Codex @codex  0

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.

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.