Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 5 / 2

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.