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For a group of isometries acting properly discontinuously on the hyperbolic plane, and a point $p$ with trivial point stabilizer, the closed Dirichlet domain isIt is the intersection of the closed half-planes bounded by perpendicular bisectors between $p$ and its other group orbit points. The group orbit is locally finite and the hyperbolic plane is a proper metric space, so every point has a nearest group orbit point. This shows that the translates of $D_G(p)$ cover the plane. Distinct interiors are disjoint, since being strictly closer to two different group orbit points is impossible. Boundary ties are allowed in the closed domain. Thus it is a geometric fundamental region.

Consider determinant-one matrices $\left(\begin{array}{cc}u& v\\ \overline{v}& \overline{u}\end{array}\right)$ over Gaussian integers with $u+v\equiv 1(\mathrm{mod}2\mathbb{Z}[i])$, modulo their common sign. They act on the Poincare disc model by fractional linear maps. Their origin-centered Dirichlet domain isFor a nonidentity matrix set $N=|v{|}^2\ge 1$ and $uv=m+in$. The group orbit point is $(m+in)/(N+1)$ and its bisector inequality is $2(mx+ny)\le N(1+x^2+y^2)$. Since $m^2+n^2=N(N+1)$, integrality gives $|m|,|n|\le N$, so the four inequalities defining $D$ imply all of them. They are necessary because the matrices with $u=1\pm i$, $v=\pm i$ realize the four combinations $m,n=\pm 1$. The four sides are arcs of the unit-radius circles centered at $\pm 1\pm i$, and the ideal vertices are $1,i,-1,-i$.

A representative of an element of $G$ is a determinant-one matrix $\left(\begin{array}{cc}a& b\\ \overline{b}& \overline{a}\end{array}\right)$ over the Gaussian integers, considered up to its common sign. It acts by a Möbius transformation of the disk. Since $T(0)=b/\overline{a}$ and $|a{|}^2-|b{|}^2=1$, we haveOnly finitely many Gaussian-integer pairs $(a,b)$ can have $T(0)$ in a fixed compact subdisk. More generally, if $T$ carries some point of a compact set $K$ into $K$, and $K$ lies in the hyperbolic ball of radius $R$ about zero, the triangle inequality gives $\rho (0,T0)\le 2R$. Thus there are only finitely many such $T$. This proves the properly discontinuous group action, and in particular discreteness. The specified $A,B$ belong to $G$, so their subgroup $H$ is discrete as well.

The fixed-point equations reduce to $-i(z-1{)}^2=0$ for $A$ and $i(z+1{)}^2=0$ for $B$. Thus their unique fixed points are respectively $1$ and $-1$ on the ideal boundary. Neither transformation is the identity; their displayed determinant-one representatives have trace two. They are therefore parabolic Möbius transformations.

To find the Dirichlet domains, use the Möbius transformationIt takes the disk to the upper half-plane, and direct substitution givesFor $w=x+iy$ and $v$ in the upper half-plane, the hyperbolic distance obeys $\cosh \rho (w,v)=1+|w-v{|}^2/(2y\mathrm{Im}v)$. Since $C(A^{k}0)=i-2k$, the distance inequalities are equivalent toThese reduce to $kx+k^2\ge 0$ for all integers $k$. The cases $k=1,-1$ imply $-1\le x\le 1$, which also suffices for every other $k$. Therefore the cyclic Dirichlet domain becomes the strip $|\mathrm{Re}w|\le 1$.

The two vertical hyperbolic geodesics pull back to arcs joining $1$ to $i$ and $1$ to $-i$. In disk coordinates they are the circles $|z-(1+i)|=1$ and $|z-(1-i)|=1$, restricted to the disk. ThusConjugating by $z\mapsto -z$ gives $B$, so its two hyperbolic geodesics join $-1$ to $i$ and $-1$ to $-i$, andThe requested drawing and their intersection are shown below.

For any $z_0$, the orbit points satisfying $\rho (0,T{z}_0)\le \rho (0,z_0)$ form a nonempty finite set, by proper discontinuity and compactness of closed hyperbolic balls. Choose a closest one, $q=T{z}_0$. For every integer $k$, its minimality givesand the same inequality with $B$. Hence $q$ lies in the ideal hyperbolic quadrilateralIts ideal vertices are $1,i,-1,-i$.

For the quotient identification, it remains to justify that this covering region is a fundamental polygon. The four open corner caps excluded by $P$ are pairwise disjoint inside the disk. The transformation $A$ carries the exterior of the lower-right cap into the upper-right cap; $A^{-1}$ reverses the pairing. Similarly $B$ carries the exterior of the upper-left cap into the lower-left cap, and $B^{-1}$ reverses this. These assertions follow either from the upper-half-plane formulas or by mapping their boundary circles and testing zero: $A0=(1+i)/2$ and $B0=(-1-i)/2$.

Apply a reduced word in $A^{\pm 1},B^{\pm 1}$ from right to left to a point in the interior of $P$. The first letter puts it into its target cap. Every subsequent letter can do the same, since its inverse cap is distinct from the preceding target cap unless the word cancels. Thus the final point lies in the target cap of the leftmost letter, never in the interior of $P$. This ping-pong lemma argument proves freeness and disjointness of distinct translated interiors. Together with orbit coverage it makes $P$ a fundamental polygon. The corresponding argument on closed caps shows that boundary identifications are generated by the stated side pairings.

The generator $A$ pairs the side from $1$ to $-i$ with the side from $1$ to $i$, fixing the ideal endpoint $1$ and sending $-i$ to $i$. The generator $B$ pairs the side from $-1$ to $i$ with the side from $-1$ to $-i$, fixing $-1$ and sending $i$ to $-i$. There are three ideal-vertex classes: $\{1\}$, $\{-1\}$ and $\{i,-i\}$. The first two give parabolic ends from $A,B$; the third does too, since $AB$ is a nonidentity parabolic transformation fixing $i$. Adding one point at each end produces an oriented closed surface with one face, two edges and three vertices. Its Euler characteristic is $1-2+3=2$, so it is a sphere. Removing the three added points gives a thrice-punctured sphere, proving

Represent a transformation byMultiplication preserves this form and determinant, and the inverse has the same form with parameters $(\overline{u},-v)$. To check the congruence condition, complex conjugation is the identity modulo $2\mathbb{Z}[i]$. Thus $u+v\equiv 1$ says exactly that $M$ fixes the column $(1,1{)}^T$ modulo two. Products and inverses preserve this property. The identity belongs to the set, and changing the common sign of $(u,v)$ does not change either the transformation or its congruence. Hence these transformations form a group.

Because $|u|>|v|$, the denominator has no zero in the disc. Direct calculation givesThe inverse also maps the disc into itself, so $g$ is a disc automorphism, and these identities show that it preserves the hyperbolic metric $2|dz|/(1-|z{|}^2)$.

The action is properly discontinuous. If $C\subset \mathbb{D}$ is compact, put $R={\max }_{z\in C}\rho (0,z)$. If $gC\cap C\ne \varnothing$, choose $z,w\in C$ with $w=gz$. ThenSince $g0=v/\overline{u}$ and $|u{|}^2-|v{|}^2=1$, its distance bound gives $|v/u|\le \tanh R<1$ and therefore $|u{|}^2=1/(1-|v/u{|}^2)\le {\cosh }^{2}R$. There are only finitely many Gaussian integers $u,v$ with these bounds, so only finitely many $g$ satisfy $gC\cap C\ne \varnothing$. This proves the required properly discontinuous group action directly.

If $g0=0$, then $v=0$ and $u$ is one of the Gaussian units $1,-1,i,-i$. The congruence $u-1\in 2\mathbb{Z}[i]$ leaves only $u=\pm 1$. Both give the identity transformation. Thus the point stabilizer of zero is trivial.

The closed Dirichlet region centered at zero is the set of $z$ with $\rho (z,0)\le \rho (z,g0)$ for every $g$. For a nonidentity element, $v\ne 0$. WriteThe distance inequality is equivalent, by pseudohyperbolic distance, to $|z|\le |(z-q)/(1-\overline{q}z)|$. Squaring and subtracting yieldsSince $|z|<1$ and $|q{|}^2=N/(N+1)$, the bisector half-plane is exactlyFor $u=1+i$ or $1-i$ and $v=i$ or $-i$, all the group conditions hold and the products $uv$ are the four numbers $\pm 1\pm i$, with $N=1$. Their four inequalities forceConversely, for any group element $m^2+n^2=|uv{|}^2=N(N+1)<(N+1{)}^2$. As $m,n$ are integers, $|m|,|n|\le N$. Thus every point satisfying those four inequalities satisfies every other one:This proves, with no missing bisectors, the Gaussian-integer disc group with an ideal-square Dirichlet domain formulaEquivalently it is the part of the unit disc outside all four open circles of radius one centered at $1+i,1-i,-1+i,-1-i$. Its sides are the inward arcs of those circles. They meet the unit circle orthogonally, and their ideal vertices are $\boxed{1,i,-1,-i}$; these vertices themselves lie outside the open disc.

The origin has trivial point stabilizer and the group orbit is locally finite by the discontinuity proof, so this closed bisector intersection has translates covering the disc with disjoint interiors, as in the definition of a Dirichlet domain. Side identifications are allowed on its boundary.