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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 $k{C}_G(P)$.

The centralizer condition defining block induction refers to a defect group of a block of the actual source block. Containment of the centralizer of an unrelated p-subgroup does not suffice. For example, in characteristic $3$ take $G=A_5$, $H=A_4$ and $D=C_3$. Then $D{C}_G(D)\le H$, but the unique defect-zero block of $k{A}_4$ occurs on bimodule restriction of both three-dimensional defect-zero blocks of $k{A}_5$. The two ordinary degree-three characters have identical restriction to $A_4$, namely its degree-three character, and defect-zero reduction gives the same simple projective module. Thus the unique-block characterization fails if $D$ is not the defect group of the source block.

If a block of a group algebra $b$ of $kH$ has defect group of a block $D$ and $D{C}_G(D)\le H\le G$, then its block induction $b^G$ is the principal block if and only if $b$ is the principal block. Equivalently, the Brauer morphism of the principal block idempotent at any p-subgroup $Q$ is the principal block idempotent of $k{C}_G(Q)$.

Let $P$ be any Sylow subgroup containing a defect group of a block $D$. Regard the block of a group algebra as a permutation module summand for $G\times G$, acting by left and right multiplication. Its vertex is $\Delta D$. Restricting to $P\times P$ retains a summand with this vertex. Transitive permutation modules for a finite p-group are indecomposable, with their point stabilizers as vertices. Thus some orbit in $G$ has stabilizer $\Delta D$ at a suitably chosen point $c$. This means $c$ centralizes $D$ and $P\cap {}^{c}P=D$. Choosing $P$ to contain a Sylow subgroup of $N_G(D)$ shows that $D=O_p(N_G(D))$; every normal p-subgroup of $G$ is contained in both Sylow factors and hence in $D$.

If $D$ is a normal p-subgroup of $N$, any indecomposable projective module of $k(N/D)$, inflated to $N$, is a summand of $k[N/D]={\mathrm{Ind}}_D^{N}k$. It has trivial source and vertex $D$. Its Brauer quotient of a module at $D$ is the entire module, since $D$ acts trivially and all proper-subgroup relative traces multiply by an index divisible by $p$. This prevents its vertex from being smaller than $D$.

The Frattini subgroup is the intersection of the maximal subgroups of $G$, with $\Phi (1)=1$. A useful equivalent property isIndeed, a proper subgroup $H$ lies in a maximal subgroup, which also contains $\Phi (G)$. The Frattini subgroup is a characteristic subgroup, since automorphisms permute the maximal subgroups.

The Fitting subgroup is the largest nilpotent normal subgroup. To prove its existence, first form the p-core $O_p(G)$: the product of all normal p-subgroups. For two such subgroups, their product is normal and has order $|AB|=|A||B|/|A\cap B|$, a power of $p$. There are only finitely many subgroups, so $O_p(G)$ exists. For distinct primes $p,r$,Thus the product of the p-cores is a nilpotent normal subgroup, a direct product of groups of its Sylow subgroups. Conversely, a finite nilpotent normal subgroup $K$ has unique Sylow subgroups, each a characteristic subgroup of $K$. Each is normal in $G$ and lies in the corresponding p-core. Consequently the required largest subgroup exists, and

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.

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 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.

The largest normal p-subgroup of a finite group. Products of normal p-subgroups are again normal p-subgroups, since $|AB|=|A||B|/|A\cap B|$. Thus the subgroup generated by all such subgroups is itself a p-subgroup. It is a characteristic subgroup.