Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 3 (Modular Representations of Finite Groups)

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 Brauer first main theorem says that, for a fixed p-subgroup $D$ and $N=N_G(D)$, there is a bijection between blocks of a group algebra of $kG$ with defect group of a block $D$ and blocks of $kN$ with that defect group. Correspondents have the same nonzero image under ${\mathrm{Br}}_D$. The integral version follows by modular reduction of block idempotents.

Here is a proof using relative traces. Put $C=C_G(D)$ andThe latter ambient algebra is commutative because invariance under $N$ includes invariance under $C$. The Brauer morphism and relative trace identity isIndeed $D$ acts on $G/D$. Its fixed cosets are exactly $N/D$. On a nonfixed orbit the terms have the same Brauer image, so the orbit contributes zero in characteristic $p$. The fixed cosets give the right-hand side. Since every element of $kC$ is $D$-fixed, this proves that ${\mathrm{Br}}_D:I_D^G\twoheadrightarrow J_D$ is a surjective algebra homomorphism of commutative ideals. The same argument for $N$ has exactly the same target.

We use the following elementary fact about primitive idempotents under a surjection of commutative Artinian ideals. Under a surjection $f:I\twoheadrightarrow J$ of ideals in finite-dimensional commutative algebras, the primitive ambient idempotents lying in $I$ and having nonzero image correspond bijectively to the primitive idempotents in $J$. To prove it, decompose the source ambient algebra into Artinian local rings $A_i$. Either $I\cap A_i=A_i$, or it is a proper, hence nilpotent, ideal. Proper components contribute no idempotents to the image. A surviving identity $e_i$ has corner $f(e_i)J=f(A_i)$, a local quotient of $A_i$, so is primitive. The surviving identities are orthogonal and exhaust the idempotents of the target. Notice that this proof does not assert that the whole kernel is nilpotent: it can kill entire block components.

By the trace criterion for defect groups proved above, primitive block idempotents in $I_D^G$ have defect conjugate into $D$. Among them, the Brauer morphism is nonzero precisely for defect $D$, again by that criterion and its Brauer characterization. The identical reasoning applies to $I_D^N$. Thus both sets are identified with the primitive idempotents of $J_D$, proving the bijection and its uniqueness.

For the more general subgroup, $D$ is understood to be a defect group of a block $b$ of $kH$. This is the implicit hypothesis needed to apply the theorem; it is not repeated in the printed sentence. Without it the printed uniqueness assertion is false: take $G=A_5$, $H=A_4$, $p=3$ and $D=C_3$. The centralizer of $D$ in $A_5$ is $D$, but the unique three-dimensional defect-zero block of $k{A}_4$ is the bimodule restriction of each of the two three-dimensional defect-zero blocks of $k{A}_5$. The two ordinary degree-three characters both restrict to the degree-three character of $A_4$, with values $3,-1,0$ on its identity, involutions and elements of order three. Defect-zero reduction gives the same simple projective $k{A}_4$-module, so both restricted block bimodules are this same matrix block. This illustrates why block induction requires the block's defect centralizer. Write $e_b$ for its block idempotent and $K=N_H(D)$. Since $D{C}_G(D)\le H$, we have $D{C}_G(D)\le K\le N_G(D)$. Apply the theorem inside $H$ to obtain the unique corresponding block $\beta$ of $kK$. Its center of an associative algebra has a residue field $F_{\beta }$, and the quotient map is its central character of a block ${\omega }_{\beta }:Z(kK)\to F_{\beta }$. No assumption that $k$ is algebraically closed is needed here. DefineThe Brauer image is in $(k{C}_G(D){)}^{N_G(D)}\subseteq Z(kK)$, so this is a unital algebra homomorphism. Among the orthogonal primitive central idempotents of $kG$, exactly one has image $1$. The block it selects is $b^G$. If $H$ contains $N_G(D)$ this is the inverse of the bijection just proved. For arbitrary $H$, the resulting block can have a strictly larger defect group. Block induction is transitive on admissible intermediate subgroups. For the normalizer chain used in the proof below, the required transitivity will be verified directly with Brauer images.

To prove the bimodule characterization, we explicitly state the standard permutation-module fact used here. For a permutation module, the Brauer quotient of a module at $P$ has as basis the permutation-basis points fixed by $P$. An indecomposable summand has nonzero quotient at $P$ exactly when its vertex of an indecomposable module contains a conjugate of $P$. For summands with vertex exactly $P$, taking this quotient, with its $N(P)/P$-action, detects the summand and detects which orthogonal idempotent projection acts as its identity. These facts hold for trivial source modules; the fixed-basis description follows by deleting proper transfers, and the detection assertion is the permutation form of the first-main-theorem argument, applied to their endomorphism algebra.

As an $H\times H$ permutation module, with $(h_1,h_2)g=h_{1}g{h}_2^{-1}$, the group algebra decomposes asThe block bimodule $b$ is an indecomposable module with vertex of an indecomposable module $\Delta D=\{(d,d):d\in D\}$. None of the other double-coset modules can contain $b$: a basis element fixed by $\Delta D$ centralizes $D$, hence lies in $C_G(D)\subseteq H$, whereas those double cosets lie outside $H$. Thus $b$ occurs exactly once as an indecomposable summand of this restriction, in the $kH$ term. Decomposing $kG$ into its global block bimodules therefore puts this occurrence in precisely one of them, say $B$.

Under the fixed-basis description, the quotient of $kG$ at $\Delta D$ is $k{C}_G(D)$, and multiplication by $e_B$ induces multiplication by ${\mathrm{Br}}_D(e_B)$. The quotient of $b$ is ${\mathrm{Br}}_D^H(e_b)k{C}_G(D)=e_{\beta }k{C}_G(D)$. Consequently the detecting projection for this occurrence of $b$ satisfieswhich is equivalent to ${\omega }_b^G(e_B)=1$. Thus $B=b^G$, proving the requested unique-block direct-summand characterization. Here the restriction is to $H\times H$; the extra subscript in the converted TeX is a transcription error, not a different subgroup.

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

The Nagao version of Brauer second main theorem is the following module statement. Let $e$ be a central idempotent of $kG$, let $P$ be a p-subgroup, and suppose $C_G(P)\le K\le N_G(P)$. If $eM=M$, put $b={\mathrm{Br}}_P(e)$, which is central in $kK$. Thenand every indecomposable summand of the second term has vertex not containing $P$. More precisely it is relatively projective for some $p$-subgroup of $K$ not containing $P$. In particular the vertex-$P$ Green correspondent lies in the block selected by the Brauer morphism. This is the precise restriction statement needed below.

Let $B$ have defect group of a block $D$, and let $b$ be its Brauer correspondent in $N=N_G(D)$. We first build a trivial-source module in $b$. The exact functoris nonzero on the trivial $kD$-module. Otherwise it would vanish on every $kD$-module, since a group algebra of a p-group in characteristic p is local and all composition factors are trivial; but $F(kD)=b\ne 0$. Choose an indecomposable summand $V$ of $F(k)$. Since $D◃N$, its restriction to $D$ is a nonzero direct sum of trivial modules. It is $D$-projective. It cannot have a proper vertex $Q<D$: restriction would make a trivial $D$-module relatively $Q$-projective, whereas on its endomorphisms the trace from $Q$ is multiplication by $[D:Q]=0$ in $k$, contrary to the D. Higman criterion. Hence $V$ has vertex $D$ and trivial source.

Take its Green correspondent $M$ in $G$. To verify that $M$ is in the intended block rather than just some block with a large defect, let $e^{\prime }$ be the global block idempotent acting on $M$. Nagao's theorem puts the vertex-$D$ summand $V$ of $M{\downarrow }_N$ in ${\mathrm{Br}}_D(e^{\prime })$. Thus $e_b{\mathrm{Br}}_D(e^{\prime })\ne 0$. The first main theorem already gives $e_b={\mathrm{Br}}_D(e_B)$; orthogonality of Brauer images of different global block idempotents forces $e^{\prime }=e_B$. Green correspondence preserves sources, so $B$ contains an indecomposable module with vertex $D$ and trivial source. The same argument proves the same-block compatibility of Green correspondence for every vertex-$D$ module used below.

For finite representation type, first suppose $D$ is cyclic of order $q$. Write $D=⟨g⟩$ and $t=g-1$. Then $kD\cong k[t]/(t^q)$, and the indecomposable modules for a cyclic p-group are $k[t]/(t^j)$, $1\le j\le q$, by Jordan normal form. Every indecomposable module in $B$ is $D$-projective by Question 1, so is a summand of an induction of one of these finitely many indecomposable $kD$-modules. Each induction has finitely many indecomposable summands. Therefore $B$ has finite representation type.

For the converse it is important to obtain infinitely many modules of full vertex $D$, and not silently assume that the field is infinite. If $D$ is noncyclic, its quotient by the Frattini subgroup is elementary abelian of rank at least two, so $D$ maps onto $C_p\times C_p$. A rank-one Frattini quotient would make any lift of its generator generate $D$, since every proper subgroup is contained in a maximal subgroup. Here is an explicit family of full-vertex modules for a noncyclic p-group valid over every field $k$. Set $x=g-1$, $y=h-1$ for the two quotient generators. Onput $x{v}_0=0$, $x{v}_i=w_i$ for $1\le i\le r$, $y{v}_i=w_{i+1}$ for $0\le i<r$, $y{v}_r=0$, and $x{w}_i=y{w}_i=0$. All products of $x,y$ are zero, so the relations $x^p=y^p=0$ and $xy=yx$ hold, including when $p=2$.

These modules are indecomposable. Their common kernel and radical are $W=⟨w_i⟩$. An endomorphism preserves $W$; let $A$ be its induced matrix on $L_r/W$ and $C$ its matrix on $W$. The equations $CX=XA$, $CY=YA$ for the two displayed shift maps force $A$ and $C$ to be the same scalar: the first and last columns give the boundary zeroes, and the recurrence $a_{j,i+1}=a_{j-1,i}$ forces all off-diagonal entries to vanish and all diagonal entries to agree. The remaining endomorphisms map the top into $W$ and vanish on $W$, so form a square-zero ideal. Thus the endomorphism ring is local. Inflate $L_r$ to $D$ and choose the infinitely many $r$ for which $p\nmid 2r+1$. Such an $L_r$ has vertex $D$: if $1={\mathrm{Tr}}_Q^D(\alpha )$ for a proper $Q<D$, taking ordinary matrix traces would givea contradiction. Their distinct dimensions ensure distinct isomorphism classes.

Each $F(L_r)=e_b{\mathrm{Ind}}_D^N{L}_r$ is nonzero, by exactness and the already proved $F(k)\ne 0$. Its restriction to $D$ is a summand of a sum of conjugates of $L_r$, by normality of $D$ and Mackey decomposition. Consequently every indecomposable summand chosen from $F(L_r)$ has vertex $D$, and has a source conjugate to $L_r$. If $b$ contained only finitely many indecomposables, their restrictions to $D$ would contain only finitely many possible sources, contradicting this family. Green correspondence and Nagao's theorem then transfer infinitely many vertex-$D$ indecomposables in $b$ to distinct indecomposables in $B$. We have proved

Finally, for $C_p=⟨g⟩$, the algebra $k{C}_p=k[t]/(t^p)$ has precisely the indecomposablesOn $M_j$, $g$ acts as a single size-$j$ Jordan block with eigenvalue $1$. Its submodules are exactly its ideals, hence the powers of $t$: an ideal has a generator of least degree, and multiplying that generator by a unit makes it a power of $t$. Therefore its unique composition series of a module iswith $j$ trivial one-dimensional composition factors. This is uniqueness of the actual chain, stronger than uniqueness of the multiset of factors. The algebra is local, so its sole principal indecomposable module is its regular module $M_p=k{C}_p$, of dimension and composition length $p$.

The principal block $B_0(G)$ is the unique block of a group algebra containing the trivial $kG$-module. Its central character of a block is augmentation $\epsilon :Z(kG)\to k$, and its block idempotent $e_0$ is the unique primitive central idempotent with $\epsilon (e_0)=1$. Its defect groups of a block are Sylow subgroups: if $e_0={\mathrm{Tr}}_D^G(a)$, augmentation gives $1=[G:D]\epsilon (a)$, so $p\nmid [G:D]$.

The Brauer third main theorem states that, if $b$ is a block of $kH$ with defect group $D$ and $D{C}_G(D)\le H\le G$, thenWe give a proof, including the normal-subgroup ingredient needed for the difficult implication.

First, block idempotents centralize a normal p-subgroup. If $P◃L$ is a p-subgroup, the ideal $J(kP)kL$ is nilpotent: conjugation preserves $J(kP)$, so its powers are $J(kP{)}^{m}kL$, eventually zero. For a central idempotent $e$ of $kL$, the difference $e-{\mathrm{Br}}_P(e)$ is a sum of nonfixed $P$-conjugation orbit sums. Such an orbit maps to a single element of $L/P$ multiplied by its size, which is zero in $k$. Thus the difference belongs to $J(kP)kL$. Since $P$ is normal, its centralizer is normal too, and the Brauer image is $L$-invariant; consequently it is itself a central idempotent of $kL$. Commuting central idempotents congruent modulo a nilpotent ideal coincide: their off-diagonal products are nilpotent idempotents, hence zero. ThereforeIt follows that central block idempotents of $kL$ are exactly the sums over $L$-orbits of primitive central idempotents of $k{C}_L(P)$. The principal block idempotent of $k{C}_L(P)$ is invariant under every automorphism preserving augmentation, so its singleton orbit is the principal block idempotent of $kL$.

In particular, if $P{C}_G(P)\le K\le L\le N_G(P)$, the groups $K,L$ have the same centralizer $C_G(P)$. Their principal block idempotents are both the principal block idempotent of $k{C}_G(P)$. By the definition using the central character of a block of block induction, a block of $K$ inducing to the principal block of $L$ must be principal: its idempotent, a sum of centralizer block idempotents, has nonzero overlap with this single principal idempotent, and is therefore precisely that idempotent. The forward implication also holds, by the same observation.

For an arbitrary subgroup $H$ in the theorem, use the Brauer first main theorem inside $H$ to replace $b$ by its correspondent $\beta$ in $K=N_H(D)$. This preserves principalness. Indeed a Brauer image preserves augmentation on central elements, since nonfixed $D$-orbits have sizes divisible by $p$; hence the principal block of $H$, when it has defect $D$, corresponds to the principal block of $K$. Conversely, if $\beta$ is principal, its induced central character of a block isso ${\beta }^H$ is principal. Transitivity gives ${\beta }^G=b^G$. Thus we may work with $D{C}_G(D)\le K\le N_G(D)$ and a block $\beta$ of defect $D$.

The forward implication is already proved by this augmentation calculation, now with $H$ replaced by $G$. For the converse, fix $G$ and suppose there is a counterexample with $\beta$ nonprincipal but ${\beta }^G$ principal. Choose one whose defect group $D$ has maximal order. Put $N=N_G(D)$ and $c={\beta }^N$. To verify the transitivity needed here, write $B={\beta }^G$. Its defining character gives $e_{\beta }{\mathrm{Br}}_D(e_B)=e_{\beta }$. The image ${\mathrm{Br}}_D(e_B)$ is $N$-invariant and central in $kN$, so the definition of $c$ gives $e_c{\mathrm{Br}}_D(e_B)=e_c$. If $E\ge D$ is a defect group of $c$, applying ${\mathrm{Br}}_E$ gives ${\mathrm{Br}}_E(e_c){\mathrm{Br}}_E(e_B)={\mathrm{Br}}_E(e_c)$, since $C_G(E)\subseteq C_G(D)$. The character defining $c^G$ therefore selects $e_B$. Thus $c^G={\beta }^G$. The normal-subgroup observation shows that $c$ is nonprincipal, since otherwise $\beta$ would be principal. Every defect group of $c$ contains the normal $p$-subgroup $D$ up to conjugacy: ${\mathrm{Br}}_D(e_c)=e_c\ne 0$. Choose a defect group $E\ge D$ of $c$. Then $C_G(E)\le C_G(D)\le N$, so $(G,N,c)$ is another admissible counterexample, now with defect group $E$. Maximality forces $E=D$.

The first main theorem applied to $N=N_G(D)$ therefore implies that the principal block $c^G$ has defect $D$. Thus $D$ is a Sylow subgroup of $G$, and also of $N$. The principal block of $N$ then has defect $D$ and, by the forward implication, induces to the principal block of $G$. Injectivity in the first main theorem gives $c=B_0(N)$, contradicting the established nonprincipalness of $c$. This proves the third main theorem.

A useful equivalent consequence, with its justification, isLet $a$ be a block idempotent of $k{C}_G(Q)$ occurring in the left side, and let $E$ be a defect group of $a$ containing the central $p$-subgroup $Q$. Then $C_G(E)\le C_G(Q)$, so $a^G$ is defined. Applying ${\mathrm{Br}}_E$ to $a{\mathrm{Br}}_Q(e_0)=a$ shows that the central character of a block defining $a^G$ selects $e_0$: deleting coefficients outside $C_G(E)$ can be done after deleting those outside $C_G(Q)$. Hence $a^G=B_0(G)$, and the theorem makes $a$ principal. The image is nonzero because its augmentation is $1$, so it is exactly the principal centralizer idempotent.

Now put $C=C_G(x)$ and $P=O_p(C)$, the p-core of a finite group. The hypothesis $O_{p^{\prime }}(C)=1$, using the p-prime core of a finite group, and the definition of a p-constrained group say $C_C(P)\le P$. The normal-subgroup observation puts every block idempotent of $kC$ inside $k{C}_C(P)\subseteq kP$. A group algebra of a p-group in characteristic p is local, so its only idempotents are $0,1$. Thus $kC$ has a single block, its principal block.

Write $Q=⟨x⟩$. By the trace and Brauer characterization of defect in Question 1,For the forward direction one can conjugate $Q$ into $D$; the nonzero image at $D$ then ensures the image at $Q$ is nonzero. The reverse direction follows because every subgroup with nonzero image is conjugate into a defect group. Since $kC$ has only one block, a nonzero image must equal $1=e_0(C)$. The preceding local-block argument then gives $B=B_0(G)$ by the third main theorem. Conversely, if $B$ is principal, its defect group is a Sylow subgroup and every $p$-element is conjugate into it. ThereforeThis includes $x=1$: in that case the hypotheses force $kG$ itself to have only its principal block.

Choose a maximal Brauer pair $(D,b_D)$ for $B$, and let $N_G(D,b_D)$ be the subgroup normalizing $D$ and stabilizing the local block $b_D$. The inertial quotient of a block and its order, the inertial index of a block, areThe Brauer extended first main theorem says that maximal local blocks of defect $D$ are parametrized by the normalizer orbits of defect-$D$ blocks of $k(D{C}_G(D))$, with stabilizer quotient of order prime to $p$. For abelian $D$, conjugation embeds $E(B)$ in $\mathrm{Aut}(D)$. When $D=C_{p^n}$, its prime-to-$p$ subgroups are cyclic of order dividing $p-1$, so $e\mid p-1$.

We use two precise structural results for a block with cyclic defect group, with $k$ an algebraically closed field. First, the basic algebra is a Brauer tree algebra with $e$ edges and exceptional multiplicity $m=(p^n-1)/e$; its simple modules are indexed by its edges. Second, its stable Auslander–Reiten quiver has shapeThe vertices are the isomorphism classes of nonprojective indecomposable modules. These are the cyclic-defect structure and indecomposable-classification results, rather than a claim that all Brauer trees are Morita equivalent to a star. The tube description is also recorded in Hiss–Lassueur, Section 1.

The first result gives $l(B)=e$. In the second, there are $p^n-1$ positions in each slice and $e$ slices modulo the translation. Thus there are $e(p^n-1)$ nonprojective indecomposables. A finite-dimensional split algebra has one principal indecomposable module that is a projective cover for each simple module, so there are another $e$ projective indecomposables. ThereforeThe total includes projectives; it is different from the count of ordinary irreducible characters. This is the indecomposable count for a cyclic defect block.

Now let $p=2$ and $D=C_2\times C_2$, the Klein four-group. The extended theorem makes $E(B)$ a subgroup of order prime to $2$ of $\mathrm{Aut}(D)\cong {\mathrm{GL}}_2({\mathbb{F}}_2)\cong S_3$. Its order divides $6$ and is odd, soFor the remaining conclusion we use the central-defect quotient fact: if $Z\le Z(L)$ is a central $p$-subgroup, blocks of $kL$ correspond to blocks of $k(L/Z)$ under the quotient, and their defect groups quotient by $Z$. Central block idempotents lift uniquely here because the kernel is generated by central nilpotent elements; an off-diagonal corner lies in its own product with that nilpotent ideal and is therefore zero. Defect groups quotient by $Z$ by the relative-trace, or equivalently Brauer-pair, characterization. In particular a block with central defect group $D$ has a defect-zero block as its quotient by $J(kD)$.

First suppose $D\le Z(G)$, and write $A=B$. Since $k$ is an algebraically closed field, the defect-zero quotient is $\overline{A}\cong {\mathrm{Mat}}_r(k)$ for some $r$. The algebra $A$ is a direct summand of $kG$ as a $kD$-module, hence is projective and therefore free over the local ring $kD$. Its rank is $r^2$, since reduction modulo $J(kD)$ gives the $r^2$-dimensional quotient. Lift a primitive idempotent $\bar{f}$ of $\overline{A}$ to $f\in A$. The corner $fAf$ is a $kD$-linear direct summand of $A$ and is free over $kD$. Its quotient by $J(kD)$ is $\bar{f}\overline{A}\bar{f}\cong k$, so its rank is one. The central mapis surjective by Nakayama's lemma, since $f$ generates its reduction, and is injective because both sides have dimension $|D|$. Thus $fAf\cong kD$ as algebras. Orthogonal primitive idempotents and their equivalences in the matrix quotient lift modulo the nilpotent ideal; lifting the equivalence maps and correcting their products by units lifts the matrix units. ConsequentlyThis proves the actual algebra isomorphism of central defect blocks are matrix algebras, as an algebra.

Finally suppose $D◃G$ but is not central. Put $C=C_G(D)$, which is normal in $G$ and contains $D$ since $D$ is abelian. The normal-subgroup argument in Question 5 puts the block idempotent $e_B$ in $kC$. It is therefore the sum of the $G$-conjugates of a block idempotent $c$ of $kC$. The extended first main theorem chooses such a $c$ with defect group $D$; since $D$ is normal, its pair stabilizer is exactly $I=\{g\in G:{}^{g}c=c\}$. Nowso $I=C$. Within $C$, the defect group $D$ is central, and the first case gives $kCc\cong {\mathrm{Mat}}_r(kD)$.

We can see the Fong–Reynolds matrix decomposition explicitly in this case. Choose representatives $g_1,\dots ,g_t$ for $G/C$, where $t=[G:C]$, and putDifferent conjugates of $c$ are orthogonal, so $u_{ij}{u}_{\ell m}={\delta }_{j\ell }{u}_{im}$ and ${\sum }_i{u}_{ii}=e_B$. Moreover $cBc=ckGc=kCc$: a term $ckCgc$ vanishes unless $g$ stabilizes $c$, and that stabilizer is $C$. These matrix units therefore identify $B$ with ${\mathrm{Mat}}_t(kCc)$, whenceThe inertial index condition concerns the stabilizer of the local block; it does not require every element of $G$ to centralize $D$. That distinction is why the second case is necessary.