Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 5 (Modular Representations of Finite Groups)

Take finite-dimensional modules over $k$ and finite $R$-free lattices over the complete valuation ring, the usual integral modular representation theory convention. If a completion has several local factors, use the factor supporting the modular block. A block of a group algebra is an indecomposable two-sided ideal $B=RGb$, where $b$ is a primitive central idempotent. The identity decomposes as the sum of orthogonal block idempotents, and $M={⨁}_{b}bM$ for every module. Hence an indecomposable module belongs to $B$ exactly when $bM=M$.

A defect group of a block in characteristic $p$ is a maximal p-subgroup $D$ with ${\mathrm{Br}}_D(b)\ne 0$. Equivalently, it is minimal with $b\in {\mathrm{Tr}}_D^G((kG{)}^D)$; the trace criterion for defect groups and conjugacy are proved below. For a complete discrete valuation ring, use the corresponding reduced modular block; modular reduction of block idempotents preserves its defect groups. The equivalent bimodule description says that the module vertices of $B$, with $(x,y)$ acting by $a\mapsto xa{y}^{-1}$, are conjugates of $\Delta D=\{(d,d):d\in D\}$. This basic characterization follows from the relative-projectivity trace criterion for the bimodule action.

Fix a Sylow subgroup $P$ containing $D$. The regular group algebra is the permutation module $R[G]$ for $G\times G$, and multiplication by $b$ is its block projection. Restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex: a module source at that module vertex is a summand of the module restriction, and the Mackey restriction formula and minimality prevent all the restricted summands from having smaller module vertices. Thus $B{\downarrow }_{P\times P}$ has an indecomposable summand with module vertex $\Delta D$.

For a finite p-group $Q$, the transitive permutation module $k[Q/L]$ is indecomposable: the group algebra of a p-group in characteristic p is local, and this module has a one-dimensional top, incompatible with a sum of two nonzero modules. Its module vertex is $L$, because it is induced from the trivial $L$-module and its Brauer quotient of a module at $L$ has the nonzero fixed coset $L$. The fixed-basis rule for this quotient follows by removing nonsingleton orbit sums, which are proper-subgroup traces. It excludes a smaller module vertex. Reduction gives the same statements for permutation lattices. Therefore Krull-Schmidt decomposition identifies the chosen block summand with an entire transitive $P\times P$ orbit in $G$.

Choose a point $c$ in that orbit whose stabilizer is exactly $\Delta D$. Being fixed by every $(d,d)$ gives $c\in C_G(D)$. Its full stabilizer isEquality with $\Delta D$ gives $P\cap c^{-1}Pc=D$. Conjugating, and using that $c$ centralizes $D$, provesThis is defect groups are centralizer-conjugate Sylow intersections.

If $Q◃G$ is a p-subgroup, then $QP$ is a p-subgroup for every Sylow $P$, so $QP=P$. It lies in both Sylow factors above. Thus every normal p-subgroup of $G$ is contained in $D$.

Finally put $N=N_G(D)$, choose $S\in {\mathrm{Syl}}_p(N)$, and extend it to a Sylow $P$ of $G$. Then $D\le S$ and $P\cap N=S$. Apply the proved intersection result to this $P$. Its element $c$ belongs to $C_G(D)\le N$, so $cP{c}^{-1}\cap N=cS{c}^{-1}$ is Sylow in $N$ too. Hence $D=S\cap cS{c}^{-1}$. The p-core $O_p(N)$ is contained in both factors, whereas normality of $D$ in $N$ gives the opposite inclusion. Therefore $D=O_p(N_G(D))$, so $D$ is a p-radical subgroup. The centralizer condition on the conjugating element is important for this deduction.

The Brauer morphism retains precisely the fixed group-basis elements under conjugation by $D$:It is unital and surjective because $k{C}_G(D)\subseteq (kG{)}^D$. For multiplicativity, fix $x\in C_G(D)$. Its coefficient in a product is ${\sum }_{gh=x}{a}_g{b}_h$. Conjugation by $D$ permutes these pairs and the summand is constant on each orbit. Nonsingleton orbits have p-divisible size and contribute zero. Singleton pairs have both entries in $C_G(D)$, so they give exactly the coefficient in ${\mathrm{Br}}_D(a){\mathrm{Br}}_D(b)$. This proves the ring-homomorphism assertion.

Orbit sums form a basis of $(kG{)}^D$. A nonfixed orbit sum of $g$ is ${\mathrm{Tr}}_{C_D(g)}^D(g)$, a proper-subgroup relative trace. Conversely any proper-subgroup trace has zero coefficient on a fixed basis element, since that coefficient is multiplied by $[D:Q]=0$ in $k$. ThusThere is no assertion that this whole kernel is nilpotent.

Brauer first main theorem bijects the modular blocks of $kG$ with defect $D$ and those of $k{N}_G(D)$ with defect $D$, matching their nonzero Brauer images. Here is a proof with its auxiliary ingredients.

Write $N=N_G(D)$, $C=C_G(D)$ and defineThese are ideals in the indicated commutative algebras, since an invariant multiplier passes through a relative trace. The Brauer morphism and relative trace identityfollows by letting $D$ act on $G/D$. Fixed cosets have representatives in $N$; projection onto $C$ gives equal contributions on each other orbit, whose p-divisible size makes the contribution zero. Thus ${\mathrm{Br}}_D$ surjects $I_D(G)$ onto $J_D$.

To prove the trace criterion for defect groups, choose a p-subgroup $E$ of least order with a given block idempotent $b$ in $I_E(G)$. Such a choice exists because a Sylow $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_E^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_E(b)=0$, the kernel formula and trace transitivity give $b\in {\sum }_{Q<E}{I}_Q(G)$. But $bZ(kG)$ is an Artinian local ring, and a finite sum of proper ideals cannot contain its identity. Multiplication by $b$ therefore forces $b\in I_Q(G)$ for some smaller $Q$, a contradiction. Hence ${\mathrm{Br}}_E(b)\ne 0$.

If $F$ is not conjugate into $E$, its action on $G/E$ has no fixed coset and the same orbit argument gives ${\mathrm{Br}}_F(I_E(G))=0$. Every subgroup with nonzero Brauer image of $b$ is therefore conjugate into $E$. Its maximal such subgroups are exactly the conjugates of $E$, proving the definition's conjugacy assertion andAmong block idempotents in this ideal, the Brauer image at $D$ survives precisely when their defect is $D$.

We also use primitive idempotents under a surjection of commutative Artinian ideals. Decompose the ambient algebras into local factors. The component of an ideal in a local factor is either the whole factor, possessing its identity, or a proper ideal in its nilpotent radical, possessing no nonzero idempotent. A nonzero quotient of a local algebra is local. A surjection of these ideals consequently bijects its surviving primitive idempotents with those of the target. Entire factors may be killed; the whole kernel need not be nilpotent.

Apply this result to $I_D(G)\to J_D$. Its surviving primitive idempotents are exactly the modular blocks of $G$ with defect $D$. Apply it also to $I_D(N)\to J_D$, where the target is identical since $N_N(D)=N$ and $C_N(D)=C$. Its surviving primitive idempotents are exactly the modular blocks of $N$ with defect $D$. Matching images in $J_D$ proves Brauer first main theorem, in both directions and with uniqueness.

For the last construction, the intended hypothesis is that $D$ is a defect group of the source modular block $b$. Put $L=N_H(D)$ and use the theorem inside $H$ to obtain the corresponding modular block $c$ of $kL$. The centralizer condition gives $C_H(D)=C_G(D)$, so ${\mathrm{Br}}_D$ maps $Z(kG)$ into $Z(kL)$. Let ${\omega }_c:Z(kL)\to F_c$ be the central character of a block, retaining its residue field $F_c$ rather than assuming it equals $k$. The unital compositesends exactly one global block idempotent to one, because their orthogonal idempotent images sum to one in a field. Define $b^G$ to be this unique modular block. This is block induction, independent of the chosen $H$-conjugate of $D$. It needs no containment of $N_G(D)$ in $H$, and its defect may be larger than $D$. With that containment it agrees with the first-main-theorem correspondence.

The printed PDF does not require $D$ to be a defect group of $b$. Its literal conditions do not always define the usual induced modular block. For example, over an algebraically closed field of characteristic three, take $G=A_5$, $H=A_4$ as a point stabilizer, and $D=⟨(123)⟩$. Then $D{C}_G(D)=D\le H$. The ordinary degree-three irreducible of $A_4$ gives a defect-zero modular block $b$ on reduction. The two distinct ordinary degree-three irreducibles of $A_5$ give distinct defect-zero modular blocks $B_{+}$ and $B_{-}$. Both restrict to that same character of $A_4$, with values $3,-1,0$ on its identity, double transpositions and three-cycles.

The standard defect-zero lifting theorem says that these reductions are simple projective modules and their block algebras are full matrix algebras. Thus both module restrictions $B_{+}{\downarrow }_{H\times H}$ and $B_{-}{\downarrow }_{H\times H}$ contain the same modular block bimodule $b$; indeed each module restriction is isomorphic to it here. The actual defect of $b$ is 1, whose centralizer is $G$, not a subgroup of $H$. The unrelated subgroup $D=C_3$ cannot select between the two modular blocks. This is block induction requires the block's defect centralizer; the qualified construction above answers the intended request.

Nagao module theorem states that if $e\in Z(kG)$ is a central idempotent, $eM=M$, and $C_G(Q)\le K\le N_G(Q)$, then ${\mathrm{Br}}_Q(e)$ is central in $kK$ andEvery indecomposable summand of the second term is relatively projective for some subgroup of $K$ not containing $Q$. In particular, if $Q\le K$, the error term has no summand with module vertex containing $Q$. This is the Nagao version of the second main theorem; it identifies the modular block of a module vertex-$Q$ Green correspondent.

Let $e_B$ be the identity of $B$, $N=N_G(D)$, and $b$ its Brauer correspondent of a block in $kN$. Because $D◃N$, the ideal $J(kD)kN$ is nilpotent: normality allows moving $kN$ past $J(kD)$ and the augmentation ideal of the p-group is nilpotent. Hence $b$ has nonzero image in $k(N/D)$. Choose an indecomposable summand $U$ of $bk[N/D]$. It is projective over $k(N/D)$ and belongs to modular block $b$ after inflation.

Since $k[N/D]={\mathrm{Ind}}_D^{N}k$ is a permutation module, its summand $U$ has trivial module source and module vertex contained in $D$. As $D$ acts trivially, all its proper-subgroup traces vanish and its Brauer quotient of a module is $U(D)=U\ne 0$. The module vertex must also contain $D$, so it is exactly $D$. This is inflated quotient projectives have trivial source.

Let $M$ be the Green correspondent of $U$ in $G$, using the theorem proved in the next solution. It has module vertex $D$ and the same trivial module source. For a normal p-subgroup, each central block idempotent of $kN$ equals its Brauer image: the two central idempotents agree modulo the nilpotent ideal $J(kD)kN$, because nonfixed conjugation-orbit sums map to zero in $k(N/D)$; central idempotents congruent modulo a nilpotent ideal coincide. The first main theorem therefore gives ${\mathrm{Br}}_D(e_B)={\mathrm{Br}}_D(b)=b$.

Let $e_C$ be the block identity acting on $M$. Nagao's theorem forces ${\mathrm{Br}}_D(e_C)U=U$, since $U$ has module vertex $D$. But $bU=U$, and for $C\ne B$ orthogonality would give ${\mathrm{Br}}_D(e_C)b={\mathrm{Br}}_D(e_C{e}_B)=0$. Therefore $M$ belongs to $B$, has module vertex $D$ and has trivial module source. For $D=1$, this supplies an indecomposable projective in the defect-zero modular block.

For the representation-type deduction, modules in a block are projective relative to its defect group: the block trace identity acts as a relative trace equal to the module identity, so D. Higman criterion gives split induction from $D$. If $D$ is cyclic of order $d$, then $kD\cong k[t]/(t^d)$, whose indecomposable modules are the nilpotent Jordan chains $k[t]/(t^j)$, $1\le j\le d$. Each indecomposable in $B$ is a summand of the module induction of its module restriction to $D$, and hence occurs among the summands of these finitely many fixed induced modules. Their finite Krull-Schmidt decompositions give finite representation type.

Conversely, the regular defect-group bimodule inside a block givesIndeed this module restriction is a permutation summand. Its Brauer quotient at $\Delta D$ is nonzero, because on the fixed group-basis elements $C_G(D)$ the block projection becomes multiplication by ${\mathrm{Br}}_D(e_B)\ne 0$. Every transitive $D\times D$ orbit in $G$ has stabilizer of order at most $|D|$. A summand with a nonzero quotient at $\Delta D$ must therefore have stabilizer exactly $\Delta D$ up to conjugacy, and is the regular $kD$ bimodule. Tensoring this split inclusion on the right with an arbitrary $kD$-module $V$ yieldsIf $B$ has only finitely many indecomposable modules, their restrictions contain only finitely many indecomposable summands. The displayed inclusion puts every indecomposable $V$ on that finite list, so $kD$ has finite representation type too.

A noncyclic finite p-group has a quotient $C_p\times C_p$, by the Burnside basis theorem for its Frattini quotient. Here is an explicit infinite family over any field, including a finite field. For $s\ge 1$, take basis $v_0,\dots ,v_s,w_1,\dots ,w_s$ and operators $x=g-1$, $y=h-1$ defined byand $x{w}_i=y{w}_i=0$. All products of $x,y$ vanish, so $1+x,1+y$ satisfy the group relations in characteristic $p$. The radical is $W=⟨w_1,\dots ,w_s⟩$. An endomorphism preserves $W$, and commuting with the two shift maps forces its induced maps on $V/W$ and $W$ to be the same scalar. To see this, write the shift matrices as $X=(0I_s)$ and $Y=(I_{s}0)$: the equations $CX=XA$, $CY=YA$ force the off-diagonal entries of $A,C$ to vanish and all diagonal entries to agree, by comparing successive rows, including the zero first column of $X$ and zero last column of $Y$. The remaining linear map sends the top into $W$ and has square zero. Thus the endomorphism ring is a scalar algebra with square-zero radical, hence local; the module is indecomposable. Inflation to $D$ preserves that ring. Its dimensions $2s+1$ are unbounded, contradicting finite representation type. Therefore $B$ has finite representation type if and only if $D$ is cyclic, including trivial defect.

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.

Choose a maximal Brauer pair $(D,b_D)$ for $B$. Define its inertial quotient of a block and inertial index of a block byConjugacy of maximal pairs makes the index well-defined. The standard maximal-pair property used here is that $E(B)$ is a $p^{\prime }$-group. Since $D$ is cyclic, its faithful conjugation action embeds it into $\mathrm{Aut}(D)$, so $e\mid p-1$. Write $d=p^n$ and $m=(d-1)/e$.

We use two precise structural results. Over an algebraically closed field, the basic algebra of a block with cyclic defect group is a Brauer tree algebra with $e$ edges and exceptional multiplicity $m$. For a Brauer tree algebra with $t$ edges and exceptional multiplicity $m$, all nonprojective indecomposable modules form the stable Auslander–Reiten quiverThese are the cyclic-defect structure theorem and the stable-quiver classification for tree algebras; we are not taking the requested numerical count as an assumption. Morita equivalence preserves simple, projective and indecomposable isomorphism classes.

There is one simple module per edge, so $l(B)=e$. The vertices of $\mathbb{Z}{A}_s$ can be indexed by $(j,i)$, $j\in \mathbb{Z}$, $1\le i\le s$. The quotient by ${\tau }^e$ leaves $e$ choices of $j$ for each $i$. With $s=em=d-1$, there are $e(d-1)$ nonprojective indecomposable modules. There are also $e$ indecomposable projective modules, one projective cover per simple. ThereforeThis counts isomorphism classes, not multiplicities in the regular module. In the one-edge case it agrees with the $d$ Jordan chains for $k[t]/(t^d)$.

For the definition, take a finite connected tree with a cyclic order of the incident edges at every vertex and positive integer multiplicities equal to one except possibly at a distinguished vertex. Its Brauer tree path presentation has a quiver vertex for each tree edge, and successor arrows around each tree vertex, including loops for valency one. If edge $i$ has endpoints $v,w$, let $C_{i,v}$ and $C_{i,w}$ be the cycles around these endpoints. The relations identifykill $C_{i,v}^{m_v}{\alpha }_{i,v}$, and kill paths switching between successor cycles belonging to different tree vertices. Redundant valency-one loops may be eliminated. This quotient of the path algebra is the basic Brauer tree algebra; allowing an arbitrary algebra with Morita equivalence to it does not change the module assertions. In particular, the cyclic orders and multiplicity, not merely the underlying graph, are part of the data.

For the construction, use right modules with paths multiplied left to right; equivalently use the opposite quiver for left modules, choosing orientation so radical factors follow the specified order. The indecomposable projective $P_1$ with top $S_1$ has simple socle $S_1$ and two uniserial radical branches $U_v,U_w$ meeting in that socle. The relations show that $U_v$ has length $mr$ and factors $S_2,\dots ,S_r,S_1$ repeated $m$ times. The other branch has length $m_w\mathrm{val}(w)$. Thus $W_v=P_1/U_w$ is a uniserial module of length $mr$, with factors $S_1,S_2,\dots ,S_r$ cyclically repeated. For $J$ the Jacobson radical,is the desired uniserial branch of a Brauer tree algebra. If an endpoint is a nonexceptional leaf, its branch is just the socle; the formula still applies.

For uniqueness, length one gives just $S_1$. If $q>1$ and $r>1$, the next factor $S_2$ chooses the successor arrow at $v$ uniquely, because two distinct edges cannot meet at both endpoints in a tree. Subsequent factors force the same cyclic succession. Any arrow into an edge not incident with $v$ acts as zero, because that simple does not occur. An opposite nonexceptional leaf loop is the full socle path around $v$, of length $mr$, hence zero on modules of length at most $mr$. If an opposite leaf is exceptional, then $v$ has multiplicity one and each incident simple occurs at most once; its nilpotent self-loop is zero on that one-dimensional vertex space as well.

The remaining action is one nilpotent successor chain. Choose a top generator $z_0$ and let $z_j$ be its successive path images for $0\le j<q$. Uniseriality makes each image nonzero in its respective radical layer; these images form a basis in the basic algebra description. Their labels are $S_{1+(j\mathrm{mod}r)}$, the successor arrow sends $z_j$ to $z_{j+1}$, and all other actions vanish. Any nonzero scalar choices are absorbed by rescaling this basis. This determines the module solely by its top and length.

If $r=1$ and $q>1$, then $m>1$ and $v$ is the exceptional leaf. The successor is a single loop and its action is the unique nilpotent Jordan chain of length $q$. The other branch either enters a different edge or is a redundant socle loop and contributes nothing. Hence the prescribed uniserial module is unique up to isomorphism for every positive integer $q\le mr$.