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For a split block of a group algebra with cyclic defect group of a block of order $q=p^n$ and inertial index of a block $e$, its Brauer tree algebra has $e$ simple modules and exceptional multiplicity $(q-1)/e$. Its stable Auslander–Reiten quiver is $\mathbb{Z}{A}_{q-1}/⟨{\tau }^e⟩$, giving $e(q-1)$ nonprojective indecomposables. Adding the $e$ indecomposable projectives gives $eq$ indecomposables in total.

For a maximal Brauer pair $(D,b_D)$ of a block of a group algebra, the inertial quotient is the displayed quotient of its pair stabilizer by $D{C}_G(D)$. It has order coprime to $p$, and its order is the inertial index of a block. For abelian $D$, conjugation embeds it in $\mathrm{Aut}(D)$.

Choose a maximal Brauer pair $(D,b_D)$ for $B$. Define its inertial quotient of a block and inertial index of a block byConjugacy of maximal pairs makes the index well-defined. The standard maximal-pair property used here is that $E(B)$ is a $p^{\prime }$-group. Since $D$ is cyclic, its faithful conjugation action embeds it into $\mathrm{Aut}(D)$, so $e\mid p-1$. Write $d=p^n$ and $m=(d-1)/e$.

We use two precise structural results. Over an algebraically closed field, the basic algebra of a block with cyclic defect group is a Brauer tree algebra with $e$ edges and exceptional multiplicity $m$. For a Brauer tree algebra with $t$ edges and exceptional multiplicity $m$, all nonprojective indecomposable modules form the stable Auslander–Reiten quiverThese are the cyclic-defect structure theorem and the stable-quiver classification for tree algebras; we are not taking the requested numerical count as an assumption. Morita equivalence preserves simple, projective and indecomposable isomorphism classes.

There is one simple module per edge, so $l(B)=e$. The vertices of $\mathbb{Z}{A}_s$ can be indexed by $(j,i)$, $j\in \mathbb{Z}$, $1\le i\le s$. The quotient by ${\tau }^e$ leaves $e$ choices of $j$ for each $i$. With $s=em=d-1$, there are $e(d-1)$ nonprojective indecomposable modules. There are also $e$ indecomposable projective modules, one projective cover per simple. ThereforeThis counts isomorphism classes, not multiplicities in the regular module. In the one-edge case it agrees with the $d$ Jordan chains for $k[t]/(t^d)$.

For the definition, take a finite connected tree with a cyclic order of the incident edges at every vertex and positive integer multiplicities equal to one except possibly at a distinguished vertex. Its Brauer tree path presentation has a quiver vertex for each tree edge, and successor arrows around each tree vertex, including loops for valency one. If edge $i$ has endpoints $v,w$, let $C_{i,v}$ and $C_{i,w}$ be the cycles around these endpoints. The relations identifykill $C_{i,v}^{m_v}{\alpha }_{i,v}$, and kill paths switching between successor cycles belonging to different tree vertices. Redundant valency-one loops may be eliminated. This quotient of the path algebra is the basic Brauer tree algebra; allowing an arbitrary algebra with Morita equivalence to it does not change the module assertions. In particular, the cyclic orders and multiplicity, not merely the underlying graph, are part of the data.

For the construction, use right modules with paths multiplied left to right; equivalently use the opposite quiver for left modules, choosing orientation so radical factors follow the specified order. The indecomposable projective $P_1$ with top $S_1$ has simple socle $S_1$ and two uniserial radical branches $U_v,U_w$ meeting in that socle. The relations show that $U_v$ has length $mr$ and factors $S_2,\dots ,S_r,S_1$ repeated $m$ times. The other branch has length $m_w\mathrm{val}(w)$. Thus $W_v=P_1/U_w$ is a uniserial module of length $mr$, with factors $S_1,S_2,\dots ,S_r$ cyclically repeated. For $J$ the Jacobson radical,is the desired uniserial branch of a Brauer tree algebra. If an endpoint is a nonexceptional leaf, its branch is just the socle; the formula still applies.

For uniqueness, length one gives just $S_1$. If $q>1$ and $r>1$, the next factor $S_2$ chooses the successor arrow at $v$ uniquely, because two distinct edges cannot meet at both endpoints in a tree. Subsequent factors force the same cyclic succession. Any arrow into an edge not incident with $v$ acts as zero, because that simple does not occur. An opposite nonexceptional leaf loop is the full socle path around $v$, of length $mr$, hence zero on modules of length at most $mr$. If an opposite leaf is exceptional, then $v$ has multiplicity one and each incident simple occurs at most once; its nilpotent self-loop is zero on that one-dimensional vertex space as well.

The remaining action is one nilpotent successor chain. Choose a top generator $z_0$ and let $z_j$ be its successive path images for $0\le j<q$. Uniseriality makes each image nonzero in its respective radical layer; these images form a basis in the basic algebra description. Their labels are $S_{1+(j\mathrm{mod}r)}$, the successor arrow sends $z_j$ to $z_{j+1}$, and all other actions vanish. Any nonzero scalar choices are absorbed by rescaling this basis. This determines the module solely by its top and length.

If $r=1$ and $q>1$, then $m>1$ and $v$ is the exceptional leaf. The successor is a single loop and its action is the unique nilpotent Jordan chain of length $q$. The other branch either enters a different edge or is a redundant socle loop and contributes nothing. Hence the prescribed uniserial module is unique up to isomorphism for every positive integer $q\le mr$.

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.

A finite-dimensional group algebra has finite representation type of a group algebra when there are only finitely many isomorphism classes of finite-dimensional indecomposable modules. The definition concerns all dimensions, not merely finitely many modules in each fixed dimension.

Let $e_B$ be the block idempotent and choose a maximal Brauer pair $(D,b_D)$: $b_D$ is a block idempotent of $k{C}_G(D)$ with $b_D{\mathrm{Br}}_D(e_B)\ne 0$. The inertial index of a block iswhere $N_G(D,b_D)$ is the stabilizer of the chosen pair. Changing the pair by conjugation does not alter this number. For cyclic $D$, it is a prime-to-$p$ divisor of $p-1$. Here $n\ge 1$, as required for the order-$p$ subgroup $Q$ to exist.

Put $H=N_G(Q)$. Since $Q$ is characteristic in $D$, $N_G(D)\le H$, so Green correspondence applies. In fact the nonprojective modules in $B$ have a particularly simple correspondence. If $g\notin H$, a nontrivial intersection $D\cap {}^{g}D$ would contain the unique order-$p$ subgroup of both cyclic groups, giving $Q={}^{g}Q$ and hence $g\in H$, a contradiction. Thus every such intersection is trivial. The error modules in induction are projective. In restriction, an intersection $H\cap {}^{g}D$ cannot contain a conjugate of a nontrivial subgroup of $D$ by an element of $H$: such a conjugate contains $Q$, and that would again force ${}^{g}Q=Q$. The general Green correspondence therefore pairs all indecomposable modules with nontrivial module vertex contained in a conjugate of $D$. In particular it pairs the nonprojective modules of $B$ with the nonprojective modules of its Brauer correspondent $b$ in $H$; the block-compatible form of the correspondence selects $b$. Explicitly,where $P$ is projective and the restriction error family is $\mathcal{Y}=\{H\cap {}^{g}D:g\notin H\}$. One should not replace this restriction error term by projectives without an additional argument about $H$.

For the final structure calculation we use the general cyclic-defect structure theorem, which is permitted here as a general fact: the basic algebra of a split block with cyclic defect group is a Brauer tree algebra. Its tree has $e(B)$ edges, one simple module per edge, and exceptional multiplicityFor a one-edge tree the basic algebra presentation is $k[t]/(t^{m+1})$. This is the local presentation of the tree algebra, not an assertion that an arbitrary block with one simple module is automatically uniserial.

With $e(B)=1$ the tree has one edge and $m=p^n-1$. Thus the basic algebra of $B$ isIt has exactly one simple module, $k=A_0/(t)$. Its ideals are precisely $(t^i)$: for any nonzero element, factor out its lowest power of $t$, after which the remaining factor is a unit. The regular projective therefore has the unique submodule chainwith each consecutive factor one-dimensional and simple. A Morita equivalence from $A_0$ to $B$ preserves simple modules, projective covers, submodule lattices and composition length. HenceIn the split setting one may also write $B\cong M_{{\dim }_{k}S}(A_0)$. The projective cover has $p^n$ composition factors all isomorphic to $S$, so its vector-space dimension is $p^n{\dim }_{k}S$, not generally $p^n$.

There is also an elementary reason why the uniserial conclusion follows once the one-simple-module conclusion and finite type are known. If the basic local algebra had $\dim J/J^2\ge 2$, quotienting its radical square and all but two generators would give $k[x,y]/(x,y{)}^2$. The two-dimensional parameter family from Question 3(b) would then give infinitely many indecomposables. Thus $J/J^2$ is one-dimensional. Nakayama lemma makes $J$ generated by one element $t$, so its powers yield a chain and the basic algebra is $k[t]/(t^r)$. The cyclic-defect calculation above identifies $r=p^n$. This explains why finite representation type rules out branching in this projective cover.