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Let $C_{p^n}=⟨g⟩$ and let $k$ have characteristic $p$. With $u=g-1$,Its finite-dimensional indecomposable modules areEach is a uniserial module, with radical $u{M}_r$ and one-dimensional socle $u^{r-1}{M}_r$.

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 $q\le mr$.

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

Put $q=p^n$. A finitely generated module over $A=k[X]/(X^q)$ is finite-dimensional over $k$, and multiplication by $X$ is a nilpotent linear map $T$ with $T^q=0$. Its Jordan normal form exists over $k$ because its minimal polynomial is a power of $X$. Each Jordan block has size $r\le q$ and is the cyclic module $M_r=k[X]/(X^r)$. HenceA submodule of $M_r$ is an ideal of $k[X]/(X^r)$, and its inverse image is an ideal of the principal ideal domain $k[X]$ containing $(X^r)$. The only possibilities are $(X^j)$ with $0\le j\le r$. Thus all submodules form the chainEvery successive quotient is the one-dimensional simple module on which $X$ acts as zero. This is its unique composition series, so $M_r$ is a uniserial module and is an indecomposable representation: two nonzero direct summands would be incomparable submodules. Only $M_1$ is simple.

For $G=⟨g⟩$ of order $q$, the group algebra satisfiessince $(T-1{)}^q=T^q-1$ in characteristic $p$. Consequently, up to isomorphism, $\boxed{M_1,\dots ,M_{p^n}}$ are precisely the finite-dimensional indecomposable modules, and $\boxed{M_1\text{ is the unique simple module}}$. The finite-dimensional convention follows from the finite-generation hypothesis; “exactly” counts isomorphism classes.

An $A$-module is a semisimple module when it is a direct sum of simple modules, equivalently when every submodule has a complementary submodule. The finite-dimensional algebra $A$ is a semisimple algebra when its left regular module ${}_{A}A$ is semisimple.

Let $C_{p^n}=⟨g⟩$, put $N=p^n$, and suppose $\mathrm{char}k=p$. The freshman's dream givesso, with $u=g-1$,The indecomposable modules of a cyclic p-group in characteristic p are preciselyIndeed, a module is a vector space with a nilpotent operator $u$, and its decomposition into Nilpotent Jordan blocks gives these modules. Each $M_r$ is a uniserial module, with unique chainTherefore

Let $i$ be an edge at $v$. Kill the other radical branch of its indecomposable projective module. The resulting uniserial module has length $m_v\mathrm{val}(v)$ and its factors follow the cyclic edge order beginning at $i$. Quotienting its radical powers gives all shorter such modules. The edge succession uniquely fixes the path, and its nonzero arrow maps can be rescaled to one, giving uniqueness up to isomorphism. At an exceptional leaf this reduces to a nilpotent single Jordan chain; the other leaf loop, when present, is a socle path and vanishes in these shorter quotients.