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A block of a finite-dimensional associative algebra $R$ is a two-sided ideal $Re$ associated with a primitive idempotent $e$ in the center of an associative algebra. Its identity is $e$, and $R$ is the product of these blocks. An $R$-module belongs to this block if $eM=M$. For a semisimple algebra, the Artin–Wedderburn theorem says the blocks are its full matrix algebras over division rings. For a group algebra this specializes to a block of a group algebra.

Let $A$ be a finite-dimensional semisimple algebra over an algebraically closed field $k$, acting faithfully on a finite-dimensional vector space $T$. Write its isotypic decomposition as $T={⨁}_i{S}_i{\otimes }_k{M}_i$. The image of $A$ acts by the full matrix algebra ${\mathrm{End}}_k(S_i)$ on each simple factor, while its commutant of an operator algebra acts by ${\mathrm{End}}_k(M_i)$ on the multiplicity factors. Taking the commutant twice therefore recovers the original image. This is the algebraic mechanism behind Schur–Weyl duality.

The splitting assumption matters for this matrix-factor description. Over a general field, simple factors can involve division algebras, and multiplicity spaces and endomorphisms must be taken over those division algebras. For example, the real algebra of quaternions acting on itself by left multiplication has the opposite quaternion algebra as its commutant, rather than an ordinary full real matrix algebra.

In characteristic $p>0$, the displayed map on the cocenter of a finite-dimensional algebra is additive and Frobenius-semilinear. Expand a $p$th power into words: nonconstant cyclic-word orbits have length $p$ and cancel modulo commutators. Also $(ab{)}^p$ and $(ba{)}^p$ have the same cocenter class, so the map kills additive commutators and is well defined.

Over a perfect field with split semisimple quotient, the stable image of this map has dimension equal to the number of simple modules of the algebra. Indeed, the kernel of $C(A)\to C(A/J(A))$ has representatives in the nilpotent Jacobson radical, so an iterate kills it. On the semisimple algebra quotient, each matrix-block coordinate is a field trace, and $\mathrm{tr}(a^p)=\mathrm{tr}(a{)}^p$. For a split semisimple quotient, Frobenius is bijective on every coordinate. In a group algebra over an algebraically closed field, the stable image has the p-regular element conjugacy classes as its basis, proving the simple-module counting theorem.

For the chain $S_0\subset \cdots \subset S_n$, the Gelfand–Tsetlin algebra in $\mathbb{C}{S}_n$ is generated by the centers of $\mathbb{C}{S}_r$ for $0\le r\le n$. Multiplicity-free restriction makes its joint eigenspaces one-dimensional in each irreducible representation. Products of the central primitive idempotents along restriction paths give its minimal projections. Thus it is a maximal commutative subalgebra and a semisimple algebra, isomorphic to a finite product of copies of $\mathbb{C}$.

For a finite group $G$, the group norm element in the group algebra $kG$ is $N_G={\sum }_{g\in G}g$. It spans the invariant subspace of the left regular group representation, and $N_G^2=|G|N_G$. If the characteristic divides $|G|$, a projection from $kG$ onto $k{N}_G$ would send $N_G$ to both itself and zero, proving that $kG$ is not a semisimple algebra. Its action on indecomposable modules is described by group norm element detects the trivial projective cover.

For a finite-dimensional symmetric algebra over any field, an indecomposable projective module $P=Ae$ has isomorphic simple head of a module and socle. A symmetrizing functional pairs $eA$ with $Ae$ nondegenerately, identifying $Ae$ with the left dual module $D(eA)$. Taking its socle gives $D(eA/eAJ(A))$. This dual of the right simple top and the left simple top $Ae/J(A)e$ belong to the same matrix block of the semisimple algebra $A/J(A)$, so they are isomorphic. The result does not require an algebraically closed ground field.

For a left module $M$ over a unital associative algebra $A$, right multiplication in $A$ gives a right module structure on ${\mathrm{Hom}}_A(M,A)$ by $(f\cdot a)(m)=f(m)a$. This remains a module homomorphism because $f(bm)a=bf(m)a$; associativity and the unit verify the action axioms. When $A=M_d(\mathbb{C})$ and $M={\mathbb{C}}^d$ is its simple module of columns, every such map is $m\mapsto mr$ for a unique row $r$, making the Hom space the simple row right module. This illustrates how left and right simple modules correspond for a semisimple algebra.

The left regular module of a unital associative algebra $A$ is its underlying additive group with $A$ acting by left multiplication. Its submodules are exactly the left ideals of $A$. The associative algebra is a semisimple algebra precisely when this module is a semisimple module.

If $I$ is a nilpotent ideal and $J(A/I)=0$, then $I=J(A)$. Every $1-ax$, $x\in I$, has a finite geometric-series inverse, giving $I\subseteq J(A)$. The radical's image lies in $J(A/I)=0$, giving the reverse inclusion. A semisimple algebra as quotient supplies the required zero radical.

A simple module is a nonzero module with no proper nonzero submodule. For a finite-dimensional simple complex module $S$, any endomorphism $\phi$ has a complex eigenvalue $\lambda$. The nonzero kernel of $\phi -\lambda I$ is an $A$-submodule, hence all of $S$. Thus $\phi =\lambda I$, proving Schur's lemma.

More generally a nonzero map between two simple modules is an isomorphism, because its kernel is zero and its image is the whole target. The endomorphism algebra of $S_1\oplus S_2$ is the block algebra with entries ${\mathrm{Hom}}_A(S_j,S_i)$. For nonisomorphic summands its off-diagonal entries vanish and its diagonal entries are scalars, giving $\mathbb{C}\times \mathbb{C}$. For isomorphic summands, choose one fixed isomorphism to identify them; every block is scalar and block composition is matrix multiplication, giving $M_2(\mathbb{C})$.

A finite-dimensional algebra is a semisimple algebra when its left regular module is a direct sum of simple modules. The complex Artin–Wedderburn theorem gives $A\cong {\prod }_i{M}_{d_i}(\mathbb{C})$, with the multiset of positive integers $d_i$ uniquely determined. Its dimension is ${\sum }_i{d}_i^2$. For dimension eight only sizes one and two can occur; the number of size-two blocks is zero, one, or two. Thus there are exactly three isomorphism types:They are distinct by the uniqueness clause, or by their numbers of one-dimensional simple modules. If $G$ is nonabelian, $\mathbb{C}G$ is noncommutative because its group elements form a basis, so the first type is excluded. The augmentation $\sum a_{g}g\mapsto \sum a_g$ provides a one-dimensional simple module, excluding the third type. ConsequentlyThere is one simple module for each matrix block, of dimension the block size, so exactly four are one-dimensional. Every nonabelian group of order eight therefore has this same complex group algebra.

The Maschke theorem says that for a finite group over a field whose characteristic does not divide its order, every finite-dimensional group representation is semisimple; equivalently its group algebra is a semisimple algebra. This applies over $\mathbb{C}$.

The Artin–Wedderburn theorem says that a finite-dimensional complex semisimple algebra is a product ${\prod }_i{M}_{d_i}(\mathbb{C})$, with one simple module ${\mathbb{C}}^{d_i}$ for each factor, up to isomorphism. Applying it to $\mathbb{C}G$ yieldsThe dimension of the left-hand side is $|G|$, while a matrix algebra $M_d(\mathbb{C})$ has dimension $d^2$. Consequently

A semisimple module is a direct sum of simple modules. An algebra is semisimple when all its modules are semisimple. Suppose the left regular module $A$ is semisimple. Every module $M$ is a quotient of a free module ${⨁}_{j\in J}A$, by sending its basis elements to a generating set of $M$. This free module is a direct sum of simple modules.

A quotient of a direct sum of simple modules is semisimple: the images of its simple summands are simple or zero and span the quotient. Choose a maximal direct sum among these simple images, allowing all simple submodules in the choice. If it does not span the whole quotient, some simple image is not contained in it. Its intersection with that direct sum is zero by simplicity, so it could be added, contradicting maximality. For infinite index sets, Zorn's lemma supplies the maximal direct sum; in finite dimension the construction terminates directly.

Hence every $M$ is semisimple, and $A$ is a semisimple algebra. This proves semisimplicity from the regular module.

Let $w={\sum }_{g\in G}g$. This is nonzero since its basis coefficients are all one, and $gw=w$ for every $g\in G$. Hence $kw$ is a one-dimensional left ideal of $kG$.

If the regular module is semisimple, $kw$ is a direct summand. Let $\pi :kG\to kw$ be the corresponding $kG$-linear projection, so $\pi (w)=w$. Write $\pi (1)=cw$ for $c\in k$. ThenBecause $w\ne 0$, $c|G|=1$ in $k$. ThusTogether with the Maschke theorem, this gives the exact semisimple algebra criterion for a group algebra.

A semisimple module is a direct sum of simple modules. A finite-dimensional semisimple algebra is one whose left regular module is semisimple; equivalently, all its finite-dimensional modules are semisimple, or every submodule of a finite-dimensional module has a complementary submodule.

The Maschke theorem states that, for a finite group $G$ and a field $k$ with $|G|\ne 0$ in $k$, every finite-dimensional $kG$-module is semisimple, so $kG$ is a semisimple algebra. The assumption is equivalent to saying that the characteristic of a field does not divide $|G|$. The averaging construction makes a projection onto a submodule equivariant by replacing $P$ with $|G{|}^{-1}{\sum }_{g\in G}gP{g}^{-1}$.

The Artin–Wedderburn theorem says that a finite-dimensional semisimple algebra over $\mathbb{C}$ is a finite product ${\prod }_j{M}_{d_j}(\mathbb{C})$: the division algebras in the general theorem reduce to $\mathbb{C}$ because the base field is algebraically closed. Each factor has center of an associative algebra $\mathbb{C}{I}_{d_j}$. Consequently a central element corresponds to $({\lambda }_j{I}_{d_j}{)}_j$. If it is nilpotent, then ${\lambda }_j^n=0$ for every $j$, whence every ${\lambda }_j=0$. Therefore $\boxed{z=0}$. The centrality assumption matters: a full matrix algebra has nonzero nilpotent elements.

The Jacobson radical is $J=\mathbb{C}{e}_{12}$. Indeed $J^2=0$, so $J$ is contained in the Jacobson radical, and $A/J\cong \mathbb{C}\times \mathbb{C}$ is a semisimple algebra, giving the reverse inclusion. The radical of a module of the left regular module is thereforeIts socle is the sum of its simple modules. The displayed left action shows that its $S_1$-submodules span $\mathbb{C}{e}_{11}\oplus \mathbb{C}{e}_{12}$. There is no $S_2$-submodule: if $av=tv$ for every $a$, applying $e_{11}$ forces the first row of $v$ to vanish, and applying $e_{12}$ then forces the remaining $e_{22}$ coefficient to vanish. HenceA homomorphism from $S_1$ is determined by any image $\alpha e_{11}+\beta e_{12}$ of $s_1$. Thus $V_{S_1}$ is the two-dimensional row right module with actionIts line $\mathbb{C}{e}_{12}$ is a nonzero proper submodule, so $V_{S_1}$ is not simple. Since $S_2$ has no copy in the socle, any homomorphism $S_2\to A$ is zero. Thus $V_{S_2}=0$ is not simple either, since a simple module is required to be nonzero.

Since $A$ is a nonzero associative algebra of finite dimension, it has a nonzero left ideal $I$ of least positive dimension. Such a minimal left ideal is a simple module under left multiplication.

For $a\in A$, right multiplication defines the module homomorphismIts kernel is a submodule of $I$, so its image $Ia$ is either zero or a simple module isomorphic to $I$. Each $Ia$ is a left ideal, since $c(xa)=(cx)a$.

ConsiderIt is a left ideal. It is also stable under right multiplication, because $(Ia)b=I(ab)$, so it is a two-sided ideal. Moreover $I=I1\subseteq J$, giving $J\ne 0$. The assumption that $A$ is a simple ring therefore forces $J=A$.

Only finitely many summands are needed: express each member of a finite basis of $A$ as a finite sum of elements of the $Ia$, and collect the finitely many indices used. After discarding zero summands, $A$ is a finite sum of simple modules. Part (a)(ii) makes its left regular module a semisimple module. This is exactly the definition of a semisimple algebra, proving that a finite-dimensional simple algebra is semisimple. The finite dimension assumption is essential to this assertion about a simple ring.

Write $A=kG$ and $J=J(A)$, its Jacobson radical. First suppose the characteristic is $p>0$. We prove the counting statement through the cocenter of a finite-dimensional algebrawhere $[A,A]$ is the linear span of commutators, not an ideal. In a group algebra, this span is exactly the span of differences of conjugate group elements: $gh-hg$ is such a difference, and $xg{x}^{-1}-g=[x,g{x}^{-1}]$. Consequently $C(A)$ has one basis vector for each conjugacy class.

There is a well-defined Frobenius map on an algebra cocenterHere $F$ is additive and $F(c[a])=c^{p}F([a])$. To prove additivity, expand $(a+b{)}^p$ into words of length $p$. Cyclic rotations of a word are equal modulo commutators. Every nonconstant word has a rotation orbit of length $p$, so its orbit contributes zero; only $a^p,b^p$ remain. Alsobecause the two products are cyclic rotations. Additivity then shows that the whole commutator subspace maps to zero, proving well-definedness on the quotient.

Let $s$ be the number of simple modules up to isomorphism. The semisimple algebra $A/J$ is ${\prod }_{\nu =1}^s{M}_{d_{\nu }}(k)$: over an algebraically closed field, a finite-dimensional division algebra is the field itself, since the minimal polynomial of any element splits into linear factors and a division algebra has no zero divisors. In a matrix algebra, the commutator subspace is precisely the traceless matrices. Off-diagonal matrix units are commutators, and $E_{ii}-E_{jj}=[E_{ij},E_{ji}]$; these span that subspace in every characteristic. ThusThe map induced by $F$ on each coordinate is $z\mapsto z^p$, since $\mathrm{tr}(B^p)=(\mathrm{tr}B{)}^p$. This is bijective because $k$ is a perfect field.

The natural surjection $C(A)\twoheadrightarrow C(A/J)$ has kernel $(J+[A,A])/[A,A]$. Every kernel class has a representative in $J$. Since $J$ is nilpotent, $F^N$ kills that kernel for sufficiently large $N$. On the quotient it remains bijective, so its image has dimension exactly $s$. On the conjugacy-class basis, however,For large $N$, every $g^{p^N}$ is a p-regular element; moreover the $p^N$-power map permutes the p-regular conjugacy classes, because an inverse exponent can be chosen modulo the prime-to-$p$ part of the exponent of $G$. The stable image therefore has precisely those conjugacy classes as its basis. Comparing dimensions provesThis argument uses dimensions of semilinear images: a bijective scalar Frobenius map preserves dimension, so the usual rank count applies.

In characteristic zero every group element is relevant. Averaging a linear projection onto a submodule over $G$ makes that projection $G$-equivariant, proving Maschke's theorem and semisimplicity of $kG$. The same matrix-algebra cocenter calculation now gives $s=\dim C(kG)$, the total number of conjugacy classes.

For the second request, $M$ is a relative projective module for $H$ when every $G$-equivariant epimorphism onto $M$ which splits after restriction to $H$ already splits over $G$. Equivalently $M$ is a direct summand of $kG{\otimes }_{kH}M$, the induced representation of the restriction of a representation to $H$. We can verify the defining splitting property directly: if $q:N\twoheadrightarrow M$ has an $H$-linear section $u$, choose representatives $T$ of the left cosets $G/H$ and setThe sum is independent of representatives by $H$-linearity. Reindexing $gt=t^{\prime }h$ proves $G$-linearity, while $qv(m)=[G:H{]}^{-1}{\sum }_{t}m=m$. Hence every $M$ is relatively $H$-projective when the index is invertible in $k$.

For completeness the canonical induced-module epimorphism $\epsilon (g\otimes m)=gm$ has the explicit $G$-linear sectionIt exhibits the promised direct summand and is the projectivity detected on a subgroup of invertible index construction.

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 $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

A partition of an integer $n$ is a finite sequence ${\lambda }_1\ge \cdots \ge {\lambda }_{\ell }>0$ with sum $n$; append zeros when comparing lengths. Its Young diagram has cells $(i,j)$ with $1\le j\le {\lambda }_i$. A Young tableau is a bijective filling of these cells by $1,\dots ,n$. Write $R_t,C_t$ for its row and column stabilizers, and fix the conventionPermutations act on entries on the left, with the rightmost factor acting first. This defines the Young symmetrizer used throughout. In the dictionary order on integer partitions, $\lambda >\mu$ means that at the first differing part ${\lambda }_i>{\mu }_i$; equality is allowed in $\ge$.

Suppose there is no row-column collision between $t$ of shape $\lambda$ and $u$ of shape $\mu$. Each column of $u$ contains at most one entry from each row of $t$. The first $k$ rows of $t$ therefore contain at mostentries, where ${\mu }_j^{\prime }$ is a column length. Thus ${\sum }_{i\le k}{\lambda }_i\le {\sum }_{i\le k}{\mu }_i$ for every $k$: $\mu$ dominates $\lambda$ in dominance order on partitions. If $\lambda \ge \mu$ in dictionary order on integer partitions, a first strictly larger part would contradict the corresponding prefix inequality. Hence $\lambda =\mu$.

All the bounds must now be equalities. Every column $j$ of $u$ contains exactly one entry from row $i$ of $t$ whenever $i\le {\lambda }_j^{\prime }$. Choose $r_0\in R_t$ sending that entry to the entry of $t$ in cell $(i,j)$. These prescriptions are bijections within the rows. The columns of $r_{0}u$ then have exactly the same sets of entries as the columns of $t$, so $r_{0}u=c_{0}t$ for some $c_0\in C_t$. ConsequentlyIf a collision was present instead, its two entries supply the first alternative. This proves the row-column collision lemma, including the prescribed order of the two stabilizer factors.

We next prove the Specht module classification. By Maschke's theorem, the group algebra $A=\mathbb{C}{S}_n$ is a semisimple algebra. Use the permitted basic quasi-idempotence of a Young symmetrizer,and put $e_t=h_t/H_{\lambda }$. The coefficient of $1$ in $h_t$ is one, because $R_t\cap C_t=\{1\}$, so $e_t\ne 0$.

For $\pi \in S_n$, consider $r_t\pi c_t$. A collision between the rows of $t$ and the columns of $\pi t$ gives a transposition $\tau \in R_t\cap \pi C_t{\pi }^{-1}$. Row symmetrization fixes $\tau$, whereas column antisymmetrization changes its sign, so $r_t\pi c_t=0$. If there is no collision, the proved lemma gives $\pi =rc$ with $r\in R_t,c\in C_t$, andIt follows that $h_t\pi h_t$ is zero or $\mathrm{sgn}(c)H_{\lambda }{h}_t$. Since permutations span $A$, $e_{t}A{e}_t=\mathbb{C}{e}_t$. In a semisimple algebra this means that $e_t$ is a primitive idempotent, so $A{e}_t=A{h}_t$ is an irreducible left module.

If $\lambda >\mu$ in dictionary order on integer partitions, the collision lemma applied to every $\pi u$ gives $r_{t}A{c}_u=0$, and therefore $h_{t}A{h}_u=0$. Sincethe two simple modules cannot be isomorphic. Conversely, Young tableaux of the same shape are related by a permutation, which conjugates their Young symmetrizers and gives isomorphic left ideals. Finally, the center of $\mathbb{C}{S}_n$ has the conjugacy-class sums as a basis. Conjugacy classes are indexed by cycle-type partitions, so its dimension is the number of partitions of $n$. The Artin–Wedderburn theorem gives exactly that many simple-module isomorphism classes. We have already produced one for each partition. Thus the $A{h}_{\lambda }$ form a complete set of pairwise nonisomorphic irreducible modules.