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Write $\Gamma =S{L}_2(\mathbb{Z})$ and $G=G{L}_2(\mathbb{Q}{)}^{+}$. The double-coset Hecke algebra consists of $\Gamma$-bi-invariant complex functions on $G$ supported on finitely many double cosets, with convolutionEach double coset has finitely many orbits under left multiplication by $\Gamma$, by rational conjugation of finite-index modular subgroups. Thus the sum is finite and independent of representatives. The characteristic functions of the double cosets form its basis, and its right action on invariant modular forms is $f\ast h={\sum }_{\Gamma x}h(x)f{|}_{k}x$, using the determinant-normalized slash operator.

For a positive integer $n$, let ${\mathcal{M}}_n$ be all integral two-by-two matrices of positive determinant $n$. It is $\Gamma$-bi-invariant. Define $T(n)$ to be its indicator function, equivalently the sum of the distinct double cosets it contains, each with coefficient one. This is the convention consistent with the requested formula. For composite $n$, it need not be the single double coset of $\mathrm{diag}(1,n)$: for instance $2I\in {\mathcal{M}}_4$ cannot lie in that double coset, since multiplying by unimodular integral matrices preserves the greatest common divisor of the entries.

Prove all the needed subgroup facts directly. For a finite-index subgroup $L\subseteq {\mathbb{Z}}^2$, take $a$ to be the least positive first coordinate appearing in $L$ and $d$ the least positive second coordinate on its intersection with the second axis. Euclidean division then shows that the first-coordinate projection is $a\mathbb{Z}$ and $L\cap (\{0\}\times \mathbb{Z})=\{0\}\times d\mathbb{Z}$. Positivity follows, for example, because the finite quotient group kills a nonzero multiple of each coordinate vector. Choose $(a,b)\in L$ and reduce $b$ modulo $d$ to $0\le b<d$. Every vector of $L$ has first coordinate a multiple of $a$, and subtracting that multiple of $(a,b)$ leaves a multiple of $(0,d)$. Therefore these two vectors form a basis of $L$. The parameters $a,d,b$ are unique. Reducing the first coordinate modulo $a$ and then the second modulo $d$ gives precisely $ad$ quotient representatives, so $[{\mathbb{Z}}^2:L]=ad$.

Apply this row Hermite normal form in rank two to the row lattice of an integral matrix $A$ with $\det A=n$. Its row lattice contains $n{\mathbb{Z}}^2$, since $\mathrm{adj}(A)A=nI$, so it has finite index. Its two rows and the displayed two rows are bases of the same row lattice. The two inverse change-of-basis matrices have integer entries, so their determinants are integers whose product is one. Thus the change-of-basis matrix has determinant $\pm 1$; since both orientations are positive, its determinant is one. Thus each orbit under left multiplication by $\Gamma$ has exactly one of the determinant-n matrix representatives for Hecke operatorsThere are ${\sum }_{d\mid n}d$ such representatives, proving finiteness as well as the formula. The subgroup argument is a proof of the relevant Hermite normal form, not an invocation of an unproved lattice classification.

Consequently the normalized Hecke operator isIt preserves $M_k(\Gamma )$: right multiplication by $\Gamma$ permutes the left-multiplication orbits in ${\mathcal{M}}_n$, giving invariance, and cusp holomorphy under rational slash operators gives the holomorphy of each term at every cusp.

For a triangular representative the determinant-normalized slash operator is $f{|}_{k}A=n^{k/2}{d}^{-k}f((az+b)/d)$. Summing the Fourier expansion of a modular form over $b$ kills every index not divisible by $d$, by finite exponential orthogonality. ThusThe Fourier coefficients of a composite-index Hecke operator are thereforeIn particular $a_1(T_{n}f)=a_n(f)$. If $T_{n}f=\lambda f$, comparison of the $q$ coefficients gives

Finally suppose $a_0(f)\ne 0$ and $f$ is a simultaneous eigenfunction. Constant coefficients force ${\lambda }_n={\sigma }_{k-1}(n)$, so $a_n(f)=a_1(f){\sigma }_{k-1}(n)$ for all $n\ge 1$. Weight two cannot occur, by vanishing of weight-two level-one modular forms. For even $k\ge 4$, use the normalized Eisenstein series with its Fourier expansion of a normalized Eisenstein serieswhere $B_k$ is the Bernoulli number. Then $h=f-a_0(f)E_k$ is a cusp form with coefficients $(a_1(f)-a_0(f)c_k){\sigma }_{k-1}(n)$. The Fourier coefficient bound for a cusp form bounds these by $O(n^{k/2})$, but at arbitrarily large primes $p$ their magnitude is $|a_1-a_0{c}_k|(1+p^{k-1})$. Since $k-1>k/2$, the coefficient factor must vanish. All coefficients of $h$ then vanish, including its constant coefficient, so $h=0$ by its cusp expansion and the identity theorem. This proves the noncuspidal level-one Hecke eigenform characterization:

For a normal subgroup $H⊴G$, the quotient group $G/H$ is the set of cosets $gH$ with multiplication $(gH)(kH)=gkH$. If $gH=g^{\prime }H$ and $kH=k^{\prime }H$, normality moves the intervening elements of $H$ past $k$, proving that the product is independent of representatives.

If $g$ has finite order $m$, then $(gH{)}^m=H$, so $\mathrm{ord}(gH)$ divides $m$. For finite $G$, the greatest element order in $G/H$ is therefore no greater than that in $G$.

Every kernel is normal because $\phi (gh{g}^{-1})=\phi (g)\phi (h)\phi (g{)}^{-1}=1$ whenever $h\in \ker \phi$. Conversely, if $H$ is normal, the quotient map $G\to G/H$ has kernel $H$. Thus $H⊴G$ exactly when it is the kernel of a group homomorphism.

A group is metacyclic when it has a cyclic normal subgroup with cyclic quotient. A dihedral group has its cyclic rotation subgroup normal and quotient $C_2$, so every dihedral group is metacyclic.

Up to isomorphism the groups of order eight areAll except $C_2^3$ are metacyclic: use a cyclic subgroup of order four in the three noncyclic examples that have one. In $C_2^3$, every cyclic subgroup has order at most two and its quotient is not cyclic. The normal subgroups of $A_4$ are $1,V_4,A_4$; none gives the required cyclic subgroup and quotient, so $A_4$ is not metacyclic. Finally, $S_1,S_2,S_3$ are metacyclic, while $S_4$ and $S_5$ are not, by their normal-subgroup structures. Hence

Because $L$ is a simplicial subcomplex of $K$, every face of a simplex of $L$ also belongs to $L$. The simplicial boundary operator therefore satisfiesso $C_{\bullet }(L)$ is a chain subcomplex of the simplicial chain complex $C_{\bullet }(K)$.

Define a map on the quotient groups byIf $c^{\prime }=c+\ell$ with $\ell \in C_k(L)$, then $\partial c^{\prime }-\partial c=\partial \ell \in C_{k-1}(L)$, so this definition is independent of the representative. Moreover,Thus $C_{\bullet }(K,L)$ is the relative simplicial chain complex.

A subgroup $N\le G$ is normal whenIts left and right cosets then agree, and multiplicationis well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the quotient group $G/N$.

For a group homomorphism $\theta :G\to H$, its kernel and image areThe kernel is normal becausewhenever $k\in \ker \theta$. Conversely, if $K⊴G$, the quotient mapis a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the mapis well defined and bijective and preserves multiplication. This is the first isomorphism theorem.

DefineEuler's formula shows that $\Phi$ is a homomorphism, its image is the complex unit circle, and its kernel is $\mathbb{Z}$. The first isomorphism theorem therefore givesThe image of $\mathbb{Q}/\mathbb{Z}$ consists exactly of the roots of unity: if $t=p/q$, then $\Phi (t{)}^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

Let $x$ and $y$ denote the two oriented noncontractible cycles of the torus. Their transport operators intersect once, so the mutual braiding phase gives the Wilson-loop algebra of an Abelian Chern--Simons theoryThe integer charge vectors are identified modulo local electrons: $q$ and $q+K\ell$ have identical braiding with every quasiparticle. The distinct topological charges therefore form the finite quotient group

Choose a ground state that diagonalizes the mutually commuting $x$-cycle operators. Acting with $T_y(q)$ multiplies their eigenvalues by the character $q^{\prime }\mapsto e^{2\pi i{q}^{\prime T}{K}^{-1}q}$. This pairing is nondegenerate on $\mathcal{A}$: if the character is trivial for every $q^{\prime }$, then $K^{-1}q$ is integral and $q$ is the trivial class. Thus the $|\mathcal{A}|$ states obtained from inequivalent $q$ have distinct loop eigenvalues and are orthogonal. The torus ground-state degeneracy of an Abelian Chern--Simons theory is consequently at least $|\det K|$.

The all-ones matrix has eigenvalue $n$ along $t$ and eigenvalue zero on the $(n-1)$-dimensional perpendicular subspace. Hence $K=I_n+pM$ has eigenvalues $1+np,1,\dots ,1$, and

The map $(a,b)\mapsto \mathrm{sgn}(a)$ is a surjective group homomorphism with kernel $H_0$, so $H_0$ is a normal subgroup and $G/H_0\cong C_2$. The subgroup $H_1$ is not normal: for a translation $t=(1,c)$,which leaves $H_1$ when $a\ne 1$ and $c\ne 0$. Finally, $(a,b)\mapsto a$ is a surjective homomorphism onto ${\mathbb{R}}^{\ast }$ with kernel $H_2$, so $H_2◃G$ and the quotient group $G/H_2\cong {\mathbb{R}}^{\ast }$.

For a subspace $A\subseteq X$, the relative chain complex is the quotient group in each degree,with boundary operator $[c]\mapsto [\partial c]$. This is well defined because $C_{\ast }(A)$ is a chain subcomplex of $C_{\ast }(X)$.