$. ThenChoose one lift in $A^m$ of each member of $P$ and multiply those lifts by $A^2\cap N$. Distinct fibers are disjoint, while $|\pi (A)||A^2\cap N|\ge |A|$.

For a group homomorphism $\phi :G\to H$, the group isomorphism $g\ker \phi \mapsto \phi (g)$ givesThe group kernel records which inputs become indistinguishable, and the image records which outputs actually occur. The first isomorphism theorem says that identifying precisely those indistinguishable inputs gives a group isomorphic to the image. It is one of the general isomorphism theorems. ^[ref].

For example, reduction modulo three gives a surjective group homomorphism from the additive cyclic group $\mathbb{Z}/6\mathbb{Z}$ to $\mathbb{Z}/3\mathbb{Z}$. Its group kernel is $\{0,3\}$. The three cosets $\{0,3\}$, $\{1,4\}$, and $\{2,5\}$ become the three outputs zero, one, and two, respectively. ThusThe quotient group diagram shows both the identification and the resulting operation. More generally, fibres of a group homomorphism are cosets of its kernel explains why this map is well-defined and injective; it is surjective onto the image by definition.

A group records reversible operations that can be combined consistently. Group theory studies their subgroups, group homomorphisms, group actions, and quotient groups. A symmetry group makes the abstraction visible: transformations may change individual points while preserving a specified object or its structure. An orbit of a group action records which positions can be exchanged, while a stabilizer subgroup records which operations leave a selected feature fixed.

Given a quotient group map $\pi :G\to G/N$ and a representation $\rho$ of $G/N$, its inflation to $G$ is the representation $\rho \circ \pi$. Conversely, a representation of $G$ factors through $G/N$ exactly when its kernel contains $N$.

The isomorphism theorems relate homomorphisms, their kernels and images, and quotient constructions in algebra. For groups, the first isomorphism theorem identifies a homomorphism's image with the quotient group by its group kernel. Analogous statements include the first isomorphism theorem for rings and the First isomorphism theorem for vector spaces: identifying precisely the inputs with the same output preserves the remaining algebraic structure.