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Write the finite abelian group additively. A character of a finite abelian group is a group homomorphism $\chi :G\to \{z:|z|=1\}$. We first prove that there are exactly $|G|$ such characters of a finite abelian group, without assuming a structure theorem.

Use extension of a character across a cyclic quotient. Given a subgroup $H$ and $a\notin H$, let $k$ be the least positive integer with $ka\in H$. The subgroup $K=H+⟨a⟩$ has $k$ cosets of $H$. If $\varphi$ is a character of a finite abelian group on $H$, choose any of the $k$ roots ${\lambda }^k=\varphi (ka)$ and defineThis is well-defined: two representations differ by an integer multiple of $ka$, and the root equation exactly cancels that difference. It is a character of a finite abelian group, and every extension arises from one of the $k$ choices of $\lambda$. Build a chain from the trivial subgroup to $G$ by adjoining elements. The character of a finite abelian group count multiplies by the same factor as the subgroup order at every step, so $|\hat{G}|=|G|$.

For a nontrivial character of a finite abelian group $\eta$, choose $a$ with $\eta (a)\ne 1$. Translating the group sum showsso the sum is zero. Applied to $\chi \overline{\psi }$, this provesThe $|G|$ orthogonal nonzero characters of a finite abelian group therefore form a basis of all complex functions on $G$, a vector space of dimension $|G|$.

With the unnormalized Fourier coefficients $\hat{f}(\chi )={\sum }_{x\in G}f(x)\overline{\chi (x)}$, expansion in that basis gives the Fourier inversion on a finite groupUnder a normalized forward-transform convention the prefactor would instead be one. Thus the unspecified constant is determined by the convention, and the inversion itself follows directly from character of a finite abelian group counting and orthogonality.

Let $R$ be rotation through $2\pi /3$ about $z$, and let $S$ be rotation through $\pi$ about $x$. These generate the order-six dihedral group with relationsnamely $\{I,R,R^2,S,RS,R^{2}S\}$. We specify its order explicitly because dihedral notation has two conventions. It is a proper finite subgroup of the full axial rotation group.

To see that it is sufficient, write a general matrix in blocks $T=\left(\begin{array}{cc}B& u\\ v^T& c\end{array}\right)$. Commutation with $R$ forces $u=v=0$, since a nontrivial planar $120$-degree rotation fixes no nonzero planar vector. Its planar part is $\cos (2\pi /3)I+\sin (2\pi /3)J$, with nonzero sine, so $B$ must commute with $J$ and has the form $aI+bJ$. Commutation with $S$ again kills $b$. Hence this six-element subgroup has precisely the required invariant tensors.

It is also smallest by order. Groups of orders $2,3,5$ are cyclic: a nonidentity element has order dividing the prime group order by Lagrange theorem. For a cyclic rotation subgroup, the nonzero antisymmetric tensor representing cross product with its rotation-axis direction is invariant, so it cannot force the symmetric form above. The trivial subgroup plainly imposes no constraint. A group of order four is either cyclic or has three nonidentity elements of order two; in the latter case they commute, since $(AB{)}^{-1}=AB$ and also equals $BA$. Commuting distinct half-turns in three dimensions have perpendicular axes: conjugation by one half-turn must preserve the other's axis, and a distinct preserved axis lies in its perpendicular plane. In the specified axial group, their three axes are consequently $z$ and two perpendicular horizontal directions. In coordinates along them every diagonal matrix is invariant, including one with unequal horizontal entries. Such a group again does not force the required form.

Thus the smallest subgroup has six elements, generated by the axial $120$-degree rotation and one horizontal half-turn. It is the smallest axial rotation group forcing second-rank transverse isotropy; conjugating the horizontal axis gives equally valid choices.

Choose a positive common denominator $d$ so that $x=a/d$ and $y=b/d$ with nonzero integers $a,b$. Put $h=\gcd (a,b)>0$. Every element of the generated subgroup has the form $(ma+nb)/d$, so it lies in $(h/d)\mathbb{Z}$. Conversely, Bezout identity gives integers $u,v$ with $ua+vb=h$, so $h/d$ belongs to the generated subgroup. ConsequentlyThis map is a bijective additive group homomorphism. It is the two-generator instance of a finitely generated subgroup of the rational additive group being an infinite cyclic group.

For the real example, take $x=1$ and $y=\sqrt{2}$. The group homomorphism $(m,n)\mapsto m+n\sqrt{2}$ from ${\mathbb{Z}}^2$ onto the generated subgroup is injective because $\sqrt{2}$ is an irrational number. Thus this subgroup is isomorphic to ${\mathbb{Z}}^2$. More directly, if a single real number $t$ generated it, then $1=at$ and $\sqrt{2}=bt$ for integers $a\ne 0,b$, giving $\sqrt{2}=b/a$, a contradiction. The choice $1,\sqrt{2}$ generates a noncyclic additive subgroup.

The Möbius transformations are bijections, so they preserve which coordinates of a triple are equal. Conversely, part (i) implies transitivity within each equality pattern. For a pattern with fewer than three distinct entries, extend the chosen distinct points to a triple before applying part (i); this works also for $p=2$, when the projective line has exactly three points. There are therefore five orbits of a group action.

We first count $|G{L}_2({\mathbb{F}}_p)|$. Its first column can be any nonzero vector, giving $p^2-1$ choices. The second must be outside the first column's one-dimensional span, giving $p^2-p$ choices. HenceFor the all-equal pattern take $(\infty ,\infty ,\infty )$. Its stabilizer subgroup consists of upper triangular matricesof order $p(p-1{)}^2$. Its group orbit has order $p+1$.

The three exactly-two-equal patterns have representatives $(\infty ,\infty ,0)$, $(\infty ,0,\infty )$ and $(0,\infty ,\infty )$. Each stabilizer subgroup fixes $0$ and $\infty$ individually, so it is the diagonal subgroup of order $(p-1{)}^2$. Each group orbit has order $p(p+1)$; their repeated-coordinate positions keep these three orbits distinct.

For the all-distinct pattern use $(0,1,\infty )$. Its stabilizer subgroup is the scalar subgroup of order $p-1$, and the group orbit has order $(p+1)p(p-1)$. Stabilizers of arbitrary triples are conjugates of the displayed representative stabilizers. In summary the equality-pattern orbits of projective triples haveThe sizes add to $(p+1{)}^3$, and each orbit size times its stabilizer order is $|G|$, as in the orbit-stabilizer theorem.

A subgroup $N$ is a normal subgroup of $G$ if $gN{g}^{-1}=N$ for every $g\in G$. Define multiplication on its left cosets byTo check that this is well-defined, replace $g,h$ by $g{n}_1,h{n}_2$ with $n_1,n_2\in N$. Thenbecause normality gives $h^{-1}{n}_{1}h\in N$. Thus the product depends only on the two cosets. Associativity follows from that of $G$, the identity is $N$, and the inverse of $gN$ is $g^{-1}N$. These are all the group axioms, so $G/N$ is a group under coset multiplication. The natural projection $g\mapsto gN$ is a surjective group homomorphism with kernel $N$.

The set is the ideal quotient $K=(I:J)$. It contains zero. If $r,s\in K$, then $(r-s)J\subseteq I$, since $I$ is an subgroup. If $a\in R$ and $r\in K$, then $(ar)J=a(rJ)\subseteq I$ because $I$ is an ideal. Thus $K$ is an ideal.

Write $J=(a)$ and $K=(b)$. Because $I\subseteq J$, each $x\in I$ has the form $ar$. Then $rJ=r(aR)=xR\subseteq I$, so $r\in K$, and $x\in aK$. Conversely $aK\subseteq I$ by the definition of $K$. Thereforeso $I$ is a principal ideal. This argument uses only the commutative ring convention, not cancellation or an integral domain hypothesis.

The central matrices $\pm I$ act identically on the complex upper half-plane, so the effective group is ${\mathrm{PSL}}_2(\mathbb{Z})$. Away from points with nontrivial effective stabilizer, properly discontinuous action supplies ordinary quotient-disc charts. At a fixed point $z_0$, the coordinate $u=(\tau -z_0)/(\tau -{\overline{z}}_0)$ identifies the stabilizer action with a rotation; its invariant coordinate is $u^e$, where $e$ is the effective stabilizer order. These charts give the quotient its Riemann surface structure.

The elliptic stabilizers of the modular group occur only in the orbits of $i$ and $\omega =e^{2\pi i/3}$. They have effective orders two and three. Consequently the analytic ramification indices of the map from the half-plane areThe stabilizers in ${\mathrm{SL}}_2$ have orders four and six, but the central factor does not double these indices.

All rational boundary points, including infinity, lie in one cusp of a modular group: a primitive column $(a,c)$ can be completed to a determinant-one integral matrix taking infinity to $a/c$. The stabilizer of infinity is generated effectively by $T:\tau \mapsto \tau +1$, and the coordinate $q=e^{2\pi i\tau }$ identifies its high horodisc quotient with a punctured disc. Adding $q=0$ fills that disc. This constructs the compactified modular curve from the extended half-plane; it does not use the ordinary subspace topology on the rational boundary.

For compactness use the standard fundamental domain of the modular group. Its part below a fixed height $Y>1$ is compact, since its imaginary part is at least $\sqrt{3}/2$. The part above $Y$, modulo translation and with the modular cusp added, is a closed disc in the $q$-coordinate. Their images cover the quotient, so it is compact. The same argument with finitely many translates proves compactness for every finite-index subgroup.

Reduction modulo $p$ maps ${\mathrm{SL}}_2(\mathbb{Z})$ onto ${\mathrm{SL}}_2({\mathbb{F}}_p)$, of order $p(p^2-1)$. The image of ${\Gamma }_1(p)$ is the upper unipotent subgroup of order $p$. Thus its index in ${\mathrm{SL}}_2$ is $p^2-1$, and, since $-I\notin {\Gamma }_1(p)$ for $p\ge 5$, its effective index isEvery element of ${\Gamma }_1(p)$ has trace congruent to two. An effective elliptic element of order two or three has trace zero or $\pm 1$ in a lift, so neither is possible for $p\ge 5$. Hence $r_2=r_3=0$.

Represent a modular cusp by a primitive column $(a,c)$, modulo sign. Its reduction is a nonzero vector in ${\mathbb{F}}_p^2$ modulo sign, and the unipotent subgroup acts by $(a,c)\mapsto (a+bc,c)$. For $c=0$, the nonzero values of $a$ give $(p-1)/2$ orbits. For $c\ne 0$, $a$ varies freely and $c$ modulo sign gives another $(p-1)/2$ orbits. Thus there are $p-1$ modular cusps. The reduction classification is sufficient as well as necessary: completing two primitive columns to determinant-one matrices and adjusting the second columns by a translation makes congruent columns related by $\Gamma (p)$.

More explicitly, if a determinant-one matrix has first column $(a,c)$, conjugating $T^w$ givesIts least allowable cusp width in ${\Gamma }_1(p)$ is one for $c\equiv 0$ and $p$ otherwise. Both types therefore number $(p-1)/2$, with cusp widths one and $p$; their cusp width sum is $d$. There are no sign-twisted modular cusp periods here, because trace two cannot be congruent to minus two for these primes. Substitution givesThese are the prime Gamma 1 cusp counts and widths.