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Let $N_G(D)\le H\le G$. Set $\mathcal{X}=\{D\cap {}^{g}D:g\notin H\}$ and $\mathcal{Y}=\{H\cap {}^{g}D:g\notin H\}$. Green correspondence pairs indecomposable modules with vertex $D$ for $G$ and $H$: restriction has a unique vertex-$D$ summand, with the other summands relatively projective for $\mathcal{Y}$; induction has a unique vertex-$D$ summand, with the other summands relatively projective for $\mathcal{X}$. The two assignments are inverse and preserve sources. Its more general form applies to vertices contained in $D$ but not conjugate into any member of $\mathcal{X}$.

If $N_G(D)\le H\le G$, Brauer correspondence pairs the blocks of a group algebra of $RG$ and $RH$ having defect group of a block $D$. This is Green correspondence for their bimodules, with vertex of an indecomposable module $\Delta D$ in $G\times G$. The normalizer of $\Delta D$ lies in $H\times H$, and the corresponding restriction and induction have unique summands with that vertex.

The Green correspondent is the unique indecomposable summand of restriction or induction having the selected vertex of an indecomposable module, in the setting of Green correspondence. The two correspondence maps are inverse and preserve sources of an indecomposable module up to the appropriate normalizer conjugacy.

For an indecomposable $kK$-module $V$ with vertex $D$ and ${\mathrm{Br}}_D(e)V=V$, every indecomposable summand of $(1-e){\mathrm{Ind}}_K^{G}V$ has vertex conjugate into $D\cap {}^{g}D$ for some $g\notin N_G(D)$. The refinement keeps the intersection subgroups from Green correspondence, rather than asserting only that the remaining vertices are smaller than $D$.

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.

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

Let $D$ be a $p$-subgroup and let $N_G(D)\le H\le G$. PutA module relatively projective for a family means a summand of a sum of modules induced from its members, allowing subgroups and conjugates. The Green correspondence gives inverse bijections $f,g$ between the indecomposable $RG$-modules and $RH$-modules with module vertex $D$, up to isomorphism, such thatThe error terms are respectively relatively $\mathcal{Y}$-projective and $\mathcal{X}$-projective. In particular the displayed correspondents are the unique summands of module vertex $D$, occur with multiplicity one, and have the same module source as the original module, up to normalizer conjugacy.

We will also use the general form of Green correspondence: replace module vertex $D$ by any module vertex $Q\le D$ not conjugate into a member of $\mathcal{X}$. Restriction and induction again have unique summands with that module vertex, and the same error-family description. This is important in the final induction argument: merely knowing that an error term has a proper module vertex would not give the required intersection bound.

Take $N=N_G(D)$ and $b={\mathrm{Br}}_D(e)$. If $eM=M$, the Nagao module theorem with $K=N$ puts every vertex-$D$ summand of $M{\downarrow }_N$ inside $bM$. In particular its Green correspondence summand $M^{\prime }$ satisfies $b{M}^{\prime }=M^{\prime }$.

Conversely a central idempotent acts as zero or one on the indecomposable module $M$. If $eM=0$, apply the same theorem to $1-e$. Its Brauer image is $1-b$, so $(1-b)M^{\prime }=M^{\prime }$, incompatible with $b{M}^{\prime }=M^{\prime }$. Thus

For the unheaded induction request, write $\mathcal{X}=\{D\cap {}^{g}D:g\notin N\}$. We prove the sharper intersection bound, not just a drop in module vertex size. We use Green correspondence on permutation bimodules and the usual module vertex test for a permutation-module summand: its Brauer quotient of a module at $P$ is nonzero precisely when a module vertex contains a conjugate of $P$. This test follows by restricting to the $p$-group $P$, decomposing the permutation basis into its orbits, and noting that proper-orbit sums are precisely the proper-subgroup transfer terms.

Consider $kG$ as a $k[G\times D]$-module, with $(g,d)a=ga{d}^{-1}$. It is the transitive permutation module induced from $\Delta D$. Set $L=G\times D$ and $L_0=N\times D$; the latter contains $N_L(\Delta D)$. Its restriction decomposes asEach outside term has stabilizer a conjugate of a subgroup of $\Delta D$. If a diagonal subgroup $\Delta Q\le \Delta D$ embeds into one of these stabilizers, then $Q\le D\cap {}^{g}D$ for an outside representative. Thus its diagonal module vertices are in the intersection error family.

Every indecomposable $L_0$-summand of $kN$ has module vertex $\Delta D$. To justify this, its endomorphisms are right multiplications by elements of $(kN{)}^D$. The coefficient projection ${\mathrm{Br}}_D^N$ on this algebra has nilpotent kernel: every nonsingleton $D$-orbit sum lies in the kernel of $kN\to k(N/D)$, by part 1(b), and that kernel is nilpotent by part 1(a). Hence no nonzero idempotent endomorphism can have zero Brauer image. Its image summand has nonzero Brauer quotient of a module at $\Delta D$, and its module vertex, already contained in a conjugate of $\Delta D$, must be $\Delta D$.

Apply the general Green correspondence from part (a) to $L,L_0$. If an indecomposable summand of $kG$ has module vertex outside the intersection error family, its correspondent must be a summand of $kN$, since the outside double-coset modules have only error-family module vertices. It therefore has module vertex $\Delta D$. Consequently every summand of $kG$ not having module vertex $\Delta D$ has module vertex conjugate into an outside diagonal intersection. An intersectionwith $(g,d)\notin L_0$ projects in its first coordinate into $D\cap {}^{g}D$ with $g\notin N$; this is the error family required in the question.

Now form the split permutation bimoduleIts Brauer quotient of a module at $\Delta D$ is zero. Indeed $kG(\Delta D)=k{C}_G(D)$, left multiplication by $e$ becomes multiplication by ${\mathrm{Br}}_D(e)=b$, and right multiplication by $b$ remains multiplication by $b$. ThereforeHere $b$ is central in $k{C}_G(D)$, so the last product is zero. Every indecomposable bimodule summand of $W$ consequently has module vertex in the outside intersection family, by the preceding paragraph, and has trivial module source since it is a permutation-module summand.

Since $V$ has vertex $D$, necessarily $D\le K$. Let $S$ be a module source of $V$ at $D$. The splitting $V\mid kK{\otimes }_{kD}S$ remains a splitting after applying $b$, since $bV=V$. Inducing to $G$ and applying $1-e$ givesA trivial-source bimodule summand with module vertex conjugate into $\Delta Q$ is a summand of ${\mathrm{Ind}}_{\Delta Q}^{G\times D}k$. Tensoring with $S$ turns this into a module induced to $G$ from $Q$ (with the appropriate conjugate restriction of $S$). Thus every summand on the right is relatively projective for some $Q\le D\cap {}^{g}D$, $g\notin N$. Each $V_j$ lies in the left-hand side, and Krull–Schmidt theorem now givesThis proves Juhász induction refinement. If there are no outside elements, the error family is empty and the complementary module is zero.

A finite-dimensional group algebra has finite representation type of a group algebra when there are only finitely many isomorphism classes of finite-dimensional indecomposable modules. The definition concerns all dimensions, not merely finitely many modules in each fixed dimension.

Let $e_B$ be the block idempotent and choose a maximal Brauer pair $(D,b_D)$: $b_D$ is a block idempotent of $k{C}_G(D)$ with $b_D{\mathrm{Br}}_D(e_B)\ne 0$. The inertial index of a block iswhere $N_G(D,b_D)$ is the stabilizer of the chosen pair. Changing the pair by conjugation does not alter this number. For cyclic $D$, it is a prime-to-$p$ divisor of $p-1$. Here $n\ge 1$, as required for the order-$p$ subgroup $Q$ to exist.

Put $H=N_G(Q)$. Since $Q$ is characteristic in $D$, $N_G(D)\le H$, so Green correspondence applies. In fact the nonprojective modules in $B$ have a particularly simple correspondence. If $g\notin H$, a nontrivial intersection $D\cap {}^{g}D$ would contain the unique order-$p$ subgroup of both cyclic groups, giving $Q={}^{g}Q$ and hence $g\in H$, a contradiction. Thus every such intersection is trivial. The error modules in induction are projective. In restriction, an intersection $H\cap {}^{g}D$ cannot contain a conjugate of a nontrivial subgroup of $D$ by an element of $H$: such a conjugate contains $Q$, and that would again force ${}^{g}Q=Q$. The general Green correspondence therefore pairs all indecomposable modules with nontrivial module vertex contained in a conjugate of $D$. In particular it pairs the nonprojective modules of $B$ with the nonprojective modules of its Brauer correspondent $b$ in $H$; the block-compatible form of the correspondence selects $b$. Explicitly,where $P$ is projective and the restriction error family is $\mathcal{Y}=\{H\cap {}^{g}D:g\notin H\}$. One should not replace this restriction error term by projectives without an additional argument about $H$.

For the final structure calculation we use the general cyclic-defect structure theorem, which is permitted here as a general fact: the basic algebra of a split block with cyclic defect group is a Brauer tree algebra. Its tree has $e(B)$ edges, one simple module per edge, and exceptional multiplicityFor a one-edge tree the basic algebra presentation is $k[t]/(t^{m+1})$. This is the local presentation of the tree algebra, not an assertion that an arbitrary block with one simple module is automatically uniserial.

With $e(B)=1$ the tree has one edge and $m=p^n-1$. Thus the basic algebra of $B$ isIt has exactly one simple module, $k=A_0/(t)$. Its ideals are precisely $(t^i)$: for any nonzero element, factor out its lowest power of $t$, after which the remaining factor is a unit. The regular projective therefore has the unique submodule chainwith each consecutive factor one-dimensional and simple. A Morita equivalence from $A_0$ to $B$ preserves simple modules, projective covers, submodule lattices and composition length. HenceIn the split setting one may also write $B\cong M_{{\dim }_{k}S}(A_0)$. The projective cover has $p^n$ composition factors all isomorphic to $S$, so its vector-space dimension is $p^n{\dim }_{k}S$, not generally $p^n$.

There is also an elementary reason why the uniserial conclusion follows once the one-simple-module conclusion and finite type are known. If the basic local algebra had $\dim J/J^2\ge 2$, quotienting its radical square and all but two generators would give $k[x,y]/(x,y{)}^2$. The two-dimensional parameter family from Question 3(b) would then give infinitely many indecomposables. Thus $J/J^2$ is one-dimensional. Nakayama lemma makes $J$ generated by one element $t$, so its powers yield a chain and the basic algebra is $k[t]/(t^r)$. The cyclic-defect calculation above identifies $r=p^n$. This explains why finite representation type rules out branching in this projective cover.

A p-local module here is a finite direct sum of induced representations from normalizers $N_G(Q)$ of nontrivial p-subgroups $Q$. This usage includes induction from $G$ itself when $G$ normalizes such a subgroup. Green correspondence and induction on vertex order express every finite module, or lattice over a complete valuation ring, as a difference of p-local modules after adjoining projective modules.