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Lie theory studies continuous symmetries through Lie groups, Lie algebras, and their representations. A rotation can change by an arbitrarily small angle: its parameter varies continuously, yet composing two rotations obeys an exact group operation. The central question is how such finite transformations are controlled by their velocities at the identity element.

An infinite group need not require infinitely many independent parameters. The planar rotation group has infinitely many elements but only one local angle coordinate. More generally, an $n$-dimensional Lie group has an $n$-dimensional tangent space $\mathfrak{g}=T_{e}G$ at its identity element. A basis of this Lie algebra, together with the Lie brackets between its basis vectors, supplies finite data for infinitesimal composition. These brackets record the first correction to adding small transformation parameters. The Exponential map of a Lie group turns a chosen infinitesimal direction into an actual motion.

Finite tangent data consists of a finite basis and a finite table of Lie brackets. For planar rotation, every generator is $aJ$ with $a\in \mathbb{R}$, using the quarter-turn matrix $J$ computed below: one real coefficient specifies any angular velocity. This Lie algebra has infinitely many vectors, but its dimension is one. Linear combinations of the chosen basis vectors supply all the possible infinitesimal motions.

This passage between geometry and algebra is useful because linear maps and matrix commutators are easier to compute than an entire family of transformations. The local information does not determine every global identification: the additive real line and the circle group have isomorphic one-dimensional Lie algebras, but only the latter wraps around. For the local-versus-global viewpoint and the connection with differential equations, see Meinrenken, Lie groups and Lie algebras, introduction and Section 4.

An algebraic magma is a set equipped with a binary operation. No associative property, commutativity, or identity element is required. A semigroup adds the associative property; a monoid also adds an identity element.

The matrix ring consists of all $n\times n$ matrices over a ring $R$, using entrywise addition and matrix multiplication. Its identity element is the identity matrix. Even when $R$ is a commutative ring, this ring need not have commutative multiplication.

A left $R$-module consists of an abelian group $(M,+)$ and scalar multiplication $R\times M\to M$, written $(r,m)\mapsto rm$, where $R$ is a ring with identity element $1$. For every $r,s\in R$ and $m,n\in M$, requireThe module elements can be added to one another and multiplied by scalars; a product of two module elements is not part of the definition. These conventions agree with ^[ref]. For a noncommutative ring, the side matters: a right module instead has $(mr)s=m(rs)$. Over a commutative ring, the two conventions agree.

Over a field $k$, these are exactly the vector space axioms. A nonzero $a\in k$ has a multiplicative inverse, so $am=0$ implies $m=a^{-1}(am)=0$. Over the integers, $2$ has no multiplicative inverse: $2x=1$ has no solution in $\mathbb{Z}$. In the module $\mathbb{Z}/6\mathbb{Z}$, division can also lose uniqueness: $2\overline{1}=2\overline{4}=\overline{2}$. A ring element which is a unit still acts invertibly on every module; the difficulty lies with scalars which are not units.

A monoid is a semigroup with a two-sided identity element $e$: $e\ast a=a\ast e=a$ for every element $a$. Associativity handles the grouping of a nonempty sequence; the identity element also supplies the result for an empty sequence. This makes the same rules apply to combining zero, one, or many contributions. ^[ref].

The nonnegative integers $\{0,1,2,\dots \}$ with addition form a commutative monoid, with identity element zero. They do not form a group: a positive integer has no additive inverse in that set. All words over an alphabet with concatenation form a free monoid, whose identity element is the empty word. The full transformation monoid consists of all self-maps of a set, with function composition and the identity function.

These examples explain why monoids are useful: accumulation needs an operation and a starting value, but it need not be reversible. Adding counts, joining strings, and composing transformations all share the same grouping and identity laws. Requiring every element to have an inverse element gives a group, a stronger structure suitable for reversible operations.

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.

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 $S_3$.

If the order of a group element is $m$ for $g$ and $n$ for $h$, with both finite, thenexactly when both $m$ and $n$ divide $k$. The smallest positive such $k$ is their least common multiple, givingThis concerns the orders of the selected elements, which need not equal the orders of their entire groups. If either component has infinite order of a group element, no positive power can return the pair to the identity element, so the pair also has infinite order.

For example, in $C_4\times C_6$, the additive element $(2,3)$ has component orders $2,2$, hence pair order $2$. The element $(1,1)$ instead has order $\mathrm{lcm}(4,6)=12$. The product has $24$ elements, so even this latter element does not generate it. This distinguishes group order from order of a group element while exposing the simultaneous-return interpretation.

The order of a finite group is its cardinality, written $|G|$. The identity element is included: $|\mathbb{Z}/6\mathbb{Z}|=6$, and the trivial group has order one, rather than zero.

Keep this count distinct from the order of a group element. In the additive finite cyclic group $\mathbb{Z}/6\mathbb{Z}$, the element $1$ has order six, $2$ has order three, $3$ has order two, and $0$ has order one. Each element's order is the least positive number of repetitions needed to return to the identity element $0$. The order of an element of a finite cyclic group computes these periods directly.

The count also constrains subgroups: Lagrange's theorem gives $|G|=[G:H]|H|$ for a subgroup $H$ of a finite group $G$, where the index of a subgroup counts its cosets. For $H=\{0,2,4\}$ inside $\mathbb{Z}/6\mathbb{Z}$, the two cosets are $\{0,2,4\}$ and $\{1,3,5\}$, so $6=2\cdot 3$. Divisibility is a necessary constraint; it does not imply that every divisor occurs as the order of a group element, as the Klein four-group shows.

With $R=C_2(z)$, $S={\sigma }_{xz}$ and $T={\sigma }_{yz}$, composition givesEvery entry belongs to the set, proving closure. $E$ is the identity element; every element is its own inverse. Associativity follows because these are actual spatial transformations: composing functions is associative. Thus all group axioms hold, and the group is isomorphic to the Klein four-group.

For a concrete matrix check, use $E=\mathrm{diag}(1,1,1)$, $R=\mathrm{diag}(-1,-1,1)$, $S=\mathrm{diag}(1,-1,1)$ and $T=\mathrm{diag}(-1,1,1)$. Their matrix products reproduce the table.

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.

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.

Standards of mathematical proof changed between the seventeenth and nineteenth centuries because both the objects studied and the operations performed on them changed. A finite geometrical construction, an infinite expansion, and an abstract group raise different questions about justification. There was no universal seventeenth-century indifference to rigour followed by its sudden nineteenth-century invention. The more defensible thesis is that successful methods repeatedly exposed unstated assumptions; later work made some of those assumptions into explicit definitions, hypotheses and existence arguments.

Pierre de Fermat's adequality illustrates the distinction between an effective method and its justification. For $F(a)=a(L-a)$, comparing neighbouring expressions givesDividing by $e$ and suppressing the remaining term in $e$ produces $a=L/2$. There is an independent finite verification:Thus the result can be secure even when the general procedure raises a question: why is the increment first nonzero for division and then discarded? In the parabola construction of question 2, the same issue appears in replacing a strict geometrical comparison by adequality and extracting the subtangent. A historical explanation must describe this operation on its own terms rather than silently equate it with a modern derivative.^[ref]

Isaac Newton and Gottfried Wilhelm Leibniz supplied different languages for the new calculus. Isaac Newton's Introduction to the Quadrature of Curves by Isaac Newton, first published with the 1704 Opticks, presents quantities generated by motion and considers their rates of change. Its mature language of first and last ratios seeks exact relations as increments vanish; it should not be confused with simply asserting that a small error is harmless. Gottfried Wilhelm Leibniz's 1684 Nova methodus pro maximis et minimis by Gottfried Wilhelm Leibniz presents a calculus of differentials and operational rules. Their common problem was to explain why procedures using vanishing increments establish exact results, while their chosen objects and modes of explanation differed.^[ref][ref]

George Berkeley's The Analyst by George Berkeley (1734) exposes the logical pressure point. For the cube,The finite identity does not become a legal division at $h=0$. The modern repair is to keep $h\ne 0$ throughout the identity and prove that its right-hand side tends to $3x^2$. That limit is different from a quotient evaluated at zero. George Berkeley's criticism therefore concerns the grounds of inference, not whether the established rules often produce correct answers. His theological polemic also reminds us that standards of mathematical proof could be contested outside specialist calculation.^[ref]

The infinite introduced another burden. Joseph Fourier's 1822 use of trigonometric series to represent arbitrary functions greatly extended what calculations in mathematical physics could express. But deriving Fourier coefficients by orthogonality only shows what those coefficients must be if the proposed operations are legitimate. It does not prove convergence, equality with the initial data, or interchange of an integral and an infinite sum. The example of a constant expanded in a Fourier sine series makes a hidden qualification visible: the terms all vanish at the endpoints, even when the assigned data do not. The correct theorem must specify the domain and mode of representation.^[ref]

Peter Gustav Lejeune Dirichlet's 1829 memoir supplies a particularly precise example of changing proof obligations. It criticises a convergence argument of Augustin-Louis Cauchy's which used complex substitution into an arbitrary function and an insufficient comparison between signed series. Peter Gustav Lejeune Dirichlet supplies the instructive pairThe first series converges by the alternating series test; the second diverges because their difference is the harmonic series. Nevertheless $b_n/a_n=1+(-1{)}^n/\sqrt{n}\to 1$. Thus asymptotic equality of terms does not transfer convergence between conditionally convergent signed series. A comparison principle valid for positive terms needs additional justification when cancellation is involved. The example refutes a mode of inference rather than merely correcting a numerical answer.^[ref]

Augustin-Louis Cauchy's own foundations programme is part of this change. The 1821 Cours d'analyse by Augustin-Louis Cauchy makes limits, continuity and series convergence central to elementary reasoning. Yet its Sum theorem of Cauchy also shows why announcing rigour does not make every proof adequate. In modern terminology, continuous functions can have a discontinuous pointwise limit. An explicit series version isEach term is a polynomial and the partial sum is $x-x^{N+1}$. The convergence does not hold in uniform convergence: with $x_N=1-1/N$, the error $x_N^{N+1}$ tends to $e^{-1}$ rather than zero. The modern uniform limit theorem repairs the inference by requiring one bound on the error to work throughout the domain. This example diagnoses the missing control under a pointwise reading; it does not settle every historical interpretation of Augustin-Louis Cauchy's infinitesimal language.^[ref]

The issue is the order of dependence in quantifiers. Pointwise convergence permits the threshold $N$ to depend on both the error tolerance and the point. Uniform convergence requires a threshold depending only on the tolerance. If $f_N$ is continuous near $x_0$ and the convergence is uniform, choose $N$ so that $|f-f_N|<\epsilon /3$ nearby, then use the continuity of this fixed $f_N$ to make $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality proves $|f(x)-f(x_0)|<\epsilon$. Writing out this proof reveals what a vague statement that the tail is small leaves unproved.

Another nineteenth-century development concerns the foundations of the real numbers themselves. Richard Dedekind's Continuity and Irrational Numbers by Richard Dedekind (1872) asks for an arithmetical account of the continuity often borrowed from a drawn line. A Dedekind cut separates the rational numbers into lower and upper classes. For instance,defines a gap in the rational numbers corresponding to $\sqrt{2}$. The cut is determined without assuming that a rational number fills it. The extended number system then supports existence theorems that a geometrical picture alone had suggested.^[ref]

This matters for apparently elementary proofs. Suppose $f$ is continuous on $[a,b]$ with $f(a)<0<f(b)$. Repeated bisection constructs nested intervals whose endpoint values have opposite signs and whose lengths tend to zero. To conclude that there is a common limiting point, one needs the completeness of the real numbers. Continuity then forces its value to be zero. Over the rational numbers, $f(q)=q^2-2$ changes sign between $1$ and $2$ but has no rational zero. Thus even a proof that avoids pictures must explain the existence of its limiting object. Richard Dedekind's account makes that requirement an explicit part of the mathematical setting, rather than an intuition about a continuous line.^[ref]

There is a parallel change in algebra. Heinrich Weber's 1882 definition in question 6 describes arbitrary elements through laws of composition. To assess such a definition one proves implications between axioms. The passage's finite set hypothesis implies that cancellative multiplication has an identity element and inverse elements; without that hypothesis, the positive integers under addition supply a counterexample. This is a new kind of proof obligation: one must establish that a proposed abstract definition has the intended consequences, rather than rely on their familiarity in particular numerical or permutation examples.^[ref]

The principal change was towards explicit control of operations, hypotheses and existence. Effective symbolic algorithms persisted, and geometric reasoning remained valuable. What became increasingly demanding was the passage from success in examples to a general claim: an infinitesimal step needed justification, an infinite expansion needed a convergence theorem, a limiting point needed a complete number system, and an abstract definition needed its consequences proved. The examples also prevent a simple progress story: nineteenth-century mathematicians disagreed about admissible reasoning, and a programme devoted to rigour could itself become the object of a successful critique.

Let $A$ be a three-dimensional vector space with basis $p,q,z$. Define a bilinear map $\bullet :A\times A\to A$ bywith all other products of basis vectors zero. Equivalently,Every product belongs to $kz$, and every product with an element of $kz$ is zero. Hence both $(u\bullet v)\bullet w$ and $u\bullet (v\bullet w)$ vanish for every $u,v,w\in A$. This proves associativity, so $A$ is an associative algebra. The product is an alternating multilinear map, and each nested product in the Jacobi identity is zero, so the same product is a Lie bracket. It is nonzero because $p\bullet q=z\ne 0$.

There is no multiplicative identity element: any candidate $1_A$ would satisfy $1_A\bullet z=0$, whereas an identity element would require $1_A\bullet z=z$. ThusThis is an associative Lie product; its underlying Lie algebra is the Heisenberg Lie algebra.

Let $O=(0:1:0)$ be the point at infinity of the Weierstrass equation of an elliptic curve. It is the identity element for the elliptic curve group law. For an affine rational point $P=(x,y)$, the inverse isIndeed, the two affine intersections with the vertical line have these ordinates, since their sum as roots of the quadratic in $y$ is $-a_{1}x-a_3$. This reflection is valid for a generalized Weierstrass equation of an elliptic curve in every characteristic of a field.

For distinct $P,Q$, draw their line; for $P=Q$, draw the tangent line. Let $R$ be its third intersection with the elliptic curve, counted with intersection multiplicity. Then the sum is the reflection of the third intersection:A vertical line has third intersection $O$, so $P+(-P)=O$. If a tangent has third intersection $O$, then $2P=O$. Define addition with $O$ by $P+O=P$; these conventions include all exceptional cases of the affine construction.

The construction is defined over the coefficient field: after two known intersections, the remaining root is rational over that field. It gives a commutative group operation. To see why it is associative, use the elliptic curve group law from Riemann-Roch: the map $P\mapsto [P-O]$ identifies the elliptic curve with its degree-zero divisor class group. A line has intersection divisor $P+Q+R$ linearly equivalent to $3O$, the divisor of the line at infinity. Thus $[P-O]+[Q-O]+[R-O]=0$. The geometric sum agrees with addition of divisor classes, which is associative.

Compose permutations from right to left. Tracking the four symbols through the two permutation cycles gives $1\mapsto 2\mapsto 1$ and $3\mapsto 4\mapsto 3$. Thus the cycle decomposition of a permutation isThe disjoint transpositions commute, and squaring their product gives the identity element, while the product itself is nonidentity. Its order of a group element is therefore two. Each transposition has negative sign of a permutation, so the product has sign $(-1{)}^2=\boxed{+1}$: it is an even permutation.

If $y=gx{g}^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=g{x}^k{g}^{-1}$ for every positive integer $k$. Hence $y^k$ is the identity element exactly when $x^k$ is, proving that conjugate group elements have the same order of a group element. For permutations, the sign group homomorphism similarly givesThus conjugate permutations have the same sign of a permutation.

The converse is false: permutation order and sign do not determine conjugacy. In the symmetric group $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gx{g}^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their cycle types $31^3$ and $3^2$ differ. Equal order and sign are necessary, but not sufficient, for conjugacy.

Assume first that the topological group is Hausdorff. This separation condition is necessary: an indiscrete group with two elements has the one-element neighbourhood basis consisting of the whole group, but its topology cannot come from a metric.

Translations are homeomorphisms, since multiplication by a fixed element is continuous and its inverse is multiplication by the inverse element. Inversion is likewise a homeomorphism. If $W$ is an neighbourhood of the identity element, continuity of the triple product at $(e,e,e)$ supplies neighbourhoods of the identity element $A,B,C$ with $ABC\subset W$. Their intersection with their inverses contains a symmetric open neighbourhood of the identity element $V$, and $V^3\subset W$. This proves the neighbourhood shrinking property needed below rather than assuming it.

Let $(B_n)$ be a countable neighbourhood basis at the identity element, choosing open representatives inside its members. Starting with $U_0=G$, recursively choose symmetric open neighbourhoods of the identity element such thatThey are still a basis. Their intersection is $\{e\}$: for $x\ne e$, the Hausdorff condition gives an neighbourhood of the identity element not containing $x$, and some $U_n$ lies inside it.

Here is the dyadic product estimate for group neighbourhoods. If $g_j\in U_{n_j}$ and ${\sum }_j{2}^{-n_j}\le 2^{-n}$, then the ordered product lies in $U_{n-1}$, for $n\ge 1$. Prove this by induction on the number of factors. A single factor has $n_j\ge n$ and hence belongs to $U_n\subset U_{n-1}$. For a longer word, choose the factor crossing the half-total-cost point. The words preceding and following it each have cost at most half the total, hence at most $2^{-n-1}$. The induction hypothesis puts both in $U_n$, and the middle factor also lies in $U_n$. Their threefold product is in $U_{n-1}$. Empty side words cause no problem because $e\in U_n$.

Give an edge from $x_{j-1}$ to $x_j$ cost $2^{-n_j}$ whenever $x_{j-1}^{-1}{x}_j\in U_{n_j}$, and defineChains exist because $U_0=G$. Reversing a chain proves symmetry, concatenating chains proves the triangle inequality, and translating every vertex on the left leaves all edge differences unchanged. Thus $d$ is left invariant. The product estimate also givesThe first implication follows by choosing a chain with total cost below the bound and applying the estimate with index $n+1$. If $d(x,y)=0$, the first implication puts $x^{-1}y$ in every $U_n$, so $x=y$. Hence $d$ is a genuine left-invariant group metric. These two inclusions show that its balls and the original neighbourhoods of the identity element refine one another; translations then show equality of the two topologies everywhere. More explicitly, every metric ball contains a sufficiently small translated $U_n$, and every open neighbourhood contains a sufficiently small metric ball. This proves the sufficient direction of the Birkhoff-Kakutani theorem.

Conversely, if a compatible metric exists, the balls $B_d(e,1/n)$ form a countable neighbourhood basis at the identity element. Invariance is not even needed for this direction. Thus, with the necessary Hausdorff convention, a compatible left-invariant metric exists exactly when the identity element has a countable neighbourhood basis.

For the requested counterexample, take the orientation-preserving affine group of the real line, represented by $(a,b)$ with $a>0$ and product $(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b+a{b}^{\prime })$. Its topology is metrizable, for instance by the ordinary Euclidean distance in coordinates $(\log a,b)$, and the product and inverse are continuous. Let $D_a=(a,0)$ and $T_b=(1,b)$. Direct multiplication gives $D_a{T}_b{D}_a^{-1}=T_{ab}$. A compatible bi-invariant group metric would make every conjugation an isometry fixing $e$. ThereforeBut $T_{1/n}\to e$ in the group's topology, a contradiction. This is the affine conjugation obstruction to a bi-invariant group metric: the group is metrizable, and even admits a compatible left-invariant metric, but admits no compatible metric invariant on both sides.

There is an essential qualification to the printed hypothesis. If compactly generated group means merely generated algebraically by a compact subset, the claim is false. In the additive group $\mathbb{Q}$ with its usual topology,is compact, being a convergent sequence together with its limit, and generates every rational number since any integer denominator divides some factorial. The group is metrizable. However, an invariant Radon measure has a common mass $c$ on each singleton. If $c>0$, the compact set $K$ has infinite measure, contradicting finiteness on compact sets. If $c=0$, countable additivity makes the entire countable group have measure zero. Thus it has no nonzero Haar measure. This proves that compact generation does not imply local compactness and disproves the literal unrestricted statement.

The intended existence statement is valid when local compactness is included, either explicitly or in the convention that the group has a compact generating neighbourhood. Here is a construction for a locally compact, compactly generated, metrizable Hausdorff group. No invariant measure is assumed during its construction.

Enlarge a compact generating set to a symmetric compact neighbourhood of the identity element $S$. Then $G={\bigcup }_n{S}^n$, and each $S^n$ is compact. Moreover $S^n$ lies in the interior of $S^{n+1}$, since multiplication by the open neighbourhood of the identity element contained in $S$ supplies an open neighbourhood of every point of $S^n$. This gives a compact exhaustion. Every compact metric space has a countable dense subset: take finite $1/k$-nets and unite them. Their union over this exhaustion is dense in $G$, so the metric space is separable and has a countable open basis.

We use nonnegative compactly supported continuous functions and cutoffs equal to one on specified compact sets. Their existence is covered by the source's permission concerning auxiliary functions. Fix a nonzero such function $h$. For nonzero $u\ge 0$ define the weighted covering quantityThe quantities are finite: an open set where $u$ is bounded below has translates covering the compact support of $f$, so finitely many sufficiently large weighted translates dominate it. The denominator is positive, since evaluation at a point where $h>0$ gives $(h:u)\ge h(y)/\Vert u{\Vert }_{\infty }>0$.

Each $I_u$ is nonnegative, homogeneous, monotone, left invariant and subadditive; these follow respectively by rescaling, comparing or combining covers, and translating their centers. Also $I_u(h)=1$. Composing a cover of $f$ by translates of $h$ with a cover of $h$ by translates of $u$ givesThis is a bound independent of $u$. On functions supported in a fixed compact set $C$, choose a cutoff $k\ge 1$ on $C$. Monotonicity and subadditivity implyThus the covering functionals are uniformly Lipschitz on each fixed support.

Choose a nonzero $u_n\ge 0$ with support in an neighbourhood of the identity element shrinking to $e$. The nonnegative functions supported in each compact member of the exhaustion form a separable metric cone in the supremum norm, as a subspace of continuous functions on that compact metric space. Select a countable dense family from each cone and include $h$. The uniform scalar bounds permit a diagonal subsequence along which $I_{u_n}$ converges on their countable union. The displayed Lipschitz estimate extends convergence to every nonnegative member of $C_c(G)$. Denote the limit by $I$. It inherits positivity, homogeneity, monotonicity, subadditivity, invariance and $I(h)=1$.

The crucial remaining step is additivity. Fix $f_1,f_2\ge 0$ with combined support $C$, and fix a compact neighbourhood of the identity element $V_0$. Choose a cutoff $k$ equal to one on $C{V}_0^{-1}{V}_0$. For $\delta >0$, putsetting the ratio to zero off the denominator's nonzero set. These ratios are continuous: $k=1$ on a neighbourhood of the supports, and off those supports the numerator vanishes. Their sum is at most one. Compactness and continuity give, for each $\eta >0$, an neighbourhood of the identity element $V$ small enough that each $r_i$ oscillates by at most $\eta$ on any translate of $V$ meeting $C$. To justify this uniform choice, such translates lie inside the fixed compact set $C{V}_0^{-1}{V}_0$ when $V\subset V_0$; continuity of $(y,v)\mapsto r_i(yv)$ and a finite cover of that compact set give a single small right-increment neighbourhood, then choose $V^{-1}V$ inside it.

If $u$ has support in $V$ and $F\le {\sum }_j{c}_j{T}_{x_j}u$, discard translates not meeting $C$ when covering $f_i$. Let $a_{ij}$ be the supremum of $r_i$ on the remaining translate's support. Then $f_i\le {\sum }_j{a}_{ij}{c}_j{T}_{x_j}u$, and comparing both suprema with the ratios at one common point gives $a_{1j}+a_{2j}\le 1+2\eta$. Taking covering infima yieldsAlong the subsequence this holds eventually, since its supports shrink. Pass to $I$, then let $\eta \downarrow 0$ and $\delta \downarrow 0$; the uniform bound on $I_u(k)$ gives $I(f_1)+I(f_2)\le I(f_1+f_2)$. Subadditivity gives the reverse inequality. Thus $I$ is additive.

Extend $I$ linearly to real functions using positive and negative parts, and then complex-linearly to $C_c(G)$. The compact-support bounds above make it a positive, locally bounded functional. The Riesz representation on compactly supported continuous functions gives a Radon measure $\mu$ with $I(f)=\int fd\mu$. Translation invariance of $I$ and uniqueness in that representation give $\mu (xE)=\mu (E)$ for Borel $E$. It is nonzero because $I(h)=1$, and finite on compact sets because a cutoff dominating their indicators has finite integral. Finally every nonempty open set has positive measure: otherwise its translates cover the compact support of $h$, a finite subcover would give that support measure zero, contradicting $I(h)=1$. Hence $\mu$ is a Haar measure under the stated local-compactness qualification. This is Haar measure from normalized covering functionals, with the limiting additivity proved explicitly.

For a locally compact Hausdorff Abelian group $G$, Bochner's theorem states that a continuous positive-definite function $p$ has a unique representationwhere $\nu$ is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means ${\sum }_{i,j}{c}_i\overline{c_j}p(x_i{x}_j^{-1})\ge 0$ for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.

First prove existence without presupposing the representation. The elementary consequences of positive definiteness give $p(x^{-1})=\overline{p(x)}$, $p(e)\ge 0$ and $|p(x)|\le p(e)$. If $p(e)=0$, then $p=0$ and take $\nu =0$. Otherwise form finite formal linear combinations of symbols $[x]$, with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space $H$. Since $G$ is Abelian, translation $U_a[x]=[ax]$ preserves the form and extends to a unitary operator. Its inverse is $U_{a^{-1}}$, so $(U_a)$ is a unitary representation. With $v=[e]$,The translated copies of $v$ span a dense subspace. MoreoverFinite sums and density, together with $\Vert U_a\Vert =1$, give strong continuity on all of $H$. This constructs the cyclic unitary representation of a positive-definite function directly.

For $f\in L^1(G)$ define the integrated unitary representation $\pi (f)=\int f(x)U_{x}dx$, with the integral applied to each vector. Its norm is at most $\Vert f{\Vert }_1$. Fubini gives $\pi (f\ast g)=\pi (f)\pi (g)$, and inversion in the Abelian group gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions $u_V$ of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every $w\in H$. These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.

The Commutative Gelfand--Naimark theorem identifies $\mathcal{A}$ with $C_0(X)$, where $X$ is its character space. The positive vector functional $\omega (B)=⟨Bv,v⟩$ becomes a bounded positive functional on $C_0(X)$, so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure $\rho$ withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.

For each $\theta \in X$, the functional $f\mapsto \theta (\pi (f))$ is a nonzero algebra character of $L^1(G)$, since $\pi (L^1(G))$ is dense in $\mathcal{A}$. By Question 3 it has the form $\int f(x){\chi }_{\theta }(x)dx$. The resulting map $\iota :X\to \hat{G}$ is continuous: pointwise convergence of $\theta$ on $\mathcal{A}$ gives convergence on all $\pi (f)$, and the topological character identification in Question 3 gives compact-open convergence of ${\chi }_{\theta }$. Push $\rho$ forward under $\iota$ to obtain $\nu$. This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.

To recover $p(x)$, note that $U_x\pi (u_V)=\pi (T_x{u}_V)\in \mathcal{A}$. HenceThe left side tends to $p(x)$. On the right, $|{\Phi }_{\chi }(u_V)|\le 1$ and ${\Phi }_{\chi }(u_V)\to 1$ uniformly on compact subsets of $\hat{G}$. For completeness, the evaluation $(\chi ,x)\mapsto \chi (x)$ is jointly continuous: near $x_0$ restrict $x$ to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at $e$, proving the stated uniform convergence. Continuity of $\iota$ gives uniform convergence on compact subsets of $X$, and the finite Radon measure $\rho$ makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking $x=e$ also gives the required total mass $p(e)$.

Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in $x$ near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.

Finally suppose two finite Radon measures have the same transform. Multiply by $f\in L^1(G)$ and integrate over $G$. Fubini shows that the measures agree on every function $\chi \mapsto {\Phi }_{\chi }(f)$. Question 3 showed that these functions are uniformly dense in $C_0(\hat{G})$. Boundedness of both measure functionals extends equality to all of $C_0(\hat{G})$; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular $p(e)=1$ corresponds exactly to a probability measure on the dual group.