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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.

Every scalar holomorphic function on a compact connected complex manifold is constant by the maximum modulus principle. Apply this to each entry of the holomorphic Adjoint representation of a Lie group; its value is constantly the identity. Differentiation makes the complex Lie algebra abelian, hence connectedness makes the group abelian. The holomorphic Lie exponential map identifies it with ${\mathbb{C}}^n$ modulo a discrete subgroup. Compactness forces that subgroup to span the underlying real vector space, giving a full-rank Euclidean lattice and a complex torus.

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.

A discrete subgroup of Euclidean isometries is a frieze group when its translation subgroup is infinite cyclic and has finite index. Translations along a line alone give $\mathbb{Z}$; adding perpendicular mirrors gives the infinite dihedral group, while adding a mirror along the translation direction gives $\mathbb{Z}\times C_2$. These groups describe patterns periodic in one planar direction.

For a non-elementary discrete subgroup generated by $A,B\in \mathrm{SL}(2,\mathbb{C})$, the displayed inequality is a necessary trace constraint. Failure of it proves nondiscreteness when non-elementarity has independently been checked. It is not a sufficient condition for discreteness and must not be applied to an elementary pair without its hypothesis.

A complex analytic Lie group is a complex manifold whose multiplication and inversion are holomorphic maps between complex manifolds. Its identity tangent space is a complex Lie algebra, and the Adjoint representation of a Lie groupis holomorphic: differentiating the holomorphic conjugation map with respect to its second variable gives holomorphically varying matrix entries.

Every scalar holomorphic function on a compact connected complex manifold is constant. Indeed its modulus attains a maximum; the maximum modulus principle, applied in complex coordinate charts and along complex lines, makes it locally constant there, and connectedness propagates that constancy. Apply this to each entry of $\mathrm{Ad}$. Since ${\mathrm{Ad}}_e=I$, it follows that ${\mathrm{Ad}}_g=I$ for every $g$. Differentiation givesThus $\mathfrak{g}$ is abelian, and the connected group $G$ is abelian by the identity-neighbourhood argument in part (i). This proves that compact connected complex Lie groups are complex tori.

If $n={\dim }_{\mathbb{C}}G$, identify $\mathfrak{g}$ with ${\mathbb{C}}^n$. The Lie exponential map is holomorphic, additive, onto, and a local biholomorphism, by the holomorphic left-invariant differential equation and its identity derivative. Its kernel $B$ is a discrete subgroup of the additive space ${\mathbb{C}}^n$. The induced map yieldsas complex Lie groups, not merely as real groups. Compactness further forces $B$ to be a full-rank real lattice. If its real span $W$ were proper, projection would give a surjective continuous map ${\mathbb{C}}^n/B\to {\mathbb{R}}^{2n}/W$, contradicting compactness. The discrete-subgroup argument in part (i) therefore gives $B\cong {\mathbb{Z}}^{2n}$. The quotient is a complex torus; it need not split into one-dimensional complex tori.

A discrete subgroup of the Euclidean isometry group has the identity isolated in the group topology. Equivalently, no sequence of distinct group elements converges to the identity; translating neighbourhoods then isolates every element. Such a subgroup is closed, and its intersection with any compact subset is finite: otherwise two elements of a convergent sequence would have quotient tending to the identity.

Write each Euclidean isometry uniquely as $x\mapsto Qx+b$, $Q\in O(2)$. Its linear part gives the group homomorphism $G\to O(2)$, since the linear part of a composite is the product of the linear parts. Its kernel is exactly the translation subgroup $G_T$. The translation vectors form a discrete additive subgroup $L\subset {\mathbb{R}}^2$.

If $L\ne \{0\}$, choose a shortest nonzero vector $a$, possible by discreteness and finiteness in compact balls. On its line, Euclidean division of a scalar coefficient by one proves $L\cap \mathbb{R}a=\mathbb{Z}a$: a nonzero remainder would be shorter than $a$. If all vectors are collinear, this finishes the proof. Otherwise orient the perpendicular coordinate so some vector has positive height. Reduce every vector's component along $a$ modulo $a$ into a bounded interval. Among vectors with positive height at most that of a chosen vector, these reduced representatives lie in a compact rectangle and are finite. Choose $b$ with the smallest such positive height. Subtracting an integer multiple of $b$ from any vector of $L$ reduces its height into $[0,b_{\perp })$; minimality makes the remainder's height zero, so it is an integer multiple of $a$. ThusThis derives the rank-at-most-two translation lattice rather than assuming every discrete subgroup of the plane is a full lattice.

Two examples with $G_T\cong \mathbb{Z}$ are the group generated by $T(x,y)=(x+1,y)$ alone, and the infinite dihedral group generated by $T$ together with $R(x,y)=(-x,y)$. The latter has $RTR=T^{-1}$; its elements are translations $T^k$ and reflections $T^{k}R$, whose mirror lines are $x=k/2$. The first group preserves the orientation of every translated motif; the second also reverses it across these mirrors.

Extend the Möbius transformations to isometries of three-dimensional hyperbolic space. For any interior point $o$, the limit set of a Kleinian group is the set of boundary accumulation points of $Go$. Its definition is independent of $o$: two interior orbits stay at a bounded hyperbolic distance, so have the same boundary limits. The Kleinian limit set is closed and $G$-invariant. Iterating a loxodromic Möbius transformation in the positive and negative directions shows that its two fixed points belong to the Kleinian limit set.

The two fixed-point sets in this problem cannot have exactly one common point. Otherwise conjugate that point to infinity and the other fixed point of the first transformation to zero. The transformations then have the forms $f(z)=\alpha z$, with $|\alpha |>1$, and $g(z)=\beta z+b$, with $b\ne 0$. But $f^{-n}g{f}^n(z)=\beta z+b{\alpha }^{-n}$ is a sequence of distinct group elements converging in the Möbius group. Taking quotients of consecutive elements gives nonidentity elements converging to the identity, contrary to discrete subgroup. Consequently the Kleinian limit set contains at least three points.

Let $\xi$ be any point of the Kleinian limit set, and choose distinct $g_n$ with $g_{n}o\to \xi$. The rank-one convergence of divergent Möbius transformations supplies a subsequence converging to $\xi$ uniformly away from one exceptional boundary point $\eta$. To see the lemma directly, normalize determinant-one representing matrices by their norms. The norms tend to infinity, and a subsequential limit has rank one: its kernel is $\eta$ and its image is the attracting point. In the positive-Hermitian-matrix model of hyperbolic space, $g_{n}o$ has this same image limit, so the attracting point is indeed $\xi$.

Choose distinct ${\zeta }_1,{\zeta }_2$ in the Kleinian limit set other than $\eta$. Then $g_n{\zeta }_1$ and $g_n{\zeta }_2$ both tend to $\xi$, belong to the Kleinian limit set, and remain distinct because $g_n$ is injective. If $\xi$ were isolated, both would eventually equal $\xi$, an impossibility. The limit set has no isolated points.

Here a planar lattice means a discrete subgroup of the additive Euclidean plane; full rank is not required. Discreteness at zero gives an $ϵ>0$ such that distinct lattice points are at least $ϵ$ apart. Consequently every bounded set contains only finitely many lattice points, by placing disjoint small disks around them. If the group is zero, it has the stated one-generator form with $w=0$.

Otherwise choose a shortest nonzero lattice vector $w_1$. Every lattice point on its line is an integer multiple: subtract the nearest integer multiple of $w_1$ from a putative noninteger multiple to obtain a nonzero shorter vector, a contradiction. This already proves the one-generator case when all lattice points lie on that line.

For the remaining case choose the perpendicular orientation so that some point has positive height above the line of $w_1$. Reduce its parallel component modulo $w_1$ into $[0,|w_1|)$. Among reduced points of positive height there is a least height $d>0$. Indeed, heights tending to zero would give infinitely many points in a bounded strip, contrary to the finiteness just proved; bounded sets also ensure that a positive infimum is attained. Choose $w_2$ at height $d$.

Given any lattice vector $v$, subtract an integer multiple of $w_2$ so that its height lies in $[0,d)$, then reduce its parallel component modulo $w_1$. Minimality of $d$ forces the residual height to be zero, and the one-dimensional argument forces the residual to be an integer multiple of $w_1$. ThereforeTogether with the rank-one and zero cases this proves the planar discrete-subgroup basis assertion. Under the alternative convention that a Euclidean lattice must have full rank, only the two-generator case is called a lattice.

A subgroup of the Möbius group ${\mathrm{PSL}}_2(\mathbb{C})$ is a discrete subgroup when the identity is isolated in its induced topology, equivalently every group element is isolated. We prove the compact-set criterion for a properly discontinuous group action: for every compact $K\subset {\mathbb{H}}^3$, only finitely many $g\in G$ satisfy $gK\cap K\ne \varnothing$.

Fix an origin $o$ and take $K\subseteq \overline{B}(o,R)$. If $x\in K$ and $gx\in K$, the triangle inequality and invariance of hyperbolic distance giveThe set of Möbius isometries with $d(o,go)\le 2R$ is compact. To see this explicitly, represent an isometry by a determinant-one complex matrix and use singular-value decomposition:The compact factors fix $o$, and the diagonal factor moves it hyperbolic distance $s$, as seen from the vertical dilation by $e^s$ in upper half-space. Thus the bounded-displacement set is the continuous image of $\mathrm{SU}(2)\times [0,2R]\times \mathrm{SU}(2)$, modulo central signs.

By the allowed closedness fact, its intersection with $G$ is a closed discrete subset of a compact space, hence finite. Otherwise an infinite sequence would accumulate in that intersection, contradicting isolation of its limit point. This proves every discrete Möbius subgroup acts properly discontinuously on hyperbolic three-space. Finite stabilizers are allowed; discreteness alone does not imply a free action.

Set $C=AB=\left(\begin{array}{cc}0& -1\\ 1& 2\lambda \end{array}\right)$. It fixes $z=-\lambda +i\sqrt{1-{\lambda }^2}$ because $z^2+2\lambda z+1=0$. Write $\lambda =\cos \theta$, $0<\theta <\pi /2$; the eigenvalues of $C$ are $e^{\pm i\theta }$.

The stabilizer of an interior point of $\mathbb{H}$ in $PS{L}_2(\mathbb{R})$ is conjugate to the compact rotation group. Its intersection with a discrete subgroup is finite: otherwise compactness gives accumulating elements, and their quotients accumulate at the identity. Thus $C$ has finite order in the projective group. Its eigenvalues then satisfy $e^{2im\theta }=1$ for some positive integer $m$, forcing $\theta /\pi \in \mathbb{Q}$. Hence $\boxed{\lambda =\cos (\pi y)\text{ for some }y\in \mathbb{Q}}$. This is a necessary condition only.

A frieze group is a discrete subgroup of Euclidean isometries whose translation subgroup is generated by one nonzero vector and has finite index. A plane crystallographic group is a discrete subgroup of Euclidean isometries with a rank-two translation lattice of finite index; equivalently its action has a compact fundamental region.

Here are three frieze groups, with explicit generators. The translation $T(x,y)=(x+1,y)$ alone gives $\mathbb{Z}$. Adding the vertical reflection in mathematics $S(x,y)=(-x,y)$ gives $\mathbb{Z}⋊C_2$, the infinite dihedral group, since $STS=T^{-1}$. Instead adding the horizontal reflection in mathematics $H(x,y)=(x,-y)$ gives $\mathbb{Z}\times C_2$, since $HT=TH$. Their translation parts are exactly $⟨T⟩$ and their point parts have respectively one, two and two elements.

For plane crystallographic groups, let $T_1(x,y)=(x+1,y)$ and $T_2(x,y)=(x,y+1)$. These generate ${\mathbb{Z}}^2$. Adding the half-turn $R_2(x,y)=(-x,-y)$ gives ${\mathbb{Z}}^2⋊C_2$, with inversion on both translation generators. Adding the quarter-turn $R_4(x,y)=(-y,x)$ gives ${\mathbb{Z}}^2⋊C_4$, with $R_4{T}_1{R}_4^{-1}=T_2$ and $R_4{T}_2{R}_4^{-1}=T_1^{-1}$. These are the translation, half-turn and square-rotation wallpaper groups. Integer translation vectors and a finite point part make each group a discrete subgroup; the unit square meets every orbit, establishing a cocompact group action.

For the final claim, suppose a Euclidean isometry $R$ has order five. An orientation-reversing planar isometry has finite order only when it is a reflection in mathematics, so $R$ is a rotation through $\theta =2\pi j/5$, $j=1,2,3,4$, about some centre. Write $A$ for its linear rotation part. If $t_v\in G$ is a nonzero translation, conjugation gives $R{t}_v{R}^{-1}=t_{Av}$. Thus $G$ contains translations by $v$ and by $(A+A^{-1})v=cv$, where $c=2\cos \theta$.

For a primitive fifth root $\zeta$, dividing $1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4=0$ by ${\zeta }^2$ gives $c^2+c-1=0$. Hence $c$ is irrational. The pigeonhole principle, applied to $N+1$ fractional parts of $0,c,\dots ,Nc$ in $N$ equal intervals, gives integers $m,n$ with $0<|nc-m|\le 1/N$. Consequently $G$ contains nonidentity translations by $(nc-m)v$ tending to the identity. This contradicts the discrete subgroup hypothesis. There is no element of order five. This argument needs only the given nonzero translation, not a rank-two lattice.

Consider the Möbius transformationsTheir group lies in ${\mathrm{PSL}}_2(\mathbb{Z})$, which is a discrete subgroup of the Möbius group: integer entries cannot approach the identity matrix except by eventually equalling it, and passing to the quotient by $\{I,-I\}$ preserves this property.

For freeness let $X=\{x\in \mathbb{R}:|x|>1\}$ and $Y=\{x\in \mathbb{R}:|x|<1\}$. For every nonzero integer $n$, $A^n(Y)=(2n-1,2n+1)\subset X$ andA reduced word involving both generators can be conjugated to a cyclically reduced word beginning with $A^n$ and ending with a nonzero power of $B$, after interchanging the generators if necessary. Repeated use of these inclusions shows that this word sends $X$ into the proper subset $A^n(Y)$ of $X$, so it cannot be the identity. Nonzero pure powers of either generator are also nonidentity. This is the ping-pong lemma in this particular action, and proves that $\boxed{⟨A,B⟩\cong F_2}$ is a free group of rank two as well as a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$.

Suppose that $H$ is not a discrete subgroup. The identity is then an accumulation point of nonidentity subgroup elements: translate any failure of isolation to the identity. Choose $h_j\in H\setminus \{e\}$ with $h_j\to e$, and use a local exponential chart to write $h_j=\exp X_j$, where $X_j\ne 0$ and $X_j\to 0$ in the real Lie algebra $\mathfrak{g}$.

Choose a norm on this finite-dimensional vector space. By compactness of its unit sphere, pass to a subsequence with $X_j/\Vert X_j\Vert \to X$, where $\Vert X\Vert =1$. For an arbitrary real $t$, choose an integer $m_j$ nearest to $t/\Vert X_j\Vert$. Then $m_j\Vert X_j\Vert \to t$ andEvery left-hand element belongs to $H$, including for negative $m_j$. Closedness gives $\exp (tX)\in H$. Since $t$ was arbitrary and $X\ne 0$, the limit directions of a closed subgroup provide a nontrivial one-parameter subgroup in $H$. Its derivative at zero is $X$, so it cannot be the constant subgroup.