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Identify the Möbius group with ${\mathrm{PSL}}_2(\mathbb{C})$, the orientation-preserving isometry group of three-dimensional hyperbolic space. Here a properly discontinuous group action permits finite stabilizers: for each compact set $K$, only finitely many $g$ satisfy $gK\cap K\ne \varnothing$.

Fix $o\in {\mathbb{H}}^3$. The set of isometries with $d(o,go)\le R$ is compact. To see this, an orientation-preserving isometry is uniquely determined by the image of an oriented orthonormal frame at $o$. The image point lies in a compact closed hyperbolic ball and its frame lies in a compact rotation-group fiber. These choices parametrize the indicated set of isometries continuously.

A discrete subgroup of a topological group is closed: otherwise distinct subgroup elements accumulating at the same point would have quotients tending to the identity, contradicting isolation of the identity. Its intersection with a compact set is therefore finite. If $K\subseteq B(o,R)$ and $gK\cap K\ne \varnothing$, choosing $x,gx\in K$ gives $d(o,go)\le d(o,gx)+d(gx,go)\le 2R$. Hence discreteness implies a properly discontinuous group action. Conversely, if the subgroup is not discrete, distinct $g_n$ tend to the identity. For a compact ball $K$ with $o$ in its interior, eventually $g_{n}o\in K$, so infinitely many $g_{n}K$ meet $K$, contradicting proper discontinuity. This proves the equivalence.

Matrices of determinant one over the Gaussian integers are closed under multiplication and inversion, with inverse $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. Quotienting by $\{\pm I\}$ gives the specified group of Möbius transformations. The Gaussian integers form a discrete set in the complex plane, so these matrices form a discrete subgroup of ${\mathrm{SL}}_2(\mathbb{C})$, and the finite quotient remains discrete. Thus this group acts properly discontinuously on ${\mathbb{H}}^3$.

Explicit representatives of all four types areThe first is $z\mapsto -1/z$ and has order two; the second is the nonidentity translation $z\mapsto z+1$. The third has positive real reciprocal eigenvalues $(3\pm \sqrt{5})/2$, giving translation along an axis without rotation. For the last, a nonreal trace excludes both unit-modulus and real eigenvalues. Its multiplier therefore has modulus different from one and a nonzero rotation angle: it is a loxodromic Möbius transformation.

A Kleinian group is a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting by Möbius transformations on the Riemann sphere, equivalently by orientation-preserving isometries of hyperbolic space ${\mathbb{H}}^3$.

Consider the classes ofThey generate a discrete subgroup because their matrix representatives have integer entries: a sequence of such matrices converging to $I$ or $-I$ is eventually constant. On the real projective line set $X=\{x:|x|>1\}\cup \{\infty \}$ and $Y=\{x:|x|<1\}$. For every nonzero integer $m$,The last inclusion follows from $|2mx+1|\ge 2|x|-1>|x|$ when $|x|>1$; also $B^m(\infty )=1/(2m)\in Y$. The ping-pong lemma therefore says that the subgroup is the free product $⟨A⟩\ast ⟨B⟩$. Both generators have infinite order, soa free group on two generators. The ping-pong inclusions ensure that no nonempty reduced word can act as the identity, which is the essential freeness argument.

Fix a real angle fraction $\alpha$, and repeatedly multiply by $g=e^{2\pi i\alpha }$ in the circle group. The generated cyclic subgroup is $H_{\alpha }=\{g^k:k\in \mathbb{Z}\}$. Equivalently, the angle fractions are the fractional parts of $k\alpha$, with one full turn identified with zero.

If $\alpha =p/q$ is a rational number in lowest terms, with $q>0$ and $p,q$ coprime, then $g^q=1$ and no smaller positive power is $1$. Hence $H_{\alpha }$ is a finite cyclic group of order $q$, consisting of equally spaced points. For $\alpha =2/7$, the successive angle fractions, starting at $k=0$, areThe subgroup is finite, so it is a closed subgroup of the circle group; it is also a discrete subgroup. It is not an open subgroup, since every ambient neighbourhood of any of its points contains an arc, rather than only these seven points.

If $\alpha$ is an irrational number, no nonzero integer $k$ satisfies $k\alpha \in \mathbb{Z}$, so the powers never repeat. The stronger statement is that $H_{\alpha }$ is a dense set in the circle group: every nonempty open arc contains a power $g^k$. Thus its closure is all of $\mathbb{T}$, while $H_{\alpha }$ itself is a countable proper subgroup and is not a closed subgroup or a discrete subgroup. See irrational rotation and everywhere interval frequency under an irrational rotation for the corresponding motion and its long-term frequencies.

For $\alpha =\sqrt{2}-1$, direct computation of the fractional parts of $k\alpha$ gives, to six decimal places,For example, $\{3\alpha \}=3\sqrt{2}-4\approx 0.242641$ and $\{5\alpha \}=5\sqrt{2}-7\approx 0.071068$. These calculations specify exactly which points the diagram displays.

This example shows why algebraic generation and closure must be distinguished: one irrational rotation generates a countable cyclic subgroup, yet its closure is the whole uncountable compact group.

A translation lattice is the additive group generated by linearly independent vectors $v_1,\dots ,v_r$ in a Euclidean space. It defines a discrete subgroup of translations. A full-rank translation lattice has a compact parallelepiped as a fundamental region and therefore a cocompact group action. This geometric usage is distinct from an order-theoretic lattice.