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If $k$ has characteristic $p$, then $kG$ has finite representation type exactly when a Sylow p-subgroup of $G$ is cyclic. Restriction and induction reduce the property to the Sylow subgroup; a cyclic p-group has the finitely many indecomposables $k[u]/(u^r)$, while a noncyclic p-group has a quotient $C_p\times C_p$ and hence infinitely many indecomposables.

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.

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 $p$.

For a block idempotent $b$ with defect $D$, the Brauer quotient of a module of its modular block bimodule at $\Delta D$ is nonzero: on the fixed basis $C_G(D)$, the block projection acts by ${\mathrm{Br}}_D(b)$. Restriction to $D\times D$ splits into transitive permutation modules; their stabilizers have order at most $|D|$. A nonzero quotient at $\Delta D$ therefore forces a summand with stabilizer exactly $\Delta D$, which is the regular $kD$ bimodule. Tensoring on the right with any $kD$-module $U$ gives $U\mid {\mathrm{Res}}_D^G(b{\mathrm{Ind}}_D^{G}U)$. Consequently finite representation type of the modular block forces finite representation type of $kD$.