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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 character group of a topological group that is an abelian group consists of its continuous group homomorphisms to the circle group, with pointwise multiplication. For a finite group, all characters are continuous and their values are roots of unity, so this agrees with the character group of a finite abelian group.

For a topological group $G$, a universal principal bundle $EG\to BG$ has contractible total space and classifies numerable principal $G$-bundles by pullback. Over a paracompact base, isomorphism classes correspond to homotopy classes of maps to $BG$. For a discrete group, its classifying space of a discrete group is a $K(G,1)$; for the circle, $B{S}^1\simeq {\mathbb{CP}}^{\infty }$.

A closed subgroup is a subgroup which is a closed set in the ambient topology. It is a topological group in its subspace topology, but need not be an open subgroup.

The additive subgroup $\mathbb{Z}\subset \mathbb{R}$ of integers is a closed subgroup: every noninteger lies in an open interval $(n,n+1)$ avoiding $\mathbb{Z}$. It is also a discrete subgroup, since $(n-1/2,n+1/2)\cap \mathbb{Z}=\{n\}$. Nevertheless it is not an open subgroup of $(\mathbb{R},+)$: every open interval containing an integer contains nonintegers too. Thus having a discrete topology internally says nothing about being an open set in the ambient topology.

A compact group is a topological group whose underlying topological space is a compact Hausdorff space. The circle group is a compact group: the unit circle is a closed bounded subset of the complex plane, so the Heine-Borel theorem gives compactness. The additive group $(\mathbb{R},+)$ is a locally compact group and a Hausdorff space, but is not a compact group: the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. A finite group with the discrete topology is another compact group; compactness does not require a smooth manifold structure.

The great advantage of a compact group is that one can average over every element without choosing a preferred origin. Precisely, every locally compact group that is a Hausdorff space admits a nonzero left Haar measure $\mu$, a regular Borel measure finite on compact sets, unique up to a positive scalar. Left invariance means $\mu (gA)=\mu (A)$ for every Borel set $A$ and every $g\in G$. There is also a right Haar measure; in general left and right Haar measures can differ. On a compact group, $0<\mu (G)<\infty$, and normalizing to $\mu (G)=1$ gives a unique regular Borel probability measure invariant under both left and right translation. See Tornier, Haar Measures, Theorem 2.2 for existence and uniqueness, and Ram's notes on left and right Haar measure for the distinction.

For a finite group $G$, normalized Haar measure assigns each element mass $1/|G|$, so integration is simply $|G{|}^{-1}{\sum }_{g\in G}f(g)$. For the circle group, the corresponding integral is uniform angle averaging, computed below. On $(\mathbb{R},+)$, Lebesgue measure is a Haar measure, but its total mass is infinite: no positive scalar multiple makes it a probability measure. Thus the existence of Haar measure does not make every topological group a compact group.

Historically, Alfréd Haar introduced the general invariant-measure construction in Der Massbegriff in der Theorie der kontinuierlichen Gruppen, Annals of Mathematics 34 (1933), pp. 147-169. This bibliographic identification is also recorded in André Weil's archived seminar notes, Mesure de Haar, reference Haa33. The conceptual step is to let group symmetry itself determine the weights used in integration, rather than relying on coordinates in an ambient Euclidean space.

The nonzero complex numbers form an abelian group under multiplication, with identity one and inverse $z^{-1}$. As a topological group it is isomorphic to $\mathbb{R}\times S^1$ by logarithmic modulus and phase. It appears naturally as a Möbius pointwise stabilizer of two points.

A continuous unitary character is a continuous group homomorphism from a topological group to the unit circle with multiplication. For abelian groups, multiplication of continuous unitary characters is pointwise. Such continuous unitary characters must be distinguished from the traces of higher-dimensional group representations, called characters of a representation, and from characters of an algebra.

For every integer $n$, the map ${\chi }_n(z)=z^n$ is a continuous unitary character of the circle group: ${\chi }_n(zw)={\chi }_n(z){\chi }_n(w)$ and $|{\chi }_n(z)|=1$. By contrast, the two-dimensional group representation $\rho (z)=\mathrm{diag}(z,z^{-1})$ has character of a representation $\mathrm{tr}\rho (e^{i\theta })=2\cos \theta$. Its value at the identity element is $2$, so this trace is not a continuous unitary character.

A subgroup $G$ of a topological group is discrete when the identity has a neighbourhood containing no other element of $G$. Translation then isolates every group element. In a Hausdorff topological group, such a subgroup is closed: any accumulation at a point would make differences of distinct nearby group elements approach the identity. Consequently its intersection with a compact set is finite. For isometries of hyperbolic space, this converts compactness of the set of isometries moving one point a bounded distance into a properly discontinuous group action.

For example, $(-1/2,1/2)\cap \mathbb{Z}=\{0\}$ isolates the identity element of the additive discrete subgroup $\mathbb{Z}\subset \mathbb{R}$. Its induced subspace topology is the discrete topology, while the surrounding real numbers retain their usual topology; see closed subgroup for why this subgroup is closed but not open.

A subset of a topological space is discrete when its subspace topology is a discrete topology: every point has an ambient neighbourhood meeting the subset only in that point. A discrete subset need not be closed; for example $\{1/n:n\ge 1\}$ in the real line accumulates at zero. The group structure makes a discrete subgroup of a Hausdorff topological group closed, an additional property used in discontinuity arguments.

For $n\ge 1$, pointwise multiplication of based cube maps into a topological group induces the ordinary addition in its homotopy group ${\pi }_n(G,e)$. Concatenation and pointwise multiplication share an identity and satisfy the interchange rule, so the Eckmann-Hilton argument identifies the two operations. In particular, the $m$-th power map induces multiplication by $m$ on every homotopy group, for all integers $m$.

A locally compact group is a topological group whose underlying space is a locally compact space. In Haar measure and Fourier transform theory one normally also assumes it is a Hausdorff space. Local compactness supplies compact neighbourhoods, not just compact sets which algebraically generate the group.

A locally profinite group is a Hausdorff, locally compact, totally disconnected topological group. It has a neighbourhood basis at the identity consisting of compact-open subgroups. In smooth representation theory this is also called an $\ell$-group; that terminology is unrelated to a lattice-ordered group.

For an open identity neighbourhood $V$ in a compact topological group, a finite left covering net is a finite set $F$ such that the left translates $fV$, $f\in F$, cover the group. Compactness gives such finite sets; minimum cardinality gives a minimal one. Left translation preserves minimality because $G=FV$ implies $G=gFV$.

A topology on a group makes it a topological group precisely when its neighbourhood bases satisfy two local tests: every neighbourhood $U$ of $xy$ contains $VW$ for some neighbourhoods $V$ of $x$ and $W$ of $y$, and every neighbourhood of $x^{-1}$ contains $V^{-1}$ for some neighbourhood $V$ of $x$. These are exactly continuity of multiplication in the product topology and continuity of inversion. Equivalently, bases can be chosen as left translates of a basis at the identity, with arbitrarily small squares, inverses and conjugates inside each prescribed identity neighbourhood.

An open normal subgroup $N$ of a topological group is simultaneously open in its topology and invariant under conjugation. For a profinite group, the quotient $G/N$ is a finite discrete group: its cosets form a disjoint open cover, so compactness gives finitely many. Conversely the group homomorphism kernel of a continuous homomorphism to a finite discrete group is open and normal. Such subgroups form an identity-neighbourhood basis of every profinite group.

Every identity neighbourhood of a compact Hausdorff totally disconnected topological group contains an open normal subgroup. Choose a clopen set $U$ containing the identity inside that neighbourhood. Compactness and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$, hence $VU=U$. The setwise stabilizer $H=\{g:gU=U\}$ is an open subgroup contained in $U$. Its index is finite, so its normal core is an open normal subgroup contained in $U$. These subgroups recover the profinite group as the inverse limit of its finite quotients.

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.

An inverse system of topological spaces over a directed set $I$ consists of spaces $X_i$ and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=1$ and $f_{ik}=f_{ij}{f}_{jk}$. Its inverse limit is a space $X$ with compatible continuous projections ${\pi }_i$ such that every compatible family of continuous maps $u_i:Y\to X_i$ factors uniquely through a continuous map $u:Y\to X$. To construct it, takewith the subspace topology inherited from the product topology. The map $u(y)=(u_i(y))$ has values in $X$ and is continuous by the definition of the product topology; its coordinates force uniqueness. If $X^{\prime }$ is another inverse limit, the two universal properties give continuous maps $X\to X^{\prime }$ and $X^{\prime }\to X$ whose composites have all the same projections as the identity. Uniqueness makes these composites the identity. The inverse limit exists and is unique up to a unique projection-preserving homeomorphism. An arbitrary inverse limit may be empty; nonemptiness requires additional hypotheses.

Use the usual Hausdorff convention for a topological group. An inverse limit of finite discrete groups is a closed subgroup of their product: each compatibility condition is closed because the finite factors are Hausdorff spaces. The Tychonoff theorem gives compactness, and distinct tuples are separated by a clopen coordinate cylinder, so the limit is totally disconnected. Coordinatewise multiplication and inversion are continuous, giving a profinite group.

Conversely, let $G$ be a compact Hausdorff space and a totally disconnected topological group. We use the standard topological fact that a compact Hausdorff space that is totally disconnected has a basis of clopen sets. Given an identity neighbourhood, choose a clopen identity neighbourhood $U$ inside it. Compactness of $U$ and continuity of multiplication give a symmetric identity neighbourhood $V$ with $VU\subseteq U$: cover $U$ by finitely many neighbourhoods on which multiplication by a sufficiently small identity neighbourhood stays inside $U$, then intersect those identity neighbourhoods. Symmetry gives $vU=U$ for $v\in V$. Consequently $H=\{g:gU=U\}$ is a subgroup containing $V$, hence an open subgroup, and $H\subseteq U$ since $1\in U$. Its open cosets cover the compact group, so there are finitely many. The normal core $N={\bigcap }_{g\in G}gH{g}^{-1}$ is a finite intersection of open subgroups, hence an open normal subgroup contained in $U$. Thus the open normal subgroups form an identity-neighbourhood basis and have trivial intersection.

The canonical group homomorphism $G\to {\underleftarrow{\lim }}_{N}G/N$ is continuous and injective. For a compatible tuple of quotient elements, its inverse-image cosets in $G$ are closed and have the finite intersection property: a finite family is refined by the quotient for the intersection of its normal subgroups. Compactness gives an element in all those cosets, proving surjectivity. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The compact totally disconnected and inverse-limit definitions of a profinite group agree.

Index the finite nonempty sets $\mathcal{P}(N)$ by open normal subgroups, with the transition from $M$ to $N$ when $M\subseteq N$ given by image under $G/M\to G/N$. Images of Sylow subgroups under surjections of finite groups are Sylow subgroups. These transitions are surjective: if $S\le G/N$ is a Sylow subgroup, choose a Sylow subgroup in its full inverse image in $G/M$. It maps onto $S$, and its order has the full $p$-part of $|G/M|$, so it is a Sylow subgroup of $G/M$ too. The transitions respect composition, giving an inverse system of finite sets.

Its inverse limit is nonempty. Indeed, finitely many compatibility conditions can be satisfied by choosing one Sylow subgroup in a quotient refining all the involved quotients and taking its images; compactness of the product of the finite sets then gives a compatible family $(P_N{)}_N$. DefineThis is a closed subgroup of $G$. Its image in $G/N$ is exactly $P_N$: for a prescribed $s\in P_N$, each finite collection of conditions defining $P$ together with ${\pi }_N(g)=s$ is satisfiable in a common refining quotient, because the transitions on the chosen Sylow subgroups are surjective. The finite intersection property again supplies $g\in P$. Hence $P\cong {\underleftarrow{\lim }}_N{P}_N$ is a pro-p group and $PN/N=P_N$ for every $N$.

Finally, if a pro-p subgroup $Q$ contains $P$, its image in every finite quotient is a p-group containing the Sylow subgroup $P_N$, so that image equals $P_N$. Thus every $q\in Q$ belongs to every ${\pi }_N^{-1}(P_N)$, whence $Q\subseteq P$. The constructed $P$ is a maximal pro-p subgroup.

A topological group is a group with a topology for which $(x,y)\mapsto xy$ and $x\mapsto x^{-1}$ are continuous; we use the Hausdorff convention. Translations and inversion are then homeomorphisms.

For the final generator-extraction assertion, use the profinite Frattini subgroupFor a finitely generated pro-p group, the Burnside basis theorem says that $G/\Phi (G)$ is a finite-dimensional vector space over ${\mathbb{F}}_p$ and a subset topologically generates $G$ precisely when its image spans that vector space. In particular its dimension is the minimal number $d(G)$ of topological generators. A given $d$-element generating set shows $d(G)\le d$. The image of any other generating set spans the same finite-dimensional vector space, so it contains a basis of $d(G)$ elements. Choose their preimages in that generating set and apply the Burnside basis theorem again. Every generating set contains a generating subset of size $d(G)\le d$. This argument also applies to an infinite generating set.

Every nonzero P-adic number has a unique decomposition $z=p^{m}u$ with $m\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_p^{\times }$. The group homomorphism kernel of reduction from the unit group onto ${\mathbb{F}}_p^{\times }$ is $1+p{\mathbb{Z}}_p$, the group of principal units. Each nonzero residue class has a unique lift $\omega$ satisfying ${\omega }^{p-1}=1$, by the simple-root Hensel lemma proved in question 3. Uniqueness makes these lifts multiplicative. They form the cyclic group ${\mu }_{p-1}$, and every unit has a unique factorizationThus ${\mathbb{Q}}_p^{\times }=p^{\mathbb{Z}}\times {\mu }_{p-1}\times (1+p{\mathbb{Z}}_p)$.

For odd $p$, the p-adic logarithm and p-adic exponential function give inverse group isomorphisms $1+p{\mathbb{Z}}_p\leftrightarrow p{\mathbb{Z}}_p$. To verify the convergence domain, for $t\in p{\mathbb{Z}}_p$ the p-adic logarithm terms have valuations $n{v}_p(t)-v_p(n)\to \infty$, and the exponential terms have valuations $n{v}_p(t)-v_p(n!)\to \infty$, since $v_p(n!)\le (n-1)/(p-1)$ and $p>2$. In both series every term after the linear one has strictly larger valuation than the linear term. Consequently $\log (1+t)\in p{\mathbb{Z}}_p$ and $\exp (t)\in 1+p{\mathbb{Z}}_p$. The formal identities $\log (vw)=\log v+\log w$ and $\exp (\log v)=v$, $\log (\exp t)=t$ hold on these convergent domains, as follows by multiplying the convergent series or passing to their formal identities termwise.

Choose a generator $\zeta$ of ${\mu }_{p-1}$. An explicit isomorphism isThe inverse reads off the valuation, the root-of-unity unit factor, and $p^{-1}\log (v)$. Both directions are continuous with the standard product topology, so this is also a topological group isomorphism. The odd-prime condition is essential for using all of $1+p{\mathbb{Z}}_p$ as the p-adic logarithm-isomorphism domain.