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Let $A$ be a finite-dimensional basic algebra over an algebraically closed field and a hereditary ring. The radicals of its indecomposable projectives are projective and satisfy $J{P}_i\cong {⨁}_j{P}_j^{a_{ij}}$, where $a_{ij}$ are the Ext quiver arrow multiplicities. Hence $\dim P_i=1+{\sum }_j{a}_{ij}\dim P_j$, forcing strict dimension decrease along arrows and therefore acyclicity. Lifting an arrow basis in $J/J^2$ gives a surjective path algebra homomorphism $kQ\to A$. The number of paths starting at $i$ satisfies the same dimension recursion. Equality of total dimensions proves the homomorphism is an isomorphism. Sums of vertex idempotents over connected components are precisely the primitive central idempotents, so components give the blocks.

For a normal p-subgroup $P◃G$, every central idempotent $e$ of $kG$ satisfies $e={\mathrm{Br}}_P(e)\in k{C}_G(P)$. Indeed a nonfixed conjugation orbit sum maps to zero in $k(G/P)$, so $e-{\mathrm{Br}}_P(e)$ belongs to the nilpotent ideal $J(kP)kG$. Normality makes the Brauer image a central idempotent of $kG$, and two central idempotents congruent modulo a nilpotent ideal coincide. Consequently the blocks of a group algebra of $kG$ correspond to $G$-orbits of blocks of $k{C}_G(P)$.

A block of an associative algebra with a finite decomposition of its identity is a two-sided direct summand $Ae$ corresponding to a primitive central idempotent $e$. Its identity is $e$, and it cannot be split into two nonzero algebras by central idempotents. For a group algebra of a finite group, this agrees with a block of a group algebra. Orthogonal primitive central idempotents give $A={⨁}_{i}A{e}_i$.

Over a field of characteristic three, the group algebra $A=k{S}_3$ has a single block of an Artinian algebra. Put $r=(123)$, $s=(12)$ and $a=r-1$. Then $a^3=0$, $sas=-a+a^2$, and $J(A)=aA$: the latter is a nilpotent ideal, with semisimple ring quotient $k{C}_2$. The center of an associative algebra is $k1\oplus k{a}^2\oplus k{a}^{2}s$, whose last two summands form a square-zero ideal. A central idempotent $d1+n$ satisfies $d^2=d$ and $(2d-1)n=0$, hence is zero or one. Thus no nontrivial central block decomposition exists.

For a complete discrete valuation ring $\mathcal{O}$ with residue field $k$, class sums give bases of $Z(\mathcal{O}G)$ and $Z(kG)$, so reduction of the center of an associative algebra is surjective. Idempotent lifting in this complete commutative algebra gives a bijection of primitive central idempotents, hence of blocks of a group algebra. Corresponding blocks of a group algebra have the same defect groups of a block: relative traces reduce, and a lifted trace whose reduction is the block identity is a unit in its complete block center.

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.

For a cell $(i,j)$ in the Young diagram of $\lambda$, its hook of a Young diagram consists of that cell, the cells to its right in row $i$ and the cells below it in column $j$. Its length isA removable rim hook is a connected border strip $\lambda /\mu$ containing no $2\times 2$ square, whose removal leaves a Young diagram. Its leg length is the number of rows it meets minus one. The hook graph is the Young diagram labeled at each cell by $h_{ij}$; here “graph” means this labeled array, not a graph of possible hook-walk moves.

Choose $l\ge \ell (\lambda )$, pad with zeros and form the beta numbersThey are distinct nonnegative integers in decreasing order and recover ${\lambda }_i={\beta }_i-l+i$. For $l=\ell (\lambda )$ they equal the first-column hook lengths, since $h_{i1}={\lambda }_i+l-i$. If $l$ is increased, old beta numbers shift upward by one and a new bead at zero is inserted.

For a prime-modulus partition abacus arrange the nonnegative positions in $p$ runners: position $pq+r$ lies at level $q$ on runner $r$, $0\le r<p$. Put beads at the beta numbers and leave the other positions empty. Replacing a bead at $b$ by a bead at a vacant $b-k\ge 0$ removes a rim hook of length $k$; its leg length is the number of beads strictly between these positions. This follows by tracing the boundary of the diagram: bead positions record vertical steps and gaps record horizontal steps, and exchanging one bead and gap removes the intervening border strip. In particular a removable $p$-hook is exactly a bead that can slide up one place on its own runner.

The partition core at modulus $p$ is the result of sliding beads up until there are no gaps above any bead on a runner. If runner $r$ initially has $b_r$ beads, its final occupied levels must be $0,1,\dots ,b_r-1$, independently of slide order. Thus its packed beta set, and hence its core partition of an integer, is unique. Increasing the number of beta numbers merely re-expresses the same boundary by shifting positions and adding the new initial bead; the packed configuration represents the same partition of an integer. This proves well-definedness both with respect to removal order and to the chosen padding. The number of slides isThe partition quotient components record how far each runner's beads lie above their packed positions, giving the hook-removal interpretation used in Question 5.

A block of a group algebra in characteristic $p$ for $F{S}_n$ is a two-sided ideal $B=F{S}_{n}e$ for a primitive central idempotent $e$, meaning a nonzero central idempotent that is not a sum of two nonzero orthogonal central idempotents. The identity is the sum of these block idempotents and the group algebra is their direct product. An indecomposable module $V$ belongs to $B$ when $eV=V$, equivalently every other block idempotent acts as zero. Indeed $V={⨁}_{e}eV$; indecomposability makes exactly one summand nonzero.

The Nakayama block theorem, historically called the Nakayama Conjecture, statesEquivalently the ordinary characters assigned to one modular block have a common partition core at modulus $p$. Its simple modules are the $D^{\lambda }$ with regular $\lambda$ having that core. Only cores $\kappa$ for which $n-|\kappa |$ is a nonnegative multiple of $p$ occur. The block weight is $(n-|\kappa |)/p$; a core alone determines the block at this fixed degree.

One route to the proof separates the combinatorial core invariant from the algebraic center invariant. Let $c_r(\lambda )$ be the number of cells of residue $r=j-i(\mathrm{mod}p)$. A length-$p$ rim hook has one cell of every residue: along its border strip the contents are consecutive integers. Therefore removal subtracts one from every $c_r$. More precisely, choose the number $l$ of beta numbers divisible by $p$ and write the runner charges as $q_r=b_r-l/p$. Reading the boundary, or adding a cell and checking its two changed beta positions, givesThe charges determine the packed runners and hence the core. For fixed $n$, two equal cores have the same weight, so their full residue multisets are equal. Conversely equal residue multisets give equal charges and the same core. This establishesfor partitions of an integer of the same integer. The fixed-degree condition matters.

For the center invariant, use the commuting Jucys–Murphy elements $J_k={\sum }_{i<k}(ik)$, with $J_1=0$. In the Young seminormal form basis, $J_k$ has eigenvalue equal to the content of the cell containing $k$. A symmetric polynomial in them therefore acts on $S^{\lambda }$ as the same scalar for every Young tableau: its value on the multiset of all contents of $\lambda$. The integral Specht lattice makes this a scalar identity before modular reduction, so modulo $p$ these central scalars depend only on the residue multiset, and every composition factor inherits them.

The substantive center theorem used in this proof is that the center of the symmetric-group algebra, integrally and over a field, consists of symmetric polynomials in the Jucys–Murphy elements. Centrality of elementary symmetric polynomials follows fromthe coefficients are central class-sum combinations. The reverse inclusion requires the integral triangular class-sum argument: expand symmetric monomials in the $J_k$ as products of transpositions, group the leading terms by reduced cycle type, and eliminate these leading class sums in the resulting unitriangular order. The integral coefficients, with no division by multiples of $p$, are crucial; knowing the center only over $\mathbb{Q}$ would not prove the modular assertion. This is the central algebraic ingredient of this proof route.

Thus equal residue multisets yield equal characters of the entire modular center. Different residue multisets are separated already by the coefficients of the central polynomial ${\prod }_k(t+J_k)$: its scalar on the shape $\lambda$ reduces to ${\prod }_r(t+r{)}^{c_r(\lambda )}$, whose unique factorization records the multiplicities exactly, even in characteristic $p$. In a finite-dimensional split algebra, two simple modules are in the same block exactly when their central characters agree: the center of each block is local, while different block idempotents separate different blocks. It follows that equal cores give one block and different cores give distinct blocks. This proves both directions rather than just observing that core is constant on known blocks.

Finally, each ordinary Specht lattice has a single reduced central character, so all constituents of its reduction lie in that one block. Question 1 supplies every simple module and its label; the regular labels therefore identify all the simples in each core block. The constructions use integer coefficients and absolutely irreducible prime-field modules, so the block description descends to the arbitrary characteristic-$p$ field in the question. This completes the main steps of the block proof without presupposing semisimplicity in positive characteristic.

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.

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 $3I$.

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.

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.

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.

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

For a finite-dimensional algebra over an algebraically closed field, basic algebra means $A/J(A)\cong k^n$, so its simple modules are one-dimensional and occur once in the semisimple quotient of the regular module. A hereditary ring has every submodule of a projective module projective; equivalently its global dimension is at most one.

Choose representatives $S_1,\dots ,S_n$ of the simple modules. Our Ext quiver convention gives one vertex $i$ for $S_i$ andWith primitive orthogonal idempotents $e_i$, put $P_i=A{e}_i$ and $J=J(A)$. Applying ${\mathrm{Hom}}_A(-,S_j)$ to $0\to J{P}_i\to P_i\to S_i\to 0$ givesThe restriction map from $\mathrm{Hom}(P_i,S_j)$ is zero since every map to a simple kills $J{P}_i$, and $P_i$ is projective. Thus the arrow multiplicity is $\dim e_j(J/J^2)e_i$. This specifies the orientation rather than silently replacing the quiver by its opposite.