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If $\pi$ is the character of $\mathbb{C}[X]$, then Burnside lemma applied to the diagonal action on $X\times X$ givesFor a transitive action $X\simeq G/H$, this also equals the number of double cosets in $H\backslash G/H$.

For the order-eight dihedral group generated by $r$ of order four and $s$ of order two with $srs=r^{-1}$, let $H=K=⟨s⟩$. There are exactly three double cosets: $H$, $H{r}^{2}H$, and $HrH$, of sizes two, two, and four. Their number does not divide eight. This differs from the number of ordinary cosets, which divides group order by Lagrange's theorem.

For a group $G$ and subgroup $\Gamma$ such that every double coset has finitely many orbits under left multiplication by $\Gamma$, this algebra consists of complex $\Gamma$-bi-invariant functions on $G$ supported on finitely many double cosets. Its convolution isThe double coset indicator functions form a basis. On invariant modular functions it acts on the right by $f\ast h={\sum }_{\Gamma x}h(x)f{|}_{k}x$. For $G=G{L}_2(\mathbb{Q}{)}^{+}$ and $\Gamma =S{L}_2(\mathbb{Z})$, rational conjugation of finite-index modular subgroups verifies the finiteness condition. The determinant-$n$ Hecke operator uses the indicator of all integral determinant-$n$ matrices, not only a single double coset when $n$ is composite.

A double coset is generally not a subgroup, so Lagrange's theorem does not force its size to divide the ambient group order. In $S_3$, take $H=K=⟨(12)⟩$. Then $HeH$ has size two, while $H(13)H$ has size four, which does not divide six. The general size formula $|HgK|=|H||K|/|H\cap gK{g}^{-1}|$ explains why different double cosets can have different sizes.

Use finite-dimensional $k$-modules or finite $R$-free lattices. We use the basic relative-projectivity facts that Krull-Schmidt decomposition holds with local endomorphism rings, module vertices are conjugate p-subgroups, and a module source $S$ at a module vertex $D$ satisfies $S\mid M{\downarrow }_D$ and $M\mid S{\uparrow }^G$. Module induction is transitive and Mackey restriction formula gives module vertex containment for restrictions and inductions. In particular, a summand induced from $D$ has module vertex conjugate into $D$, and module restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex. The latter follows from the module source conditions and minimality: restricting further to $D$ must still contain its module source, whose module vertex is $D$.

The module vertex-$D$ form of Green correspondence states that for $N_G(D)\le H\le G$ there is a bijection between indecomposable $RG$-modules with module vertex $D$ and indecomposable $RH$-modules with module vertex $D$. Its maps select the unique module vertex-$D$ summands of module restriction and induction, using $H$-conjugacy for restricted module vertices and $G$-conjugacy for induced module vertices. They occur with multiplicity one and have the same module sources. Other induction summands have module vertices conjugate to proper subgroups of $D$. The module restriction remainder is relatively projective forRelative projectivity for a family means being a summand of a finite sum of inductions from its members. We prove this form; no stronger description of the induction remainder is needed for the application.

Let $U$ have module vertex $D$ over $H$. The identity double coset in Mackey restriction formula givesSince $U$ is relatively $D$-projective, a further Mackey decomposition puts every other double coset term in the relatively-$\mathcal{Y}$ family. No member of $\mathcal{Y}$ contains an $H$-conjugate of $D$. Otherwise, conjugation within $H$ would give $D\le H\cap (ha)D(ha{)}^{-1}$ with $h\in H$, $a\notin H$. Equality of orders then makes $ha\in N_G(D)\le H$, a contradiction. Therefore the module restriction above contains exactly one summand with module vertex $H$-conjugate to $D$, namely $U$ once.

Decompose $U{\uparrow }^G$ into indecomposable modules. All their module vertices are conjugate into $D$. Each one with a module vertex $G$-conjugate to $D$ can take $D$ itself as module vertex and contributes a module vertex-$D$ summand on module restriction to $H$. There can thus be at most one such summand and its multiplicity is one. There must be at least one, because smaller module vertices cannot restrict to produce $U$, with module vertex $D$, in the displayed decomposition. Define $g(U)$ to be that summand. Its module restriction contains $U$ once, with remainder a summand of $T$. All other induction summands have strictly smaller module vertices. This is Green correspondence from Mackey multiplicity.

Conversely take $M$ of module vertex $D$ over $G$ with module source $S$. Decompose $S{\uparrow }^H$ into indecomposable modules. Since $M\mid S{\uparrow }^G=(S{\uparrow }^H){\uparrow }^G$, the local endomorphism ring and Krull-Schmidt decomposition give $M\mid U{\uparrow }^G$ for one indecomposable summand $U$ of $S{\uparrow }^H$. Its module vertex is $H$-conjugate to $D$: all module vertices are conjugate into $D$, and a smaller one could not induce $M$. The preceding uniqueness makes $g(U)=M$ and identifies $U$ as the unique module vertex-$D$ summand of $M{\downarrow }_H$. Define $f(M)=U$. This proves $f(g(U))=U$ and $g(f(M))=M$.

If $S$ is a module source of $U$, then $S\mid U{\downarrow }_D\mid M{\downarrow }_D$ and $M\mid U{\uparrow }^G\mid S{\uparrow }^G$. These two conditions and the common module vertex show that $S$ is also a module source of $M$. Thus the correspondence preserves module sources, including trivial source modules, and the proof is complete under the stated coefficient conventions.

For the required expression, a p-local module is a finite sum of inductions from $N_G(Q)$ for nontrivial p-subgroups $Q$. Induct on the order of a module vertex of $M$. Vertex 1 makes $M$ projective, so choose $L_1=L_2=P_1=0$, $P_2=M$.

For module vertex $D\ne 1$, take $U=f(M)$ for $H=N_G(D)$. Its module induction is p-local andwith smaller module vertices for the $X_i$. By induction, writewith $A_i,B_i$ p-local and $P_i,Q_i$ projective. Adding these identities and substituting the induction decomposition givesHence the explicit choiceswork. All sums are finite and module vertex orders strictly decrease, so the induction terminates. If $N_G(D)=G$, induction from that normalizer is still allowed by the definition of p-local.

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

Throughout, modules are finite-dimensional over $k$, or finitely generated $\mathcal{O}$-free lattices when the coefficient ring is $\mathcal{O}$. The p-modular system has a complete discrete valuation ring $\mathcal{O}$, maximal ideal $(\pi )$, fraction field $K$ of characteristic zero and residue field $k$ of characteristic $p$.

A block of a group algebra is a nonzero, indecomposable two-sided ideal direct summand of $A=RG$. Equivalently it is $Ae$ for a primitive central idempotent $e$. To see the equivalence, suppose $A=B\oplus C$ as two-sided ideals and write $1=e+f$ with $e\in B$, $f\in C$. Products between $B$ and $C$ vanish because they lie in their zero intersection. Thus $e$ is the identity on $B$, zero on $C$, and is a central idempotent, with $B=Ae$. Conversely $A=Ae\oplus A(1-e)$. Splitting $Ae$ further is equivalent to splitting $e$ into orthogonal nonzero central idempotents; indecomposability is exactly primitivity. If $1=\sum e_i$ is the block decomposition, $M=⨁e_{i}M$. A nonzero indecomposable module therefore belongs to exactly one block of a group algebra, namely the one with $e_{i}M=M$.

For modular reduction of block idempotents, the conjugacy class sums are an $R$-basis of $Z(RG)$. Hence $Z(\mathcal{O}G)/\pi Z(\mathcal{O}G)\cong Z(kG)$. In the complete commutative algebra $Z(\mathcal{O}G)$, idempotent lifting gives a unique lift of each idempotent of $Z(kG)$; uniqueness follows because commuting idempotents congruent modulo the Jacobson radical are equal. Orthogonal decompositions lift too, so primitivity is preserved. Reduction gives a bijection between the integral and modular blocks.

Here is a definition of defect groups of a block which also proves the required relative-projectivity assertion. Give $RG$ its conjugation action and, for $H\le G$, define the relative trace and its transfer ideal of conjugation-fixed elements byA defect group of a block $RGe$ is a $p$-subgroup $D$ minimal with $e\in I_D$. Such a subgroup exists: for a Sylow subgroup $P$, $[G:P]$ is a unit in $R$ and $e={\mathrm{Tr}}_P^G(e/[G:P])$. The defect of a block is $d$ where $|D|=p^d$.

For completeness, this definition is independent of a choice up to conjugacy. Multiplying two relative traces and grouping the pairs of cosets by double cosets givesThis identity follows by writing each pair of cosets as a simultaneous $G$-orbit; its stabilizer is $D\cap {}^{g}E$, and its contribution is a trace from that stabilizer. If $D,E$ are both minimal for $e$, then $e=e^2$ belongs to the displayed sum. The commutative block center $eZ(RG)$ is a local ring: in the field case it is a commutative Artinian ring with no nontrivial idempotents; in the integral case its reduction has this property and completeness detects units. A sum of proper ideals cannot contain its identity. Consequently $e\in I_{D\cap {}^{g}E}$ for some $g$, forcing $D={}^{g}E$ by minimality. The same argument with an arbitrary $E$ shows that $e\in I_E$ precisely when $E$ contains a conjugate of $D$. This is the trace criterion for defect groups.

To relate this to the usual modular definition, the Brauer morphism deletes the coefficients outside the centralizer:It is an algebra homomorphism: in the coefficient of an element centralizing $P$ in a product, $P$ acts on the contributing pairs; nonfixed orbits have size divisible by $p$, while fixed pairs have both entries in $C_G(P)$. Its kernel is the sum of proper-subgroup transfers, because the nonfixed conjugation orbit sums form a basis of that kernel. If $D$ is minimal for $e={\mathrm{Tr}}_D^G(a)$ and ${\mathrm{Br}}_D(e)=0$, replace $a$ by $ea$. The kernel description gives $e\in {\sum }_{Q<D}{I}_Q$, contradicting the local-center argument. Also ${\mathrm{Br}}_P(I_D)=0$ unless $P$ is conjugate into $D$: group the cosets of $G/D$ into $P$-orbits, which have no fixed point otherwise. Thus the minimal transfer subgroups are exactly the maximal $p$-subgroups with nonzero Brauer morphism of $e$.

Now take $M=eM$ and write $e={\mathrm{Tr}}_D^G(a)$. Multiplication by $a$ is a $D$-linear endomorphism $\alpha$ of $M$, andHere is the splitting behind the D. Higman criterion. The natural map $\epsilon :RG{\otimes }_{RD}M\to M$, $g\otimes m\mapsto gm$, has a $G$-linear sectionChanging a representative by an element of $D$ leaves this tensor unchanged, $G$ permutes the terms, and $\epsilon s={\mathrm{id}}_M$. HenceThis is relative projectivity for $D$, proving the Green relative projectivity theorem; indecomposability was needed only to speak of a single block of a group algebra or vertex of an indecomposable module.

Finally let $\hat{e}$ lift $e$. A trace expression for $\hat{e}$ reduces to one for $e$, so a modular defect group of a block is contained, up to conjugacy, in an integral one. Conversely choose a modular defect group $D$ and $a\in (kG{)}^D$ with $e={\mathrm{Tr}}_D^G(a)$. Conjugation orbit sums lift $a$ to $\hat{a}\in (\mathcal{O}G{)}^D$. The elementreduces to its block identity, hence is a unit in that block center. Its inverse there is central, soThe integral defect is therefore contained in the modular one. The corresponding blocks have the same conjugacy class of defect groups and the same defect.

The block statement is Green correspondence for blocks: if $N_G(D)\le H\le G$, blocks of $RG$ and $RH$ with defect group $D$ correspond, and their block bimodules are Green correspondents for the diagonal vertex $\Delta D$. We prove the module correspondence underlying this statement, then specialize it to the block bimodules. This also gives the requested direct proof in the trivial-intersection situation.

Here are the elementary ingredients, including their proofs. Mackey decomposition is obtained by partitioning the tensor basis of $RG{\otimes }_{RL}U$ by the double cosets $H\backslash G/L$; the term for $g$ is induction from $H\cap {}^{g}L$ of the suitably conjugated restriction of $U$. The D. Higman criterion follows from the section constructed in Question 1. Conversely, take a splitting of the induction counit and its coefficient at the identity coset: equivariance reconstructs the section from that $H$-linear coefficient, and the splitting identity is exactly that its relative trace is $1$. Thus the trace criterion is an equivalence.

An indecomposable module here has a local endomorphism ring. Over $k$ this follows from the Fitting lemma: an endomorphism eventually splits its kernel from its image, and indecomposability makes it invertible or nilpotent. Over $\mathcal{O}$, the endomorphism ring is finite over a complete discrete valuation ring; lift orthogonal idempotents from its finite-dimensional residue algebra to see that an indecomposable lattice has a local endomorphism ring too. The Krull–Schmidt theorem follows by the usual exchange argument: composing the inclusion and projection of one indecomposable summand across a second decomposition writes its identity as a sum of endomorphisms; one is a unit in the local endomorphism ring and pairs that summand with a summand of the second decomposition. Remove the pair and continue. Existence follows by successively splitting nontrivial idempotents, which decreases dimension or rank.

A vertex of an indecomposable module is a subgroup minimal for relative projectivity. It can be taken to be a $p$-subgroup because averaging over the index of a Sylow subgroup of any candidate subgroup is invertible. If an indecomposable module is relatively projective for $P$ and $Q$, the Mackey decomposition and its induction splittings make it a summand of a sum of modules induced from $P\cap {}^{g}Q$. In its local endomorphism ring one splitting component must be invertible, so it is relatively projective for one of these intersections. This proves conjugacy of minimal vertices and the useful criterion that a module with vertex $P$ can be relatively projective for $Q$ only when $P$ is conjugate into $Q$. A source of an indecomposable module with vertex $P$ is an indecomposable summand $S$ of its restriction to $P$ from which the original module is an induced summand. Such a summand exists by the induction splitting and Krull–Schmidt theorem; its vertex is $P$, since a proper vertex would make the original module relatively projective for a proper subgroup of $P$.

SetWhen saying a module is $\mathcal{X}$- or $\mathcal{Y}$-projective, allow conjugates in the relevant ambient group and finite sums of modules relatively projective for members of that family. No conjugate of $D$ inside $H$ belongs to $\mathcal{Y}$: if $D\le {}^h(H\cap {}^{g}D)$, equality of orders gives $D={}^{hg}D$, so $hg\in N_G(D)\subseteq H$, contradicting $g\notin H$. Every member of $\mathcal{X}$ is a proper subgroup of $D$.

The two key assertions can now be proved rather than assumed. First, if $W$ is an indecomposable $D$-projective $RH$-module, thenChoose $U$ with $W\mid {\mathrm{Ind}}_D^{H}U$. Mackey decomposition gives ${\mathrm{Res}}_H^G{\mathrm{Ind}}_D^{G}U={\mathrm{Ind}}_D^{H}U\oplus T$, with $T$ $\mathcal{Y}$-projective. Write ${\mathrm{Ind}}_D^{H}U=W\oplus W^{\prime \prime }$ and apply induction and restriction. Both $\mathrm{Res}\mathrm{Ind}W$ and $\mathrm{Res}\mathrm{Ind}{W}^{\prime \prime }$ contain their original modules, by the identity double coset. Comparing multiplicities of every indecomposable which is not $\mathcal{Y}$-projective, their extra multiplicities are nonnegative and have sum zero. Therefore neither has such an extra summand, proving the assertion.

Second, suppose $W$ has vertex $D$, and choose an indecomposable $V\mid {\mathrm{Ind}}_H^{G}W$ whose restriction contains $W$; such a $V$ exists by the first assertion. Write $\mathrm{Ind}W=V\oplus V^{\prime }$. Its restriction has just one summand isomorphic to $W$ and all other summands are $\mathcal{Y}$-projective, so $\mathrm{Res}{V}^{\prime }$ is $\mathcal{Y}$-projective. Any indecomposable $Z\mid V^{\prime }$ has a vertex $P\le D$ after conjugacy. A source $S$ with vertex $P$ is a summand of ${\mathrm{Res}}_{P}Z$. Restricting a $\mathcal{Y}$-projective decomposition of ${\mathrm{Res}}_{H}Z$ to $P$ and using Mackey decomposition forces $P$ to lie in an $H$-conjugate of some $H\cap {}^{g}D$, $g\notin H$. Since $P\le D$, this puts $P$ in $\mathcal{X}$. Thus $V^{\prime }$ is $\mathcal{X}$-projective. Moreover $V$ has vertex $D$, since it is $D$-projective and its restriction contains $W$ with vertex $D$. We have provedIn particular $V$ is unique.

Conversely, let $V$ be an indecomposable $RG$-module with vertex $D$ and source $S$. Decompose ${\mathrm{Ind}}_D^{H}S$ and choose $W$ whose induction contains $V$. Its vertex is contained in $D$ and cannot be proper, since induction would then make $V$ relatively projective for a smaller subgroup. Thus $W$ has vertex $D$, and the previous assertion makes $V$ its unique vertex-$D$ induced summand. Restriction of an $\mathcal{X}$-projective module is $\mathcal{Y}$-projective: in a Mackey term a conjugating element inside $H$ preserves the containment in an $H$-conjugate of $\mathcal{Y}$, and one outside $H$ gives a subgroup of $H\cap {}^{g}D$. Comparing the two decompositions of $\mathrm{Res}\mathrm{Ind}W$ now givesThese assignments are mutually inverse. To check sources, restrict ${\mathrm{Ind}}_D^{H}S$ back to $D$. Its terms of full vertex are conjugates of $S$ by $N_H(D)$; the other terms are induced from proper intersections. Applying the source criterion and Krull–Schmidt theorem shows that the source of $W$ is one of those conjugates. The same argument in $G$, and the uniqueness just proved, matches the source classes on the two sides. Thus Green correspondence preserves sources up to normalizer conjugacy.

To pass to blocks of a group algebra, regard $RG$ as the permutation module $R[(G\times G)/\Delta G]$. A summand of a permutation module has trivial source. Here is why: over a finite p-group, every transitive permutation module is indecomposable, since its socle is the one-dimensional fixed line and every nonzero module has nonzero fixed points. The latter follows by induction on the group order, using a central element of order $p$ and the nilpotence of its difference from $1$. Over $\mathcal{O}$ indecomposability follows by reduction. Restrict a permutation summand to a vertex $P$ and apply Krull–Schmidt theorem; a source must be a transitive permutation module $R[P/Q]$. Its vertex is contained in $Q$, so minimality forces $Q=P$, and its source is the trivial module.

For a permutation module, its Brauer quotient of a module at $P$ has the fixed permutation points as basis: every nontrivial orbit sum is a proper transfer. An indecomposable trivial-source summand with vertex $Q$ has nonzero quotient only at subgroups conjugate into $Q$, by its induced-permutation realization, and its quotient at $Q$ is nonzero because its restriction contains its trivial source. Applied to a block $B=kGe$, the quotient at $\Delta P$ is ${\mathrm{Br}}_P(e)k{C}_G(P)$. Together with the maximal-subgroup characterization of defect from Question 1, this proves that its bimodule vertices are precisely $\Delta D$ for its defect groups $D$. The integral conclusion follows by orbit-sum reduction and idempotent lifting.

FinallyApply the just-proved Green correspondence with ambient group $G\times G$. Its unique vertex-$\Delta D$ summand of $B{\downarrow }_{H\times H}$ lies in $RH$: outside $H$ there is no $\Delta D$-fixed group-basis point, as in Question 2. Since the indecomposable $H\times H$ summands of $RH$ are exactly its block ideals, this summand is a block $b$ with defect group $D$. Conversely induction of $b$ has $B$ as its unique vertex-$\Delta D$ summand. The block bijection is the Brauer correspondence of Question 2, giving the claimed block form over either coefficient ring.

For the final special case put $N=N_G(D)$ with $D$ a trivial-intersection Sylow subgroup. For $g\notin N$, the intersection $N\cap {}^{g}N$ has order prime to $p$. Indeed every $p$-subgroup of it lies both in the normal Sylow subgroup $D$ of $N$ and in ${}^{g}D$, whose intersection is trivial. Mackey decomposition therefore giveswhere every other double-coset term is induced from a group of order prime to $p$. By Maschke's theorem such modules are projective, and induction preserves projectivity, so $P$ is projective.

Every nonprojective $kG$-module has nonprojective restriction to $N$: $[G:N]$ is prime to $p$, so averaging makes the module a summand of the induction of its restriction. A projective restriction would therefore make the original module projective. Decompose ${\mathrm{Ind}}_N^{G}M$ into indecomposables. Each nonprojective summand contributes a nonprojective summand on restriction, whereas the displayed Mackey formula and Krull–Schmidt theorem have exactly one nonprojective summand, $M$, with multiplicity one. There must be one, since $M$ is nonprojective. Hence

Here is an involution form of the double-coset criterion for a one-point extension. Embed $G$ in $\mathrm{Sym}(\Omega \cup \{\infty \})$ by fixing $\infty$, choose $a\in \Omega$, and put $H=G_a$. The necessary and sufficient condition is the existence of an involution $t$ such thatProducts in this part are compositions of maps, with the rightmost map applied first. The criterion is a finite test on permutations of the enlarged set; it does not presuppose that a larger group has already been found.

For sufficiency, put $K=G\cup GtG$. Multiplying a double coset by an element of $G$ on either side leaves it unchanged. A product of two elements of $GtG$ reduces to $g_{1}tgt{g}_2$. If $g\in H$, then $tgt\in H\le G$; otherwise the last condition puts it in $GtG$. Hence $K$ is closed under multiplication. It contains the identity and is closed under inverses, since $t^{-1}=t$, so it is a group.

Every element $g_{1}t{g}_2$ sends $\infty$ to $g_1(a)\in \Omega$, whereas elements of $G$ fix $\infty$. Thus $K_{\infty }=G$. Moreover the orbit of $\infty$ contains $a$ and then all of $\Omega$, so $K$ is transitive and is the desired one-point extension of a permutation group. For the order check, $G\cap tGt=H$: an element of the intersection fixes both $\infty$ and $a$, and the reverse containment follows from $tHt=H$. Consequently $|GtG|=|G{|}^2/|H|=|\Omega ||G|$, as required.

For necessity, let $K$ be an extension. It is doubly transitive group action, of order $|\Omega |(|\Omega |+1)|G_a|$, which is even. By Cauchy theorem for groups it contains a nontrivial involution, and a pair interchanged by this involution can be carried to $(\infty ,a)$ by double transitivity. Conjugating gives an involution $t$ interchanging these two points. Their common point stabilizer is $H$, so $tHt=H$. Finally $G$ has just two orbits on the enlarged set, namely $\{\infty \}$ and $\Omega$, giving $K=G\cup GtG$. More explicitly, for $k\in K\setminus G$, choose $g\in G$ with $g(a)=k(\infty )$; then $t{g}^{-1}k$ fixes $\infty$ and belongs to $G$. If $g\notin H$, the element $tgt$ moves $\infty$, so it belongs to $GtG$, completing necessity.

The Hecke operators organize modular forms simultaneously by their transformation law and their Fourier coefficients. Work with $\Gamma =S{L}_2(\mathbb{Z})$, $M_k$ its space of weight-$k$ modular forms, and $S_k$ its cusp form subspace, with even $k\ge 4$. Odd weights give zero spaces. The normalization below makes the first coefficient of $T_{n}f$ equal to the $n$th coefficient of $f$, which is the key link between operators and arithmetic.

For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with positive determinant, the determinant-normalized slash operator for modular forms isThe automorphy factor identity proves $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Positive scalar matrices act trivially. Let ${\mathcal{A}}_n$ be the integral matrices of determinant $n>0$. Left multiplication by $\Gamma$ reduces their first column, by Bezout identity, to $(a,0)$ with $a>0$; the determinant then gives $d=n/a>0$. A row shear makes $0\le b<d$. These are unique determinant-n matrix representatives for Hecke operators, so defineRight multiplication by any element of $\Gamma$ permutes these left cosets. The slash operator for modular forms identity therefore proves that $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic on the upper half-plane. Its cusp behavior follows explicitly from the coefficient calculation below, so $T_n$ preserves both $M_k$ and $S_k$.

Insert $f={\sum }_{r\ge 0}{a}_r{q}^r$ in the sum. The finite sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Writing $r=d\ell$ leaves the term $a_{d\ell }{q}^{a\ell }$ with multiplier $n^{k-1}{d}^{1-k}=a^{k-1}$. Thus the Fourier coefficients of a composite-index Hecke operator areFor $m=0$ use $\gcd (0,n)=n$. In particular the constant coefficient becomes ${\sigma }_{k-1}(n)a_0$, so cusp forms remain cusp forms, andfor a prime number $p$. The absence of negative powers proves cusp holomorphy. Notice that no primality assumption is needed in the composite-index formula.

The Hecke multiplication relations can be derived directly, rather than assumed from the coset description. On Fourier series define auxiliary maps $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. They need not separately preserve $M_k$, but $U_p{V}_p=1$. With $c=p^{k-1}$, the coefficient formula givesMultiplying on the left by $U_p+c{V}_p$, the term $j=0$ in the $U_p$ sum contributes $U_p^{r+1}$; the terms with $j\ge 1$ become $c^j{V}_p^{j-1}{U}_p^{r-j}$. Together with the $c{V}_p$ sum this is exactly $T_{p^{r+1}}+c{T}_{p^{r-1}}$. Hence the Hecke prime-power recurrence from q-shift operators isFor coprime $m,n$, substituting the coefficient formula twice gives a sum over $a\mid \gcd (m,r)$ and $b\mid \gcd (n,r)$ with coefficient $(ab{)}^{k-1}{a}_{mnr/(ab{)}^2}(f)$. The divisors $a,b$ correspond uniquely to a divisor $ab\mid \gcd (mn,r)$, so $T_m{T}_n=T_{mn}$. The prime-power recurrence, by induction on the smaller exponent, givesCombining the coprime and prime-power formulas provesIn particular the Hecke operators commute, and the modular Hecke algebra is generated by the prime operators.

The analytic reason for simultaneous diagonalization is the Petersson inner productwhich is positive definite on $S_k$ and convergent because of exponential cusp decay. Here is the adjoint argument. For a matrix $A$, set $H_A=\Gamma \cap A^{-1}\Gamma A$. The left cosets in $\Gamma A\Gamma$ are represented by $A\gamma$ with $\gamma \in H_A\backslash \Gamma$. Unfolding this finite coset sum in the Petersson inner product givesUnder $z=A\tau$, hyperbolic area is unchanged and the determinant factors in the slash operator for modular forms giveThe integration group becomes $A{H}_A{A}^{-1}=\Gamma \cap A\Gamma A^{-1}$. Folding this finite cover back to $\Gamma \backslash \mathbb{H}$ identifies the adjoint with the correspondence for $A^{-1}$. For $A\in {\mathcal{A}}_n$, the matrix $\mathrm{adj}(A)=n{A}^{-1}$ is again integral with determinant $n$; scalar matrices act trivially, and adjugation permutes the double cosets making up ${\mathcal{A}}_n$. The real normalizing factor $n^{k/2-1}$ is unchanged. Thus this proves that Hecke operators are self-adjoint for the Petersson inner product:The valence formula for the modular group makes $S_k$ a finite-dimensional vector space. Commuting self-adjoint operators on a finite-dimensional inner product space have a common orthogonal eigenbasis: diagonalize one operator, use commutativity to preserve each of its eigenspaces, and repeat within those spaces. Therefore $S_k$ has a basis of Hecke eigenforms, with real Hecke operator eigenvalues.

If $f$ is a common Hecke eigenform, say $T_{n}f={\lambda }_{n}f$, its first Fourier coefficient cannot vanish: $a_1=0$ would imply $a_n=a_1(T_{n}f)={\lambda }_n{a}_1=0$ for every $n$. Normalize $a_1=1$. Then the Hecke operator eigenvalue ${\lambda }_n$ is precisely the Fourier coefficient $a_n$. The Hecke multiplication relations becomeConsequently the coefficients are multiplicative at coprime indices, and their local generating function satisfiesThis gives the Euler product of a Hecke eigenformin a half-plane of absolute convergence. The Fourier coefficient bound for a cusp form already supplies such a half-plane; Question 4 improves it to $\mathrm{Re}s>(k+1)/2$.

The L-function of a cusp form also has an analytic description independent of the Hecke eigenform condition. Termwise Mellin transform in an initial right half-plane givesAt infinity the cusp form decays exponentially. Modular inversion gives $f(i/y)=(iy{)}^{k}f(iy)$, so at zero it is $O(y^{-k}{e}^{-2\pi /y})$. The integral therefore converges locally uniformly for every $s$, continuing $\Lambda$ to an entire function. Changing $y$ to $1/y$ provesFor a normalized Hecke eigenform, the Euler product of a Hecke eigenform and this functional equation describe the same completed L-function of a cusp form.

There is also an integral arithmetic structure. One standard consequence of the level-one dimension formula and the valence formula for the modular group is the integral triangular basis of level-one modular forms: if $k=12m+s$ with $s\in \{0,4,6,8,10,14\}$, choose $F_s=1,E_4,E_6,E_4^2,E_4{E}_6,E_4^2{E}_6$ respectively. Thenform a basis of $M_k$, and the terms $1\le j\le m$ form a basis of $S_k$. Their first nonzero terms are $q^j$ with coefficient one and all coefficients are integers. Successive elimination of coefficients of $q,q^2,\dots ,q^m$ shows that these cusp basis elements generate exactly the lattice of rational cusp forms with integral Fourier coefficients. The composite-index coefficient formula preserves this lattice, so each $T_n$ has an integral matrix on it. Its characteristic polynomial is monic with integral coefficients. This proves that Hecke eigenvalues are algebraic integers; self-adjointness also makes every root of that characteristic polynomial real. The field generated by the eigenvalues is a finite totally real number field: a finite set of operators spans the finite-dimensional rational modular Hecke algebra, and every conjugate of each eigenvalue remains a root of its real-rooted characteristic polynomial.

Two basic examples connect the operator theory to familiar modular forms. The normalized Eisenstein series $E_k$ has coefficients proportional to ${\sigma }_{k-1}(n)$. At a prime number its coefficient formula useswhere the second term is zero unless $p\mid m$. This follows by writing $m=p^{r}u$ with $p\nmid u$ and summing a finite geometric series. It proves $T_p{E}_k=(1+p^{k-1})E_k$, and the Hecke multiplication relations give $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Meanwhile $S_{12}=\mathbb{C}\Delta$ implies $T_n\Delta =\tau (n)\Delta$, because $a_1(\Delta )=1$. Thus the Ramanujan tau function is multiplicative and satisfies $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$. For example, $\tau (2)=-24$ gives $\tau (4)=(-24{)}^2-2^{11}=-1472$. This illustrates how Hecke operators turn analytic Fourier coefficients into a coherent arithmetic system.

Use the usual Krull–Schmidt theorem setting, in which indecomposable summands have local endomorphism rings; the paper's rings $\mathcal{O},k$ satisfy it for the subgroups involved. The relative projective module condition for $H$ means that $M$ is a direct summand of a module induced from $H$, equivalently, $M$ is a summand of ${\mathrm{Ind}}_H^G{\mathrm{Res}}_H^{G}M$. A vertex of an indecomposable module is an inclusion-minimal such subgroup. A source of an indecomposable module at its module vertex $D$ is an indecomposable $RD$-summand $S$ of $M{\downarrow }_D$ such that $M\mid S{\uparrow }^G$.

Existence of a module source follows by decomposing $M{\downarrow }_D$ and using Krull–Schmidt theorem on the induced decomposition: the indecomposable summand $M$ must occur in the induction of one indecomposable summand $S$. Such $S$ has module vertex $D$ as a $D$-module, since relative projectivity of $S$ for a proper subgroup would make $M$ relatively projective for that subgroup.

In the paper's modular coefficient rings, an indecomposable module is a trivial source module when its module source is the rank-one trivial module $R$ of its module vertex. It is then a summand of $R[G/D]$, a permutation module. Conversely suppose $M$ is an indecomposable summand of a permutation module. Restriction to a module vertex $D$ is a sum of modules $R[D/E]$, one for each $D$-orbit of the permutation basis. For a $p$-group $D$, each transitive permutation module is indecomposable: over $k$ it has simple head of a module over the local ring $kD$, and over $\mathcal{O}$ a nontrivial splitting would give a nontrivial splitting after reduction. A module source $S$ must be one of these summands by Krull–Schmidt theorem. Its module vertex is $D$, whereas $R[D/E]$ is relatively $E$-projective. Thus $E=D$, and $S\cong R$. Therefore trivial module source is equivalent to being an indecomposable permutation-module summand.

The rank-one formulation uses the paper's coefficient rings. If the earlier general coefficient-ring wording were applied to a disconnected ring, it would need qualification: for $R=k\times k$ and $G=1$, $(k,0)$ is an indecomposable summand of the permutation module $R$, but is not the rank-one trivial module $R$. One can instead allow indecomposable summands of the trivial coefficient module as module sources in that generality.

For the homomorphism request, take $M_i=\mathcal{O}[G/H_i]$. Frobenius reciprocity and Mackey decomposition giveThe summand for $g$ is generated by the sum of the cosets in the $H_1$-orbit of $g{H}_2$. The identical orbit-sum construction over $k$ gives one copy of $k$ per double coset. These are free bases in both characteristics, since invariance means coefficients are constant on orbits, with no division by orbit size. HenceThe same argument applies to direct sums of coset permutation modules.