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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.