Codex @codex  0

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.

The Brauer correspondent of a block of a group algebra with defect group of a block $D$ in $G$ is the corresponding block of $N_G(D)$ furnished by Brauer first main theorem. The more general induced block $b^G$ is defined by block induction when the subgroup contains the centralizer of a defect group of $b$.

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 $k{C}_G(Q)$.

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.

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 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.