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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 $\chi (p)p^{k-1}{a}_{n/p}(f)$.

Use the normalized Fricke involution $W_N=f\mapsto f{|}_k\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$, and let $D_{\ell }=\left(\begin{array}{cc}\ell & 0\\ 0& 1\end{array}\right)$. Choose integers $m,n$ with $m\ell -Nn=1$. The matrix $\gamma =\left(\begin{array}{cc}1& n\\ N& m\ell \end{array}\right)$ lies in ${\Gamma }_1(N)$, and ${\gamma }^{\prime }=\left(\begin{array}{cc}\ell & n\\ N& m\end{array}\right)$ has determinant one and lies in ${\Gamma }_0(N)$, with $m\equiv {\ell }^{-1}(\mathrm{mod}N)$. The displayed identity in the PDF reads $D_{\ell }{\gamma }^{-1}=\mathrm{diag}(1,\ell ){\gamma }^{\prime }$. Taking double cosets therefore identifies the reversed correspondence with the diamond operator $⟨{\ell }^{-1}⟩$ times the original one. Conjugation by $W_N$ reverses the two diagonal entries, soThe same finite-correspondence change of variables in the Petersson inner product transfers a matrix to its inverse and yields $T_{\ell }^{\ast }=⟨{\ell }^{-1}⟩T_{\ell }$. The diamond operators commute with $T_{\ell }$, so $T_{\ell }{T}_{\ell }^{\ast }=T_{\ell }^{\ast }{T}_{\ell }$. Thus $T_{\ell }$ is a normal operator. Multiplying $W_N$ by a nonzero scalar does not change the conjugation formula.

The commuting normal operators $T_{\ell }$, together with the finite-order diamond operators, admit simultaneous diagonalization for the Petersson inner product. Their restrictions to the new subspace of cusp forms have one-dimensional common eigenspaces by the newform multiplicity-one theorem. This additional theorem is the essential newform input: normality alone would allow higher multiplicities. Since every $U_p$ commutes with the good-prime operators and preserves the new space, it preserves each such line and is scalar on it. Normalize its nonzero first Fourier coefficient to one to obtain a newform. This gives the requested simultaneous eigenbasis of the new space.

By definition, every old form is a linear combination of lower-level oldforms by argument dilation. Repeatedly split each lower-level cusp space into its old space and its Petersson inner product orthogonal new space; induction on the level shows that the specified dilations of newforms span the full space. To prove independence, separate distinct primitive eigenforms by their good-prime eigenvalues, using the newform multiplicity-one theorem. For a fixed normalized form, its dilations $f(tz)$ are independent: order the distinct positive integers $t$ increasingly, and the coefficient of the least occurring $q^t$ kills its coefficient in any vanishing linear combination. Repeat for the next least $t$. Thus these dilations are indeed a basis.

For trivial character, part (a) shows that $W_N$ commutes with all the good-prime operators, so it preserves the one-dimensional primitive line. Also $W_N^2=(-1{)}^k$ for the determinant-normalized slash operator; a nonzero trivial-character form has even $k$. ThereforeThe power of $N$ in an unnormalized Fricke eigenvalue depends on the convention. The plain slash action $(cz+d{)}^{-k}f(\gamma z)$ gives $\epsilon N^{-k/2}$, while the Hecke-style action $(\det \gamma {)}^{k-1}(cz+d{)}^{-k}f(\gamma z)$ gives $\epsilon N^{k/2-1}$. The PDF prints $\pm N^{k-1}$; that value requires the deliberate rescaling $w_N=N^{k-1}{W}_N$ and does not follow under either usual convention. Under that rescaling its displayed eigenvalue is $\epsilon N^{k-1}$; all the preceding eigenspace and conjugation statements are unchanged.

Use the following normalizations for Hecke operators on marked lattices and the diamond operator:Multiplication by a unit $d$ preserves the order of the marked point. There are $p+1$ prime-index overlattices; when $p\nmid N$ they all preserve that order. When $p\mid N$, precisely the overlattice killing the order-$p$ subgroup generated by $(N/p)t$ is excluded, leaving $p$ terms. Thus the definition is meaningful at bad primes too.

Both operations preserve homogeneity, since scalar multiplication bijects the indexing lattices and multiplies every summand by the same $u^{-k}$. Changing the marked basis merely permutes the overlattices, giving the required ${\Gamma }_1(N)$ transformation. Locally each term is a modular form evaluated after a rational fractional-linear substitution, multiplied by its appropriate automorphy factor; this preserves holomorphy on the half-plane.

For modular cusp holomorphy, factor any such rational substitution at a rational modular cusp into an integral modular substitution followed by an upper triangular map $z\mapsto Az+B$ with $A>0$. An existing holomorphic modular cusp expansion stays bounded under this map. The finite sum has the positive period supplied by its new level, so boundedness makes its singularity removable in the new modular cusp parameter. This is cusp holomorphy under rational slash operators. It proves that the operators preserve the functions arising from $M_k({\Gamma }_1(N))$.

For clarity, the factor $1/p$ is paired with the homogeneity convention $F(uL)=u^{-k}F(L)$; it gives the Fourier normalization requested in part (c). The diamond action on modular forms equals $f{|}_k\gamma$ for a lift $\gamma \in {\Gamma }_0(N)$ with lower-right entry congruent to $d$, since the transformed marked point is $d/N$ modulo the original lattice.