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Let $Z=⟨z⟩\le Z(G)$ have order $p$ and set $t=z-1$. The nilpotent ideal $tkG$ is generated by a central element. If an idempotent lifts a central idempotent of the quotient, its off-diagonal corner satisfies $eA(1-e)\subseteq teA(1-e)$ and therefore is zero. Thus central block idempotents lift centrally and uniquely. A principal indecomposable module is free over $kZ$, and the $p$ factors of its $t$-filtration are all its reduction modulo $t$. Consequently, under the natural indexing of simple modules, $C_G=p{C}_{G/Z}$.

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 $p$.

Choose a maximal Brauer pair $(D,b_D)$ for $B$, and let $N_G(D,b_D)$ be the subgroup normalizing $D$ and stabilizing the local block $b_D$. The inertial quotient of a block and its order, the inertial index of a block, areThe Brauer extended first main theorem says that maximal local blocks of defect $D$ are parametrized by the normalizer orbits of defect-$D$ blocks of $k(D{C}_G(D))$, with stabilizer quotient of order prime to $p$. For abelian $D$, conjugation embeds $E(B)$ in $\mathrm{Aut}(D)$. When $D=C_{p^n}$, its prime-to-$p$ subgroups are cyclic of order dividing $p-1$, so $e\mid p-1$.

We use two precise structural results for a block with cyclic defect group, with $k$ an algebraically closed field. First, the basic algebra is a Brauer tree algebra with $e$ edges and exceptional multiplicity $m=(p^n-1)/e$; its simple modules are indexed by its edges. Second, its stable Auslander–Reiten quiver has shapeThe vertices are the isomorphism classes of nonprojective indecomposable modules. These are the cyclic-defect structure and indecomposable-classification results, rather than a claim that all Brauer trees are Morita equivalent to a star. The tube description is also recorded in Hiss–Lassueur, Section 1.

The first result gives $l(B)=e$. In the second, there are $p^n-1$ positions in each slice and $e$ slices modulo the translation. Thus there are $e(p^n-1)$ nonprojective indecomposables. A finite-dimensional split algebra has one principal indecomposable module that is a projective cover for each simple module, so there are another $e$ projective indecomposables. ThereforeThe total includes projectives; it is different from the count of ordinary irreducible characters. This is the indecomposable count for a cyclic defect block.

Now let $p=2$ and $D=C_2\times C_2$, the Klein four-group. The extended theorem makes $E(B)$ a subgroup of order prime to $2$ of $\mathrm{Aut}(D)\cong {\mathrm{GL}}_2({\mathbb{F}}_2)\cong S_3$. Its order divides $6$ and is odd, soFor the remaining conclusion we use the central-defect quotient fact: if $Z\le Z(L)$ is a central $p$-subgroup, blocks of $kL$ correspond to blocks of $k(L/Z)$ under the quotient, and their defect groups quotient by $Z$. Central block idempotents lift uniquely here because the kernel is generated by central nilpotent elements; an off-diagonal corner lies in its own product with that nilpotent ideal and is therefore zero. Defect groups quotient by $Z$ by the relative-trace, or equivalently Brauer-pair, characterization. In particular a block with central defect group $D$ has a defect-zero block as its quotient by $J(kD)$.

First suppose $D\le Z(G)$, and write $A=B$. Since $k$ is an algebraically closed field, the defect-zero quotient is $\overline{A}\cong {\mathrm{Mat}}_r(k)$ for some $r$. The algebra $A$ is a direct summand of $kG$ as a $kD$-module, hence is projective and therefore free over the local ring $kD$. Its rank is $r^2$, since reduction modulo $J(kD)$ gives the $r^2$-dimensional quotient. Lift a primitive idempotent $\bar{f}$ of $\overline{A}$ to $f\in A$. The corner $fAf$ is a $kD$-linear direct summand of $A$ and is free over $kD$. Its quotient by $J(kD)$ is $\bar{f}\overline{A}\bar{f}\cong k$, so its rank is one. The central mapis surjective by Nakayama's lemma, since $f$ generates its reduction, and is injective because both sides have dimension $|D|$. Thus $fAf\cong kD$ as algebras. Orthogonal primitive idempotents and their equivalences in the matrix quotient lift modulo the nilpotent ideal; lifting the equivalence maps and correcting their products by units lifts the matrix units. ConsequentlyThis proves the actual algebra isomorphism of central defect blocks are matrix algebras, as an algebra.

Finally suppose $D◃G$ but is not central. Put $C=C_G(D)$, which is normal in $G$ and contains $D$ since $D$ is abelian. The normal-subgroup argument in Question 5 puts the block idempotent $e_B$ in $kC$. It is therefore the sum of the $G$-conjugates of a block idempotent $c$ of $kC$. The extended first main theorem chooses such a $c$ with defect group $D$; since $D$ is normal, its pair stabilizer is exactly $I=\{g\in G:{}^{g}c=c\}$. Nowso $I=C$. Within $C$, the defect group $D$ is central, and the first case gives $kCc\cong {\mathrm{Mat}}_r(kD)$.

We can see the Fong–Reynolds matrix decomposition explicitly in this case. Choose representatives $g_1,\dots ,g_t$ for $G/C$, where $t=[G:C]$, and putDifferent conjugates of $c$ are orthogonal, so $u_{ij}{u}_{\ell m}={\delta }_{j\ell }{u}_{im}$ and ${\sum }_i{u}_{ii}=e_B$. Moreover $cBc=ckGc=kCc$: a term $ckCgc$ vanishes unless $g$ stabilizes $c$, and that stabilizer is $C$. These matrix units therefore identify $B$ with ${\mathrm{Mat}}_t(kCc)$, whenceThe inertial index condition concerns the stabilizer of the local block; it does not require every element of $G$ to centralize $D$. That distinction is why the second case is necessary.

Write $A=RG$. An idempotent is an element $e$ with $e^2=e$; it is central idempotent when it commutes with every element of the ring. In a commutative ring centrality is automatic. Two idempotents are orthogonal idempotents if $ef=fe=0$. A nonzero primitive central idempotent cannot be expressed as the sum of two nonzero orthogonal central idempotents; this is primitivity in $Z(A)$, not necessarily primitivity in $A$.

A block of a group algebra is an indecomposable two-sided direct-summand ideal of $A$, equivalently an ideal $Ae$ with $e$ a primitive central idempotent. Here is the equivalence. If $A=B\oplus C$ as two-sided ideals, then $BC=CB=0$. Write $1=e+f$ with $e\in B,f\in C$. Multiplication by this equality shows that $e$ is the identity of $B$, that $f$ is the identity of $C$, and that $e,f$ are orthogonal central idempotents. Conversely a central idempotent gives $A=Ae\oplus A(1-e)$. Splitting the ideal further is therefore equivalent to splitting its central idempotent further.

For uniqueness, suppose $1={\sum }_i{e}_i={\sum }_j{f}_j$ are decompositions into primitive central idempotents. For each $i$,is an orthogonal central-idempotent decomposition. Exactly one summand is nonzero, and that summand equals $e_i$. Applying the same reasoning to the corresponding $f_j$ gives $e_i=f_j$. Thus the blocks, and their identities, are unique up to ordering. Existence in the paper's coefficient rings follows from finite-dimensionality over $k$, and over $\mathcal{O}$ from completeness and idempotent lifting of the decomposition modulo its maximal ideal.

Over $k$, a principal indecomposable module is an indecomposable summand of the regular left module $A=kG$, equivalently $P=Ae$ for a primitive idempotent $e$ of $A$. Put $J=J(A)$, the Jacobson radical. The head of a module $P/JP$ is nonzero and semisimple. It is simple: projectivity gives a surjectionwhose kernel consists of maps into $JP$ and is nilpotent, since a product of $r$ such maps sends $P$ into $J^{r}P$. If the head of a module split, a nontrivial projection on it would lift by idempotent lifting to a nontrivial projection on $P$, contradicting indecomposability. Every maximal submodule contains $JP$, since $J$ annihilates simple modules. The simplicity of the head of a module now proves $J(P)=JP$ is the unique maximal submodule, and $P$ is its head of a module's projective cover.

Choose the simple modules $S_1,\dots ,S_r$, with corresponding projective covers $P_1,\dots ,P_r$. Our convention for the Cartan matrix of a group algebra iswhere brackets denote the multiplicity in a composition series.

For the final two unheaded requests, retain the additional hypothesis $G=N{C}_G(N)$. Choose a central subgroup $Z=⟨z⟩$ of $N$ of order $p$. Every element of $N$ centralizes $Z$, and so does every element of $C_G(N)$; hence $Z\le Z(G)$. Set $t=z-1$, so $t$ is central and $t^p=0$, and $A/tA\cong k(G/Z)$.

To lift a central idempotent $\bar{e}$ of this quotient, first lift it as an idempotent $e$ by idempotent refinement theorem. Since $\bar{e}$ is central, $eA(1-e)\subseteq tA$. Multiplying a representation $ea(1-e)=tb$ on the left and right by $e,1-e$ givesThe other off-diagonal corner vanishes likewise, so $e$ is central. Two central lifts of the same idempotent coincide: their mismatched products $e(1-f)$ and $f(1-e)$ are idempotents in the nilpotent kernel and therefore zero. Primitivity is preserved because orthogonal idempotent decompositions lift. Thus the central $p$-quotient gives a bijection of block idempotents.

The quotient $G/Z$ still satisfies the required hypothesis for $N/Z$, because the image of $C_G(N)$ centralizes $N/Z$. Induction on $|N|$ therefore proves $\tau$ bijects the block idempotents of $kG$ and $k(G/N)$. The centrality argument is essential: a general nilpotent algebra quotient need not give a bijection of central block idempotents.

For the Cartan matrix of a group algebra, every projective $A$-module is free over $kZ$: $A$ is free over $kZ$, a projective $A$-module is a summand of a finite free $A$-module, and finite projectives over the local ring $kZ\cong k[t]/(t^p)$ are free. If $P$ is a principal indecomposable module, $\bar{P}=P/tP$ is a principal indecomposable module over $A/tA$: an idempotent splitting of its endomorphism ring would lift and split $P$. The $p$ quotients of its filtrationare all isomorphic to $\bar{P}$ as $k(G/Z)$-modules, by multiplication by the corresponding power of $t$. All simple modules are annihilated by the nilpotent ideal $tA$. Hence $[P:S]=p[\bar{P}:S]$, with the natural indexing of simple modules, and $C_G=p{C}_{G/Z}$. Iterating over $N$ givesThis is blocks and Cartan matrices under central p-quotients. The extra hypothesis is still in force: for example $N=C_3⊴S_3$ in characteristic three does not satisfy it, and its Group algebra Cartan matrix is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ rather than $3I$.