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A finite additive group is a finite group whose operation is written as addition. Its identity element is denoted $0$, the inverse element of $x$ is denoted $-x$, and translation $y\mapsto x+y$ is a bijection.

A group packages reversible operations together with a rule for composing them. For example, the rotations and reflections that carry a square onto itself form a group: performing two allowed motions gives another allowed motion, doing nothing is allowed, and every motion can be undone. The elements are the motions themselves, rather than the vertices they move. Labels in the figure track those vertices; they are markers, rather than extra decorations that the motions must preserve.

Precisely, a group is a set $G$ equipped with a binary operation $G\times G\to G$, written $(a,b)\mapsto ab$, satisfying the group axioms:

The group operation need not be ordinary numerical multiplication. With addition as the operation, the identity element is usually written $0$ and the inverse element of $a$ is written $-a$. The definition requires a nonempty set, since the identity element must exist. See McKernan's MIT lecture 1, Definition 1.1 for the axioms.

Why insist on associativity? It makes a string of operations unambiguous without specifying parentheses. For function composition, both $(fg)h$ and $f(gh)$ send $x$ to $f(g(h(x)))$. This fixes the order of execution while allowing regrouping. Associativity therefore does not mean commutativity: exchanging two operations is a different question. Ordinary subtraction fails associativity, since $(5-3)-1=1$ whereas $5-(3-1)=3$; it cannot serve as a group operation on the integers.

An inverse element makes reversal exact and equations solvable. In any group, $ax=b$ has the unique solution $x=a^{-1}b$, since $a^{-1}(ax)=(a^{-1}a)x=x$. Similarly, $xa=b$ has solution $x=b{a}^{-1}$. The side matters in a non-abelian group. Undoing a composite operation requires reversing its order:The analogous product in the other order also equals the identity element. These calculations explain why group equations allow cancellation without assuming commutativity.

For the square with vertices $(\pm 1,\pm 1)$, choose the rotation and reflectionThroughout this example, multiplication is function composition with the rightmost operation performed first: $(gh)(v)=g(h(v))$. Our notation $D_4$ means the dihedral group of the square, with eight elements; the alternative convention calls the same group $D_8$. Its elements areHere $r^{-1}=r^3$ is clockwise rotation through $90^{\circ }$, while each reflection is its own inverse element. There are exactly eight motions: the image of one vertex has four choices, and the image of an adjacent vertex has two choices; together these determine the motion. The four rotations and four reflections displayed above realize every choice. The vertex permutations give a concrete group action. The general polygon construction is discussed in Group Theory (v4.01) by James S. Milne (section 1.17).

Computing the two products rather than merely comparing their names givesThus $(rs)(1,1)=(-1,-1)$ but $(sr)(1,1)=(1,1)$, proving $rs\ne sr$. In fact $rs$ is reflection in the line $y=-x$, whereas $sr$ is reflection in $y=x$. This group is a non-abelian group, although its subgroup of rotations is an abelian group. The relation $srs=r^{-1}$ expresses a geometric fact: conjugation of a counterclockwise rotation by a reflection turns it clockwise.

A contrasting example is the cyclic group $\mathbb{Z}/4\mathbb{Z}$ under addition in modular arithmetic. Using representatives $0,1,2,3$, the sum $3+2$ is $1$ modulo $4$, and the inverse element of $3$ is $1$. Repeated addition of $1$ runs through $0,1,2,3,0$. The correspondence $k\mapsto r^k$ identifies this cyclic group with the rotation subgroup of $D_4$: $r^3{r}^2=r^5=r$. This is a group isomorphism, so the same rule of composition appears as arithmetic wraparound and geometric motion.

By comparison, $(\mathbb{Z},+)$ is an infinite cyclic group. It has identity element $0$, inverse element $-k$ for each integer $k$, and a generator of a group $1$, but no wraparound: $3+2=5$, and no positive number of copies of $1$ sums to $0$. Both examples are abelian groups. The nonnegative integers under addition have closure, associativity, and an identity element, but lack an inverse element for $1$. Reversibility is precisely what fails.

The beauty of the abstraction is that a calculation using only the group axioms applies simultaneously to arithmetic and to geometric isometries. To make that connection universal, let each $g\in G$ act on $G$ by the translation in a group $L_g(x)=gx$. Each $L_g$ is a bijection, with inverse $L_{g^{-1}}$, and $L_g{L}_h=L_{gh}$. Distinct elements give distinct maps because $L_g(e)=g$. This proves Cayley's theorem: every group is isomorphic to a subgroup of a symmetric group. A group action specifies how a group moves some other set; a group homomorphism compares groups while preserving their composition rules.

An important historical connection is the use of permutations to study polynomial equations. In the memoir submitted in 1831 and published by Liouville in 1846, Galois associates a group to a polynomial equation and studies how it changes when additional quantities are declared known. Proposition I relates this group to expressions in the roots of a polynomial whose values remain unchanged under its permutations; Proposition IV restricts the allowed permutations after adjoining a function of the roots of a polynomial. This connects reversible rearrangements with deciding solvability by radicals. See Galois's original memoir, pp. 417–433, especially Propositions I and IV; Liouville's introduction, p. 382.

For a finite abelian group $G$, $U(h)|g⟩=|g+h⟩$ defines the regular representation on the basis labelled by $G$. The linear characters give the common eigenbasisCharacter orthogonality proves orthonormality, and changing variables by the translation in a group proves the eigenvalue formula.

For a positive integer $n$ and a congruence class $k$ in the additive finite cyclic group $\mathbb{Z}/n\mathbb{Z}$, the order of a group element isIndeed, write $d=\gcd (n,k)$, $n=d{n}^{\prime }$ and $k=d{k}^{\prime }$. Returning to zero after $m$ steps means $n\mid mk$, equivalently $n^{\prime }\mid m{k}^{\prime }$. Since $n^{\prime }$ and $k^{\prime }$ are coprime, this holds exactly when $n^{\prime }\mid m$; the least positive such $m$ is $n^{\prime }$. This argument also covers $k=0$, since $\gcd (n,0)=n$.

For $n=6$, the orders of $0,1,2,3,4,5$ are respectively $1,6,3,2,3,6$. The translation in a group $x\mapsto x+k$ has the same period from every starting element. Consequently its cycles each have length $n/d$ and there are $d$ of them. They are exactly the cosets of the cyclic subgroup generated by $k$, which explains both the lengths and the number of components in the picture.

Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

For each $g\in G$, left translation in a group $L_g:x\mapsto gx$ is a bijection, with inverse function $L_{g^{-1}}$. Thus $\theta (g)=L_g$ belongs to $\mathrm{Sym}(G)$. For $g,h,x\in G$, associativity of the group operation givesHence $\theta$ is a group homomorphism. If $\theta (g)=\theta (h)$, evaluation at the identity element $e$ gives $g=ge=he=h$. ThereforeThis is the left regular representation used in Cayley's theorem.

For $W=aH$ and $W^{\prime }=bH$, defineIf $w=ah$ then $f(w)=bh\in bH$, so this function is well defined. Its inverse function is $w^{\prime }\mapsto a{b}^{-1}{w}^{\prime }$, since successive left translation in a group by $b{a}^{-1}$ and $a{b}^{-1}$ cancels. ThusIn particular every left coset has the same cardinality as $H$.