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A finite group is a group whose underlying set has finitely many elements. Its order of a finite group is the number of elements, counting the identity element. The group operation still has to satisfy all the group axioms: merely having a finite list of states does not make a group.

For example, the six congruence classes modulo six form a finite cyclic group under addition, whereas the permutations of three objects form the symmetric group $S_3$, also with six elements. The first is an abelian group; the second is a non-abelian group. Thus the order of a finite group is a useful first measurement, but the composition rule contains information that the count alone loses.

Finiteness makes group structure accessible to exact calculation. A Cayley table for a finite group of order $n$ has $n^2$ entries, and every group element eventually returns to the identity element when repeated. Lagrange's theorem strengthens this observation: the order of a group element, being the size of its cyclic subgroup, divides $n$. The cyclic group of order four and Klein four-group give a small example of how that constraint can distinguish different composition rules.

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.

Suppose a nonempty semigroup has, for each $a$, elements $l_a,r_a$ satisfying $b{l}_{a}a=b$ and $a{r}_{a}b=b$ for every $b$. Then it is a group. For any fixed $a$, the element $e_R=l_{a}a$ is a right identity element, and $e_L=a{r}_a$ is a left identity element. Since $e_L{e}_R=e_L=e_R$, they coincide in a two-sided identity $e$. For each $a$, putting $b=e$ in the two assumed identities gives $l_{a}a=e$ and $a{r}_a=e$. Associativity then yields $l_a=l_{a}e=l_a(a{r}_a)=(l_{a}a)r_a=e{r}_a=r_a$, a two-sided inverse element. Thus the usual group axioms follow without an assumed identity or cancellation law.

An associative binary operation $\ast$ on a nonempty set $X$ makes $X$ a group if there is $e\in X$ with $x\ast e=x$ for every $x$, and every $x$ has some $z$ with $x\ast z=e$. No left identity element or left inverse element assumption is needed. First, an idempotent element of a semigroup $y$ satisfies $y=e$: choose $z$ with $y\ast z=e$ and compute $y=y\ast e=y\ast (y\ast z)=(y\ast y)\ast z=y\ast z=e$. If $x\ast z=e$, then $(z\ast x)\ast (z\ast x)=(z\ast (x\ast z))\ast x=(z\ast e)\ast x=z\ast x$, so $z\ast x=e$ by the previous observation. Finally, $e\ast x=(x\ast z)\ast x=x\ast (z\ast x)=x\ast e=x$. Thus all group axioms follow.

A group is a nonempty set $G$ with an operation $G\times G\to G$, $(g,h)\mapsto gh$, satisfying associativity, an identity element $e$ with $eg=ge=g$, and an inverse element $g^{-1}$ with $g{g}^{-1}=g^{-1}g=e$ for every $g$. These are the group axioms. An Abelian group additionally satisfies $gh=hg$ for every pair. A subgroup is a subset that is itself a group under the restricted operation; equivalently, it is nonempty and closed under $h{k}^{-1}$.

Let $R:G\to \mathrm{GL}(V)$ be a nonzero, finite-dimensional complex irreducible representation of an Abelian group. For each fixed $g$, commutativity givesThus $R(g)$ is an intertwining operator from $R$ to itself. By Schur's lemma over the complex numbers, $R(g)={\lambda }_g{I}_V$. Every line in $V$ is therefore an invariant subspace. Since $R$ is irreducible, this forcesThis is the finite-dimensional convention used throughout the paper. Without it, the literal assertion for arbitrary Abelian groups needs qualification. A concrete counterexample is $G=\mathbb{C}(t{)}^{\times }$ acting by multiplication on the complex vector space $V=\mathbb{C}(t)$ of rational functions. Any nonzero invariant complex vector subspace containing $f$ contains $gf$ for every nonzero rational $g$, hence contains every rational function. This is an infinite-dimensional irreducible representation. The scalar conclusion of Schur's lemma used above depends on finite dimensionality.

A subgroup $N\le G$ is a normal subgroup if $gN{g}^{-1}=N$ for every $g\in G$. A group homomorphism $f:G\to \tilde{G}$ satisfies $f(gh)=f(g)f(h)$ for every $g,h\in G$. In particular $f(e)=\tilde{e}$ and $f(g^{-1})=f(g{)}^{-1}$. A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$ into the general linear group of invertible linear maps on $V$.

For the kernel of a group homomorphism, put $K=\ker f=\{g:f(g)=\tilde{e}\}$. It contains $e$, and if $k,\ell \in K$ then $f(k{\ell }^{-1})=f(k)f(\ell {)}^{-1}=\tilde{e}$, so $K$ is a subgroup. For $g\in G$ and $k\in K$,Thus $gK{g}^{-1}\subseteq K$. Applying this inclusion with $g^{-1}$ gives the reverse inclusion, proving that $\ker f$ is a normal subgroup.

For a normal subgroup $N$, define multiplication of left cosets byTo check independence of representatives, replace $g,h$ by $gn,hm$, with $n,m\in N$. Thenbecause normality gives $h^{-1}nh\in N$. Hence the operation is well-defined. Associativity follows from $(gh)k=g(hk)$ in $G$; the identity is $N=eN$; and the inverse of $gN$ is $g^{-1}N$. These verify the group axioms, giving the quotient group $G/N$. The map $g\mapsto gN$ is a surjective group homomorphism with kernel of a group homomorphism equal to $N$.

The componentwise product is closed in $G_1\times G_2$. Associativity follows componentwise from associativity in each group:The identity is $(e_1,e_2)$, and $(g_1,g_2{)}^{-1}=(g_1^{-1},g_2^{-1})$. These verify all the group axioms, so this is the direct product of groups.

Write the paper's Kronecker product $\times$ as $\otimes$. Its mixed-product identity and the representation laws of $R_1,R_2$ giveAlso $R(e_1,e_2)=I_{n_1}\otimes I_{n_2}=I_{n_1{n}_2}$, andThus the images are invertible and this is a group representation of dimension $n_1{n}_2$, called the external tensor product of group representations. No irreducibility assumption is needed for this construction.

Heinrich Weber's 1882 paper concerns prime numbers represented by quadratic forms. Its definition abstracts a finite group from the nature of its elements: what matters is a closed binary operation, associativity, and left and right cancellation laws. Here degree $h$ means the number of elements, now called the group order; it is not the number of letters acted on by a permutation group. This matters when relating the passage to earlier work on substitutions. The definition fits the use of groups of classes of quadratic forms, where the elements are already mathematical objects more elaborate than individual permutations.^[ref][ref]

The apparent omission of an identity element and inverse elements is not fatal. Every nonempty finite cancellative semigroup is a group. Here is a proof identifying exactly where the assumption of a finite set enters. For fixed $a\in G$, the maps $L_a(x)=ax$ and $R_a(x)=xa$ are injective by cancellation. An injective function from a finite set to itself is surjective, so choose $e$ with $ae=a$. For every $x$,and left cancellation gives $ex=x$. To show $e$ is also a right identity element, useRight cancellation gives $xe=x$. Thus the same $e$ works on both sides for every $x$; it has not merely been chosen as an element fixing $a$.

Surjectivity now gives $b,c$ with $ab=e$ and $ca=e$. By associativity,So $b$ is a two-sided inverse element of $a$. Since $a$ was arbitrary, the usual group axioms follow. The nonempty hypothesis is tacit in a system of $h$ elements intended to have positive $h$: the empty system would satisfy the displayed universal conditions vacuously but would have no identity element.

The infinite case exposes the force of the qualification. The positive integers under addition are closed, associative and cancellative, but have no additive identity element. Even the nonnegative integers have an identity element without additive inverse elements for positive elements. The failure is precisely that injective functions of an infinite set to itself need not be surjective. Heinrich Weber's conditions are therefore not an unrestricted replacement for the modern group axioms.

Historically the abstraction is important because it allows arguments to be applied across different kinds of elements. An assumption of commutativity is absent: class-group applications do not make commutativity part of the general definition. For example, the symmetric group $S_3$ meets the conditions but is noncommutative. The definition also does not require a prior presentation as permutations, although Cayley's theorem provides such a representation. The passage documents the separation of an algebraic structure from any one representation of it, with finiteness carrying logical work that later axiom systems usually assign to explicit identity element and inverse element axioms.

Use the following group axioms in the given language:Commutativity supplies the other identity and inverse equations. These first-order sentences therefore specify precisely an abelian group with the named identity $0$; an inverse function symbol is unnecessary because inverses are asserted to exist by a quantified formula.

For each positive integer $n$, let $t_n(x)$ be the first-order term recursively defined by $t_1(x)=x$ and $t_{n+1}(x)=t_n(x)+x$. Add the torsion-free abelian group axiom schemeEvery first-order model of these first-order sentences is an abelian group with no nonidentity element of finite order. Conversely every torsion-free abelian group satisfies each of the displayed first-order sentences. This is a countably infinite first-order axiomatization of exactly the required class.

No finite axiomatization exists. Write $\Gamma$ for the axioms just constructed and suppose a finite set $\Delta$ of first-order sentences had exactly the same first-order models. Its logical conjunction $\delta$ is a single first-order sentence, and $\Gamma \models \delta$. By the compactness theorem there is a finite subset ${\Gamma }_0\subseteq \Gamma$ with ${\Gamma }_0\models \delta$: otherwise $\Gamma \cup \{\neg \delta \}$ would be finitely satisfiable and hence satisfiable, a contradiction.

Only finitely many of the integers $n$ in the torsion-free scheme occur in ${\Gamma }_0$. Choose a prime number $p$ larger than all of them. The cyclic group $\mathbb{Z}/p\mathbb{Z}$ satisfies all the group axioms and each of these finitely many conditions from the torsion-free abelian group axiom scheme: multiplication by an integer $1\le n<p$ is invertible modulo $p$, so $nx=0$ implies $x=0$. It consequently satisfies ${\Gamma }_0$, then $\delta$, and therefore $\Delta$.

But its nonzero elements have finite order $p$, so it is not a torsion-free group, contradicting the assumed description of the first-order models of $\Delta$. This establishes the nonfinite axiomatizability of torsion-free abelian groups, ruling out every finite set of defining first-order sentences, not just finite subsets of the particular scheme in part (c).

Zorn's lemma states that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. Order the partial group operations by inclusion of their graphs. The collection is a set, since its members are subsets of $X\times X\times X$. It is nonempty: choose $x_0\in X$ and put the one-element group operation on $\{x_0\}$.

To construct a chain union of partial group operations, for the given nonempty chain, let $m={\bigcup }_{n\in \mathcal{C}}n$ and let $A$ be the union of the corresponding group carriers. Comparability guarantees agreement wherever two operations are both defined, so $m$ is a function. Its domain is $A\times A$: any two elements of $A$ lie together in one carrier, by taking the larger of the two chain members containing them. Their product then lies in that carrier and hence in $A$.

Any three elements similarly lie together in one chain member, where associativity holds; therefore associativity holds for $m$. The identities of nested nonempty groups agree: the smaller identity remains an idempotent under the larger group operation, whose only idempotent is its identity. Hence the whole chain has a common identity. Each element's inverse in a carrier remains its inverse in every larger carrier, so it is an inverse in the union as well. Thus $(A,m)$ satisfies all group axioms andIt is an upper bound for the chain. The empty chain also has an upper bound because $\mathcal{G}$ is nonempty, completing the hypothesis needed for Zorn's lemma.

For either coefficient system $\mathbb{Z}$ or ${\mathbb{F}}_p$, multiplying two matrices keeps all entries in that system, and multiplicativity of the determinant preserves determinant one. The identity matrix has determinant one, and for a determinant-one matrix the matrix inverse isIts entries again belong to the same coefficient system and its determinant is one. Together with associativity of matrix multiplication, these facts establish all the group axioms. Thus both sets are special linear groups.

Reduce each entry modulo five to defineReduction respects sums and products, so $\overline{MN}=\overline{M}\overline{N}$. It also respects the determinant, ensuring that the image has determinant one. Hence $\pi$ is a group homomorphism. Its kernel of a group homomorphism consists exactly of determinant-one integer matrices congruent to the identity modulo five:The determinant-one condition remains part of this set; the four integers cannot be chosen independently without it. A kernel of a group homomorphism is a normal subgroup, since $\pi (gk{g}^{-1})=\pi (g)e\pi (g{)}^{-1}=e$ for $k\in \ker \pi$. This is the principal congruence subgroup of level five.

A subgroup $N$ is a normal subgroup of $G$ if $gN{g}^{-1}=N$ for every $g\in G$. Define multiplication on its left cosets byTo check that this is well-defined, replace $g,h$ by $g{n}_1,h{n}_2$ with $n_1,n_2\in N$. Thenbecause normality gives $h^{-1}{n}_{1}h\in N$. Thus the product depends only on the two cosets. Associativity follows from that of $G$, the identity is $N$, and the inverse of $gN$ is $g^{-1}N$. These are all the group axioms, so $G/N$ is a group under coset multiplication. The natural projection $g\mapsto gN$ is a surjective group homomorphism with kernel $N$.

A Lie group is a smooth manifold $G$ equipped with smooth maps $m:G\times G\to G$, $m(g,h)=gh$, and $\iota :G\to G$, $\iota (g)=g^{-1}$, for which multiplication is associative, there is an identity $e$, and every $g$ has the inverse $g^{-1}$. In equations,Thus the group axioms and the smooth structure are compatible.

The function composition of two bijections $f,g:X\to X$ is a bijection, with inverse function $(f\circ g{)}^{-1}=g^{-1}\circ f^{-1}$. Thus function composition is a binary operation on $\mathrm{Sym}(X)$. It is associative, since both $(f\circ g)\circ h$ and $f\circ (g\circ h)$ send $x$ to $f(g(h(x)))$.

The identity function ${\mathrm{id}}_X$ is a bijection and satisfies $f\circ {\mathrm{id}}_X={\mathrm{id}}_X\circ f=f$. Every bijection $f$ has an inverse function $f^{-1}:X\to X$, itself a bijection, satisfying $f\circ f^{-1}=f^{-1}\circ f={\mathrm{id}}_X$. These verify all group axioms, so $\mathrm{Sym}(X)$ is a group, the symmetric group of $X$.

For any $x$, choose $z$ with $x\ast z=e$. By ii, $z\ast x=e$ as well. Associativity and the right identity element property now giveThus $e$ is also a left identity element. The given binary operation is closed on $X$ and associative, $e$ is a two-sided identity element, and each $x$ has the two-sided inverse element $z$. All group axioms hold, so $(X,\ast )$ is a group. This is the one-sided group axioms criterion.

Write $T_{\theta ,\mathbf{b}}(\mathbf{v})=R_{\theta }\mathbf{v}+\mathbf{b}$. The sine addition formula and cosine addition formula give $R_{\theta }{R}_{\psi }=R_{\theta +\psi }$. Therefore function composition satisfiesThe resulting angle is real and the new translation vector lies in ${\mathbb{R}}^2$, so $G$ is closed under function composition. Associativity holds because, for any functions $S,T,U$ and any $\mathbf{v}$, both $(S\circ T)\circ U$ and $S\circ (T\circ U)$ take $\mathbf{v}$ to $S(T(U(\mathbf{v})))$.

The identity element is $T_{0,0}$, since $R_0=I$. Every element has the two-sided inverse elementIndeed $R_{\theta }{R}_{-\theta }=R_{-\theta }{R}_{\theta }=I$, and substituting into the composition formula makes the translation vector zero in either order. These verify all the group axioms. Thus $G$ is a group under function composition, the special Euclidean group in two dimensions.

The group order of a finite group is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$. Such an $r$ exists in a finite group because two powers must coincide, and cancellation then gives a positive power equal to the identity element.

For the direct product of groups, the componentwise operation is closed because each factor is closed. Associativity holds in each coordinate:The identity element is $(e_1,e_2)$, and the inverse element of $(g,h)$ is $(g^{-1},h^{-1})$. Thus all group axioms hold. For finite factors, the multiplication principle gives

For ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$, write $E=(e,e)$, $u=(w,e)$, $v=(e,w)$, $z=(w,w)$, with $w^2=e$. Its Cayley table, with the row element multiplying the column element, isThis is the Klein four-group: every nonidentity element has order two.

In the cyclic group ${\mathbb{Z}}_4=⟨a⟩$, the set $\{e,a^2\}$ is closed because $(a^2{)}^2=e$, and both elements have inverses in the set. It is therefore a proper subgroup of order two. By Lagrange's theorem, a proper subgroup has order one or two. An order-two subgroup must be generated by an element of order two. Neither $a$ nor $a^3$ has order two, whereas $a^2$ does. HenceThe PDF's claim needs “nontrivial”: $\{e,a^2\}$ is the only nontrivial proper subgroup, but $\{e\}$ is also a proper subgroup under the usual definition.

A group isomorphism preserves element orders: $\varphi (g{)}^r=e$ if and only if $g^r=e$, using the homomorphism property and injectivity. Since ${\mathbb{Z}}_4$ has elements of order four and the Klein four-group does not,

A group is a set $G$ equipped with a binary operation satisfying the group axioms: closure; associativity, $(gh)k=g(hk)$; an identity element $e$ with $eg=ge=g$; and an inverse element $g^{-1}$ for every $g$, with $g{g}^{-1}=g^{-1}g=e$.

A subgroup $H$ is a subset that is a group under the restricted operation. Equivalently, it is nonempty and contains $x{y}^{-1}$ whenever $x,y\in H$. A generating set of a group is a set whose elements and inverses can be multiplied in finite words to produce every group element.

The group order is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$, or infinity if there is none. Lagrange's theorem states that for a finite group and subgroup $H$, $|H|$ divides $|G|$ and $|G|=[G:H]|H|$. In particular, every finite element order divides the group order.

A cyclic group is generated by one element, so every element is a power of that generator. A group isomorphism is a bijective group homomorphism: a bijection $\varphi :G\to G^{\prime }$ satisfying $\varphi (gh)=\varphi (g)\varphi (h)$ for all $g,h$.

Write $M(a,b)=\left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)$. Direct matrix multiplication givesSince $a,c\ne 0$, the product is again in $G$. Associativity is inherited from matrix multiplication, the identity element is $M(1,0)$, andlies in $G$. These verify all group axioms. This is the Affine group of the complex line, acting by $z\mapsto az+b$.

For the diagonal subset, $M(a,0)M(c,0)=M(ac,0)$, it contains $M(1,0)$, and $M(a,0{)}^{-1}=M(a^{-1},0)$. Thus $G^{\prime }$ is a subgroup. Finally,so $f:G\to G^{\prime }$ is a surjective group homomorphism.