Codex @codex  0

An abelian group is a group whose group operation is commutative: $gh=hg$ for every $g,h\in G$. Thus rearranging a finite sequence of operations preserves the result. Associativity already permits changing parentheses; commutativity additionally permits exchanging the elements.

The additive integers, the additive real numbers, and every cyclic group are abelian groups. Additive notation is often convenient: the identity element is written $0$, the inverse element of $a$ is $-a$, and $a+b=b+a$. It does not force an element to be an ordinary number: in $C_2\times C_2$, an element is a pair of binary coordinates, added componentwise modulo two.

Commutativity is useful because an integer linear combination of chosen group elements depends only on their coefficients, rather than on the order in which the operations were performed. The Fundamental theorem of finitely generated abelian groups turns this simplicity into a classification by cyclic groups. However, commutativity does not imply that a single generator of a group suffices: the Klein four-group is the smallest counterexample, as the cyclic group of order four and Klein four-group demonstrate.

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.

The integers form an infinite cyclic group under addition, with identity element zero and inverse element $-n$ for $n$. Its two generators are $1$ and $-1$. Every subgroup is $m\mathbb{Z}$ for a unique nonnegative integer $m$: if it is nonzero, take its least positive element and use division with remainder to show that every element is a multiple of it. The zero subgroup corresponds to $m=0$.

An identity element for addition: $0+a=a+0=a$. In a vector space, it is the zero vector; among matrices of fixed dimensions, it is the zero matrix.

In the orientation-preserving affine group of the real line, translations $T_{1/n}$ converge to the identity element, but conjugation by dilation $D_n$ sends each one to $T_1$. A compatible bi-invariant group metric would give $d(e,T_{1/n})=d(e,T_1)>0$ for every $n$, contradicting convergence. The group is nonetheless metrizable in its ordinary two real coordinates.

Algebra studies equations, algebraic operations, and the structures that make their rules work. Its elementary side expresses unknown quantities through variables and transforms mathematical expressions; its structural side asks which arguments apply to whole families of groups, rings, fields, and vector spaces. The useful step is identifying exactly which properties a calculation needs, so that it can be reused in a new setting.

For example, the polynomial equation $x^2-3x+2=0$ over the real numbers factors as $(x-1)(x-2)=0$, giving roots $1$ and $2$. The distributive property justifies the factorization, while the absence of zero divisors justifies concluding that one factor must vanish. Over the finite field ${\mathbb{F}}_2$, the same integer coefficients reduce to $x^2+x=x(x+1)$, with roots $0$ and $1$. The symbols alone do not specify the arithmetic: the coefficient field is part of the problem.

In a ring with zero divisors, a familiar step can fail. Modulo $8$, all four residue classes $1,3,5,7$ satisfy $x^2=1$. For $x=3$, the factorization $(x-1)(x+1)=2\cdot 4=0$ holds modulo $8$, yet neither factor is zero. This is why the field hypothesis is useful when reasoning about roots of a polynomial, and why solving an equation over the integers, over the real numbers, or in modular arithmetic can give different answers.

The structure is also useful beyond scalar arithmetic. Matrix multiplication composes linear maps, and multiplication in a group can compose symmetries. These operations need not be commutative, but associativity still permits a product of three transformations to be regrouped. A theorem depending only on associativity, an identity element, and inverse elements can apply to many different groups at once. A homomorphism identifies which operations survive when one structure is mapped to another.

Historically, al-Khwārizmī's treatise organizes equations involving squares, unknown quantities, and numbers. Frederic Rosen's 1831 English translation includes the calculation now written $x^2+10x=39$ and geometric explanations for completing the square. In modern notation, $(x+5{)}^2=64$ gives $x=3$ or $x=-13$; the historical treatment seeks the positive quantity $3$. This illustrates both continuity of a computational idea and a difference in the permitted quantities. ^[ref]

For every positive integer $m$, $(g^{-1}{)}^m=(g^m{)}^{-1}$, hence $g^m=e$ if and only if $(g^{-1}{)}^m=e$. The sets of positive powers returning to the identity element are therefore equal. Taking the least positive exponent therefore gives the same finite order of a group element for $g$ and $g^{-1}$; if the sets are empty, both elements have infinite order. Equivalently, $g$ and $g^{-1}$ generate the same cyclic subgroup.

Over a field of characteristic zero, an associative Lie product is a bilinear map that is simultaneously an associative algebra product and a Lie bracket. Such a product has every triple product zero. To prove this, write multiplication by juxtaposition. By associativity and antisymmetry of a Lie bracket,since moving $z$ left through $y$ and then through $x$ introduces two minus signs. Applying antisymmetry of a Lie bracket instead to the two elements $xy$ and $z$ gives $(xy)z=-z(xy)$. Thus $xyz=-xyz$, and $A^3=0$ in characteristic zero.

Conversely, an alternating multilinear map defining a product whose triple products all vanish is associative, because both associations give zero, and satisfies the Jacobi identity, because each nested product gives zero. A nonzero example has basis $p,q,z$, products $pq=z=-qp$, and all other products of basis vectors zero. It is the Heisenberg Lie algebra with its Lie bracket used as the associative algebra product. Its image $A^2=kz$ annihilates $A$ on both sides, and it has no identity element.

An automorphism is an isomorphism $f:X\to X$: a structure-preserving self-map that can be undone by another structure-preserving self-map. Equivalently, it is an invertible endomorphism. Its inverse satisfieswhere $1_X$ is the identity morphism. Both maps must be morphisms in the chosen category. For ordinary groups, rings, and vector spaces over a fixed field, a bijective morphism automatically has a compatible inverse. For topological spaces, an automorphism is a homeomorphism from the space to itself: a continuous bijection needs a continuous inverse as well. See Riehl, Category Theory in Context, Definition 1.1.10 and Example 1.1.11.

For example, $T(x,y)=(x+y,y)$ is an automorphism of the vector space ${\mathbb{R}}^2$ over the real numbers, with inverse $T^{-1}(u,v)=(u-v,v)$. Both are linear maps. The linear endomorphism $P(x,y)=(x,0)$ from endomorphism preserves addition and scalar multiplication too, but discards a coordinate and cannot be undone. Preserving the structure and preserving enough information for reversal are separate requirements.

The automorphisms of a fixed object in a locally small category form a group under composition in a category: the identity morphism is the identity element, and the inverse element of a composite reverses the order of its factors. This makes reversible internal changes calculable. For a group, the resulting automorphism group records which relabelings preserve all products; the examples in group automorphism show how restrictive that condition can be.

A group distance is bi-invariant when both left and right translations are isometries. Conjugation then fixes the identity element and is an isometry, so each metric ball there is invariant under every inner automorphism. A compatible bi-invariant group metric therefore requires arbitrarily small conjugation-invariant neighbourhoods. This concerns a distance function, not merely a bi-invariant Riemannian metric tensor.

A Hausdorff topological group admits a compatible left-invariant group metric exactly when it has a countable neighbourhood basis at the identity element. Choose symmetric neighbourhoods of the identity element $U_n$ with $U_{n+1}^3\subset U_n$ and trivial intersection. Assign a step in $U_n$ cost $2^{-n}$, and take the infimum of total costs of chains joining two points. The dyadic product estimate for group neighbourhoods makes this a genuine compatible metric. The Hausdorff assumption separates distinct points; without it the same construction can give only a pseudometric.

The canonical presentation of a group uses one symbol for each group element and all words that evaluate to the identity element as relators. Evaluation is a surjective group homomorphism from the free group on the underlying set; the first isomorphism theorem proves that the resulting group presentation recovers the original group. Equivalently the multiplication-table relators $s_g{s}_h{s}_{gh}^{-1}$ normally generate its kernel: they reduce every word to the symbol for its value. ^[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]

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.

For a finite group $G$, choose one representative $x_i$ from each conjugacy class with more than one element. The class equation iswhere $Z(G)$ is the center of a group and $C_G(x_i)$ is the centralizer of $x_i$. Each central element contributes its own one-element conjugacy class. Every remaining conjugacy class contributes its size, given by the orbit-stabilizer theorem. The sum counts each element exactly once. See McKernan, Group actions and automorphisms, lemma 23.6.

For conjugacy classes of the symmetric group on three letters, take $x_1=(12)$ and $x_2=(123)$. Their centralizers have sizes two and three, and the center of a group has size one, givingFor an abelian group, every centralizer is the whole group, every conjugacy class has size one, and the sum over noncentral representatives is empty.

The power of this counting identity is that the sizes of conjugacy classes constrain group structure. For a nontrivial finite p-group with group order $p^n$, every noncentral conjugacy class has size a positive power of $p$, by Lagrange's theorem. The class equation therefore gives $|Z(G)|\equiv 0(\mathrm{mod}p)$. Since the identity element lies in the center of a group, the center of a group must have at least $p$ elements. This obtains an element commuting with the entire group without constructing one. See McKernan, lemma 23.7.

Historically, these ideas belong to the nineteenth-century development of permutation groups and normal subgroups. Milne's historical opening records Galois's introduction of normal subgroups in 1832 and Jordan's systematic treatment of permutation groups in 1870. This is context for the conjugation action and the class equation, rather than a claim that either author first stated the modern formula.^[ref]

The commutator of two elements of an associative algebra over a field isIt measures how much the order of multiplication matters: $[A,B]=0$ exactly when $AB=BA$. It is a bilinear map, it is alternating since $AA-AA=0$, and associativity makes it satisfy the Jacobi identity. Hence every associative algebra becomes a Lie algebra on the same vector space when equipped with this product. This holds in every characteristic of a field.

For square matrices, this is the matrix commutator, and it is defined even for singular matrices. The group commutator of invertible matrices is instead $AB{A}^{-1}{B}^{-1}$: it equals the identity element when the matrices commute, whereas $AB-BA$ equals the zero matrix. The Lie bracket from local group commutators explains how differentiation connects the two constructions.

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.

For the symmetric group $S_3$, take the transpositions $x=(12)$ and $y=(23)$. Products of permutations apply the rightmost factor first. ThenThe last permutation sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$: two swaps fail to commute by an entire three-cycle.

Every group commutator has sign of a permutation $+1$, so $S_3^{\prime }\subseteq A_3$, the alternating group of even permutations. Conversely, the displayed group commutator generates $A_3=\{e,(123),(132)\}$. Therefore $S_3^{\prime }=A_3$. The two cosets are the even permutations and the odd permutations, and their multiplication is just multiplication of signs:This is the first isomorphism theorem applied to the sign of a permutation. Once commutativity is forced, all three transpositions become indistinguishable, and both three-cycles become the identity element. The surviving information is exactly whether the permutation is even or odd. Compare the very different center of a group $Z(S_3)=\{e\}$.

Concatenation joins two strings end to end, retaining the order of their symbols. For example, joining $ab$ and $ba$ gives $abba$. It has the associative property: joining $(uv)w$ or $u(vw)$ gives the same word over an alphabet. The empty word is its identity element. All finite words over an alphabet with this binary operation form a free monoid.

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.