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Write $A,B$ for the two given congruence subgroups, to distinguish them from the notation ${\Gamma }_1(N)$. Their double coset has a finite decompositionFiniteness follows because $B\cap {\alpha }^{-1}A\alpha$ has finite index in $B$. To check this without assuming integrality of $\alpha$, clear its denominators: a sufficiently deep principal congruence subgroup lies both in $B$ and, after conjugation by $\alpha$, in $A$. Representatives for $(B\cap {\alpha }^{-1}A\alpha )\backslash B$ give the displayed decomposition.

Using the Hecke-normalized rational slash operator from the PDF, define the double-coset operator on modular forms byReplacing ${\alpha }_j$ by $\gamma {\alpha }_j$ for $\gamma \in A$ changes no term because $f[\gamma {]}_k=f$. Right multiplication by $\beta \in B$ permutes these left cosets, so $([A\alpha B]f)[\beta {]}_k=[A\alpha B]f$. Each summand is holomorphic in the complex upper half-plane. At any rational modular cusp, factor ${\alpha }_j\rho =\sigma C$ with $\sigma \in S{L}_2(\mathbb{Z})$ and $C$ upper triangular of positive slope, as in 2(i). The cusp expansion of $f[\sigma {]}_k$ has no negative terms, and substitution by $C$ preserves boundedness. The sum is periodic in a cusp coordinate for $B$, so it has a removable singularity there. This provesIf $f$ vanishes at all its cusps, the same argument gives vanishing for the image too.

Now set $\Gamma ={\Gamma }_1(N)$. For an integer $d$ prime to $N$, choose ${\sigma }_d\in {\Gamma }_0(N)$ with lower-right entry congruent to $d$ modulo $N$. Such a lift exists by choosing $a$ with $ad\equiv 1(\mathrm{mod}N)$ and using $\left(\begin{array}{cc}a& (ad-1)/N\\ N& d\end{array}\right)$. The subgroup $\Gamma$ is normal in ${\Gamma }_0(N)$, and its quotient is $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ through the lower-right entry. Define the diamond operator byDifferent lifts differ by a member of $\Gamma$, so this is well defined. The diamond operators satisfy $⟨d⟩⟨e⟩=⟨de⟩$.

For a prime $p\nmid N$, put $A_p=\mathrm{diag}(1,p)$ and define the Hecke operatorChoose $a,b\in \mathbb{Z}$ with $ap-bN=1$ and set ${\sigma }_p=\left(\begin{array}{cc}a& b\\ N& p\end{array}\right)$. A complete set of left-coset representatives isHere $A_j=A_p{T}^j$ and $B_p=A_p\left(\begin{array}{cc}ap& b\\ N& 1\end{array}\right)$, with both right factors in $\Gamma$. To prove completeness, $H=\Gamma \cap A_p^{-1}\Gamma A_p$ consists of the members of $\Gamma$ whose upper-right entry is divisible by $p$. Reduction of $\Gamma$ modulo $p$ is onto $S{L}_2({\mathbb{F}}_p)$: use the Chinese remainder theorem to impose identity modulo $N$ and any chosen determinant-one matrix modulo $p$, then lift to $S{L}_2(\mathbb{Z})$ by elementary matrices. Reduction of $H$ is the lower triangular subgroup. Its left cosets correspond to the $p+1$ lines of top rows in ${\mathbb{F}}_p^2$. The right factors $T^j$ give lines $[1:j]$, while the last factor gives $[0:b]$ with $b\not\equiv 0(\mathrm{mod}p)$. They are all the lines, proving the decomposition.

The printed slash normalization now gives $f[A_j{]}_k=p^{-1}f((\tau +j)/p)$ andThus

To prove commutation, use the marked-lattice model of a modular form. A pair $(L,t)$ consists of a lattice and a point of exact order $N$ modulo $L$, and a weight-$k$ function satisfies $F(uL,ut)=u^{-k}F(L,t)$. In a basis $L=\mathbb{Z}\tau +\mathbb{Z}$ with $t=1/N$, it represents $f(\tau )$. The diamond operator multiplies $t$ by $d$. The same Hecke operator isEvery marked point retains exact order $N$: multiplication by $p$ is invertible on its cyclic group, since $p\nmid N$. This lattice sum agrees with the displayed slash sum. Its $p$ overlattices with basis $((\tau +j)/p,1)$ give the first terms, and the remaining one with basis $(\tau ,1/p)$ has normalized marked point $p/N$, giving $p^{k-1}(⟨p⟩f)(p\tau )$. Multiplying the mark by $d$ commutes with passage to each overlattice and does not change the set of overlattices. Hence

Finally write $f={\sum }_{m\ge 0}{a}_m(f)q^m$. The first term of $T_p$ contains the finite averageIt therefore contributes $a_{np}(f)$ to the coefficient of $q^n$. The second term contributes $p^{k-1}{a}_{n/p}(⟨p⟩f)$ if $p\mid n$, and zero otherwise. This proves the good-prime and bad-prime Hecke coefficient formula at a good prime, without assuming that $f$ has a particular character:where a nonintegral coefficient index means zero. For $n=0$ both terms are retained. On a diamond-character eigenspace the second term becomes $\chi (p)p^{k-1}{a}_{n/p}(f)$.