Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 4 (Modular Representation Theory of Finite Groups)

Write $A=RG$. An idempotent is an element $e$ with $e^2=e$; it is central idempotent when it commutes with every element of the ring. In a commutative ring centrality is automatic. Two idempotents are orthogonal idempotents if $ef=fe=0$. A nonzero primitive central idempotent cannot be expressed as the sum of two nonzero orthogonal central idempotents; this is primitivity in $Z(A)$, not necessarily primitivity in $A$.

A block of a group algebra is an indecomposable two-sided direct-summand ideal of $A$, equivalently an ideal $Ae$ with $e$ a primitive central idempotent. Here is the equivalence. If $A=B\oplus C$ as two-sided ideals, then $BC=CB=0$. Write $1=e+f$ with $e\in B,f\in C$. Multiplication by this equality shows that $e$ is the identity of $B$, that $f$ is the identity of $C$, and that $e,f$ are orthogonal central idempotents. Conversely a central idempotent gives $A=Ae\oplus A(1-e)$. Splitting the ideal further is therefore equivalent to splitting its central idempotent further.

For uniqueness, suppose $1={\sum }_i{e}_i={\sum }_j{f}_j$ are decompositions into primitive central idempotents. For each $i$,is an orthogonal central-idempotent decomposition. Exactly one summand is nonzero, and that summand equals $e_i$. Applying the same reasoning to the corresponding $f_j$ gives $e_i=f_j$. Thus the blocks, and their identities, are unique up to ordering. Existence in the paper's coefficient rings follows from finite-dimensionality over $k$, and over $\mathcal{O}$ from completeness and idempotent lifting of the decomposition modulo its maximal ideal.

Over $k$, a principal indecomposable module is an indecomposable summand of the regular left module $A=kG$, equivalently $P=Ae$ for a primitive idempotent $e$ of $A$. Put $J=J(A)$, the Jacobson radical. The head of a module $P/JP$ is nonzero and semisimple. It is simple: projectivity gives a surjectionwhose kernel consists of maps into $JP$ and is nilpotent, since a product of $r$ such maps sends $P$ into $J^{r}P$. If the head of a module split, a nontrivial projection on it would lift by idempotent lifting to a nontrivial projection on $P$, contradicting indecomposability. Every maximal submodule contains $JP$, since $J$ annihilates simple modules. The simplicity of the head of a module now proves $J(P)=JP$ is the unique maximal submodule, and $P$ is its head of a module's projective cover.

Choose the simple modules $S_1,\dots ,S_r$, with corresponding projective covers $P_1,\dots ,P_r$. Our convention for the Cartan matrix of a group algebra iswhere brackets denote the multiplicity in a composition series.

For the final two unheaded requests, retain the additional hypothesis $G=N{C}_G(N)$. Choose a central subgroup $Z=⟨z⟩$ of $N$ of order $p$. Every element of $N$ centralizes $Z$, and so does every element of $C_G(N)$; hence $Z\le Z(G)$. Set $t=z-1$, so $t$ is central and $t^p=0$, and $A/tA\cong k(G/Z)$.

To lift a central idempotent $\bar{e}$ of this quotient, first lift it as an idempotent $e$ by idempotent refinement theorem. Since $\bar{e}$ is central, $eA(1-e)\subseteq tA$. Multiplying a representation $ea(1-e)=tb$ on the left and right by $e,1-e$ givesThe other off-diagonal corner vanishes likewise, so $e$ is central. Two central lifts of the same idempotent coincide: their mismatched products $e(1-f)$ and $f(1-e)$ are idempotents in the nilpotent kernel and therefore zero. Primitivity is preserved because orthogonal idempotent decompositions lift. Thus the central $p$-quotient gives a bijection of block idempotents.

The quotient $G/Z$ still satisfies the required hypothesis for $N/Z$, because the image of $C_G(N)$ centralizes $N/Z$. Induction on $|N|$ therefore proves $\tau$ bijects the block idempotents of $kG$ and $k(G/N)$. The centrality argument is essential: a general nilpotent algebra quotient need not give a bijection of central block idempotents.

For the Cartan matrix of a group algebra, every projective $A$-module is free over $kZ$: $A$ is free over $kZ$, a projective $A$-module is a summand of a finite free $A$-module, and finite projectives over the local ring $kZ\cong k[t]/(t^p)$ are free. If $P$ is a principal indecomposable module, $\bar{P}=P/tP$ is a principal indecomposable module over $A/tA$: an idempotent splitting of its endomorphism ring would lift and split $P$. The $p$ quotients of its filtrationare all isomorphic to $\bar{P}$ as $k(G/Z)$-modules, by multiplication by the corresponding power of $t$. All simple modules are annihilated by the nilpotent ideal $tA$. Hence $[P:S]=p[\bar{P}:S]$, with the natural indexing of simple modules, and $C_G=p{C}_{G/Z}$. Iterating over $N$ givesThis is blocks and Cartan matrices under central p-quotients. The extra hypothesis is still in force: for example $N=C_3⊴S_3$ in characteristic three does not satisfy it, and its Group algebra Cartan matrix is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ rather than $3I$.

Let $I$ be the augmentation ideal of $kN$, generated by the elements $n-1$. First prove its nilpotence. A nontrivial finite $p$-group has a central element $z$ of order $p$. Put $t=z-1$. In characteristic $p$, $t^p=z^p-1=0$, andInduction on $|N|$ says that the image of $I$ in this smaller group algebra is nilpotent. Thus $I^m\subseteq tkN$ for some $m$, and $I^{mp}\subseteq t^{p}kN=0$. The trivial group starts the induction. This proves nilpotence of a p-group augmentation ideal.

The quotient homomorphism kills exactly the differences of elements in each $N$-coset, soNormality of $N$ makes $I$ stable under conjugation by $G$. Moving group elements past factors from $I$ therefore gives $(IkG{)}^r=I^{r}kG$. Consequently

Partition the conjugacy class into its orbits under conjugation by $N$. Since none of its elements centralizes $N$, each orbit has size $[N:C_N(x)]>1$, a power of $p$. All elements of such an orbit have the same image in $G/N$, because conjugation by an element of $N$ does not alter an $N$-coset. Its orbit sum therefore has imagein characteristic $p$. Summing over the orbits proves $[\mathcal{C}]\in \ker \tau$. This does not require $G=N{C}_G(N)$; that additional assumption is used in the block and Cartan conclusions proved in the root Solution.

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.

If $M$ is a trivial module source $kG$-module, choose a permutation lattice $X$ with $M\mid \bar{X}$. Its projection is an idempotent $f\in {\mathrm{End}}_{kG}(\bar{X})$. The endomorphism reduction is surjective by the preceding calculation. Its kernel is $\mathfrak{p}{\mathrm{End}}_{\mathcal{O}G}(X)$, and the endomorphism algebra is complete over $\mathcal{O}$. By idempotent refinement theorem, $f$ lifts to an idempotent $\hat{f}$. The lattice $\hat{f}X$ is a permutation-module summand, hence a trivial module source lattice, and reduces to $M$.

To prove uniqueness within the class of trivial module source lattices, let $L,L^{\prime }$ be two such lifts. They are summands of permutation lattices. Lift a modular isomorphism and its inverse between them by the same homomorphism-surjectivity argument, inserting the summand projections on either side. The lifted composites are identity modulo $\mathfrak{p}$ and therefore invertible over $\mathcal{O}$, by Nakayama lemma or the determinant criterion. The lifted maps are consequently inverse up to an invertible composite and give $L\cong L^{\prime }$. Thus each trivial module source module has a unique trivial module source lift up to isomorphism. This is permutation homomorphisms lift through modular reduction; uniqueness is not claimed among arbitrary lattice lifts.

Let $D,E$ be module vertices of $M$, and choose a module $T$ such that $M\mid {\mathrm{Ind}}_E^{G}T$. Relative $D$-projectivity then givesApply Mackey decomposition to the inner restriction and use induction transitivity. The resulting module is a direct sum of modules induced to $G$ from subgroups $D\cap {}^{g}E$, for representatives $g\in D\backslash G/E$. By Krull–Schmidt theorem, the indecomposable module $M$ is a summand of one of these induced representations. It is therefore relatively projective for one intersection contained in $D$. Minimality of $D$ forces $D=D\cap {}^{g}E$, so $D\le {}^{g}E$.

Interchanging $D,E$ gives $E\le {}^{h}D$ for some $h$. The two inclusions force equal orders, and henceThus all vertices of an indecomposable module are conjugate in $G$.

Let $S,T$ be module sources at $D$. Since $S\mid M{\downarrow }_D$ and $M\mid T{\uparrow }^G$, Mackey decomposition givesThe module source $S$ has module vertex $D$. By Krull–Schmidt theorem it is a summand of one term, and minimality of its module vertex forces $D\cap {}^{g}D=D$. As both subgroups have order $|D|$, $g\in N_G(D)$. The corresponding term is simply ${}^{g}T$, which is indecomposable. ThusThis proves conjugacy of sources of an indecomposable module at the specified module vertex.

Choose a Sylow $p$-subgroup $P$ of a module vertex $D$. Its index $[D:P]$ divides the $p^{\prime }$-part of $|G|$, so it is a unit in $R$ under the hypothesis. For every $RD$-module $U$, the induction counit from $P$ splits by averaging: equivalently, the relative trace satisfiesThe D. Higman criterion therefore makes every $RD$-module relatively $P$-projective. Since $M$ is relatively $D$-projective, transitivity of induction makes it relatively $P$-projective. Minimality of the module vertex now forces $D=P$. Thus every module vertex is a $p$-subgroup.

Put $q=p^n$ and choose a generator $h$ of $H$. In characteristic $p$, $(h-1{)}^q=h^q-1=0$, soA module is a vector space with a nilpotent operator $t$ whose nilpotence index is at most $q$. The nilpotent Jordan normal form, valid over every field, splits it into Jordan blocks. One block of size $i$ isIt is indecomposable: its endomorphism ring is $k[t]/(t^i)$, a local ring with no nontrivial idempotents. Distinct $i$ have distinct dimensions. Conversely a direct sum of at least two Jordan blocks is decomposable. Thus there are exactly $q$ indecomposable isomorphism classes.

A homomorphism $V_i\to V_j$ is determined by the image of $1$, which can be any element of $V_j$ annihilated by $t^i$. If $i\ge j$ this is all of $V_j$, of dimension $j$; if $i<j$ it is the span of $t^{j-i},\dots ,t^{j-1}$, of dimension $i$. HenceThis realizes the indecomposable modules for a cyclic p-group.

For generators $g,h$ of $P$, write $x=g-1$, $y=h-1$. ThenFor each $\lambda \in k$, define a two-dimensional module $M_{\lambda }$ on basis $v,w$ by $xv=w$, $xw=0$, $yv=\lambda w$, $yw=0$. Every product of two of these operators is zero, so the defining relations hold even when $p=2$.

The action of $x$ is one nonzero nilpotent Jordan block. A decomposition of $M_{\lambda }$ would decompose its restriction to $k[x]/(x^p)$, which is already indecomposable. Thus $M_{\lambda }$ is indecomposable. An isomorphism $F:M_{\lambda }\to M_{\mu }$ commutes with $x$ and $y$, and consequentlySince $F$ is invertible and $x\ne 0$, $\lambda =\mu$. The infinite field supplies infinitely many parameters, proving $kP$ has infinite representation type. This is the parameter family for an elementary abelian p-group.

Let $B=kGe$ have defect group $D$. Use two vertex-and-source facts, with their relevant content made explicit. Every indecomposable module in $B$ has a module vertex conjugate into $D$. Also $B$, as a $k[G\times G]$-module with $(g,h)a=ga{h}^{-1}$, is an indecomposable permutation-module summand with module vertex $\Delta D=\{(d,d):d\in D\}$ and trivial module source. The second fact is the bimodule characterization of a defect group of a block; the first follows by applying its relative-$\Delta D$ induction splitting to $B{\otimes }_{kG}M=M$.

If $D$ is cyclic, every module source for every module vertex $Q\le D$ occurs among the finitely many indecomposable modules for a cyclic p-group. Each module in the block is a summand of an induction of one of these finitely many module sources. There are finitely many subgroups $Q$ and each induced representation has finitely many indecomposable summands by Krull–Schmidt theorem. Thus cyclic defect implies finite representation type of the block.

For the converse, the restriction $B{\downarrow }_{D\times D}$ has the regular $kD$-bimodule as a direct summand. To see this from module vertices and module sources, its Brauer quotient of a module at $\Delta D$ is nonzero. Its summands are permutation modules for the $p$-group $D\times D$ and have module vertices of order at most $|D|$. A summand detected at $\Delta D$ therefore has module vertex $\Delta D$ and trivial module source, hence isHere a transitive permutation module for a $p$-group is indecomposable: it has one-dimensional head of a module over the local group algebra and cannot split into two nonzero modules. This also explains the Brauer-quotient detection used in the argument.

For any $kD$-module $U$, tensor that split bimodule inclusion with $U$. It giveswhere $\mid$ denotes being a direct summand. If $B$ has only finitely many indecomposable modules $M_1,\dots ,M_s$, each $B{\otimes }_{kD}U$ is a sum of these. Therefore every indecomposable $U$ must occur among the summands of the finite list $M_i{\downarrow }_D$. Thus $kD$ has finite representation type.

A noncyclic $p$-group has a quotient $C_p\times C_p$: its Frattini quotient has rank at least two, since a one-generator Frattini quotient forces the group itself to be cyclic. Inflation along this quotient embeds all the indecomposable modules of that elementary abelian group, preserving isomorphisms and indecomposability. Part (b) rules out finite type when $k$ is infinite.

The conclusion remains valid if the splitting field $k$ is finite. For completeness, replace the scalar parameter in part (b) by an $m\times m$ companion matrix $T_f$ of an irreducible polynomial $f$. On $k^m\oplus k^m$ useAn endomorphism commuting with these has form $\left(\begin{array}{cc}A& 0\\ C& A\end{array}\right)$ with $A{T}_f=T_{f}A$. Its endomorphism ring has a square-zero ideal of the $C$-matrices and quotient $k[T_f]\cong k[t]/(f)$, a field, so it is local and the module is indecomposable. Finite fields have irreducible polynomials of arbitrarily large degrees, giving infinitely many dimensions. This also excludes noncyclic $D$ in that case. Altogether,

For the conjugation action, fixed elements have coefficients constant on $G$-conjugacy classes. Therefore $(\mathcal{O}G{)}^G=Z(\mathcal{O}G)$. More generally an $\mathcal{O}$-basis of $(\mathcal{O}G{)}^H$ consists of sums of the $H$-conjugation orbits on $G$.

For $a\in (\mathcal{O}G{)}^H$ define the relative trace, or transfer, byThe expression is independent of the coset representatives and is $G$-fixed. If $z\in Z(\mathcal{O}G)$, then $z{\mathrm{Tr}}_H^G(a)={\mathrm{Tr}}_H^G(za)$; hence its image is a transfer ideal of conjugation-fixed elements in the center of an associative algebra.

Let $P$ be a Sylow $p$-subgroup of $H$. Trace transitivity gives ${\mathrm{Tr}}_P^G={\mathrm{Tr}}_H^G{\mathrm{Tr}}_P^H$, proving inclusion of the $P$-image in the $H$-image. Conversely $[H:P]$ is a unit of $\mathcal{O}$, and for an $H$-fixed element $a$,Thus ${\mathrm{Tr}}_H^G(a)={\mathrm{Tr}}_P^G(a/[H:P])$, proving the reverse inclusion and

Use the basis of $D$-conjugation orbit sums in $kG$. A singleton orbit is exactly an element of $C_G(D)$, so these basis vectors span the subring $k{C}_G(D)$. For a nonsingleton orbit with representative $x$, its stabilizer is the proper subgroup $E=C_D(x)<D$, and its orbit sum is ${\mathrm{Tr}}_E^D(x)$. Thus every nonsingleton basis vector belongs to the sum of proper-subgroup transfer images.

Conversely the coefficient at a $D$-fixed basis element $x\in C_G(D)$ in ${\mathrm{Tr}}_E^D(a)$ is $[D:E]a_x=0$ in characteristic $p$ whenever $E<D$. Hence the two spans intersect trivially, givingFor $u\in (kG{)}^D$, multiplication by a transfer element satisfies $u{\mathrm{Tr}}_E^D(a)={\mathrm{Tr}}_E^D(ua)$ and ${\mathrm{Tr}}_E^D(a)u={\mathrm{Tr}}_E^D(au)$. Therefore the second summand is a two-sided ideal, as required; it need not be nilpotent in this general setting.

Projection onto the subring in part (b), along the two-sided ideal, is the Brauer homomorphism:It is a unital algebra homomorphism because it is the quotient by that ideal followed by the displayed subring identification.

Put $N=N_G(D)$. The projection is $N$-equivariant, since $N$ preserves the centralizer. Its restricted kernel isEquivalently it is spanned by the sums of the $N$-conjugation orbits on $G$ disjoint from $C_G(D)$. Normality of $D$ in $N$ ensures that an $N$-orbit either lies wholly inside that centralizer or wholly outside it.

Write $N=N_G(D)$, $C=C_G(D)$, and use the commutative transfer idealsThe target fixed algebra is commutative, because $C\le N$ makes every $N$-fixed element of $kC$ central in $kC$.

The Brauer morphism and relative trace identity isIndeed let $D$ act on $G/D$ by left multiplication in the trace sum. Its fixed cosets are exactly $gD$ with $g\in N$; every other orbit has size divisible by $p$. After Brauer projection the contributions on each such nonfixed orbit are equal, so they cancel. Fixed cosets give precisely the right-hand side. Since the Brauer projection of $(kG{)}^D$ is all of $kC$, this proves a surjection $I_D\twoheadrightarrow J_D$.

We use the trace criterion for defect groups: a block idempotent $e$ belongs to $I_D$ exactly when its defect group is conjugate into $D$; among these, ${\mathrm{Br}}_D(e)\ne 0$ exactly when its defect group is $D$ up to conjugacy. The criterion can be read from the class-sum basis: if $E=C_D(x)$ thenThis coefficient is nonzero exactly when $E$ is a Sylow $p$-subgroup of $C_G(x)$. Thus $I_D$ is spanned by classes whose class defect is conjugate into $D$, and the kernel of its Brauer projection is the sum of the corresponding proper-subgroup transfer ideals. The block version follows by refinement of the central idempotents in these ideals.

For clarity, the refinement fact needed here is primitive idempotents under a surjection of commutative Artinian ideals: a surjection between ideals of commutative Artinian algebras bijects the primitive idempotents in the module source whose images are nonzero with the primitive idempotents in the target. Decompose the algebras into local factors. An ideal in a local Artinian factor either contains the identity and is the whole factor, or lies in its nilpotent maximal ideal and has no nonzero idempotents. A surviving local factor has a local quotient, so its identity remains primitive. This proves the stated refinement fact without imposing nilpotence on the entire kernel.

The primitive idempotents in $I_D$ are the block idempotents with defect conjugate into $D$, and the surviving ones have defect $D$. Applying refinement to the surjection above givesThis is the Brauer morphism on a defect transfer ideal; the target is the transfer ideal $J_D$, not the entire fixed center of an associative algebra.

If $N_G(D)\le H\le G$, then $N_H(D)=N_G(D)=N$ and $C_H(D)=C_G(D)=C$. Applying part (d) to $G$ and to $H$ therefore gives the same target idealCompose the two bijections. A block idempotent $e$ of $G$ with defect $D$ corresponds to the unique block idempotent $f$ of $H$ with defect $D$ satisfyingThis proves the requested form of Brauer first main theorem, including uniqueness and the equality of the Brauer images.

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.

Put $b={\mathrm{Br}}_D(e)$. By equivariance of the Brauer homomorphism, $b$ is a central idempotent of $kK$, since $C_G(D)\le K\le N_G(D)$. The Nagao module theorem asserts that, if $eM=M$, thenand every indecomposable summand of $(1-b)M$ is relatively projective for a $p$-subgroup of $K$ not containing $D$. Because $D$ is normal in $K$, its module vertices cannot contain $D$ even after $K$-conjugation.

Here is a direct proof. As an element of $(kG{)}^K$, $e-b$ is a linear combination of $K$-conjugation orbit sums supported outside $C_G(D)$. An orbit represented by $x$ has sumIts stabilizer $C_K(x)$ does not contain $D$. On $T=(1-b)M$, the element $e-b$ acts as the identity. Map the orbit-sum expression into ${\mathrm{End}}_k(M)$ by left multiplication, and compose on both sides with the $K$-linear projection onto $T$. It follows thatReplace each centralizer by a Sylow $p$-subgroup of it using trace transitivity and its invertible prime-to-$p$ index. Project this identity further onto an indecomposable summand $U$ of $T$. Its endomorphism ring is local by Krull–Schmidt theorem. A sum of nonunits cannot be the identity in a local ring, so one of the trace terms is a unit $u$ in ${\mathrm{End}}_{kK}(U)$. Multiplying the endomorphism inside that trace by $u^{-1}$ gives ${\mathrm{id}}_U$ as a single relative trace. The D. Higman criterion makes $U$ relatively projective for that Sylow subgroup $Q\le C_K(x)$, and $D\nleqq Q$. This proves the assertion rather than just invoking the theorem's name.

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.