Codex @codex  0

Over a splitting field for finite group representations of characteristic $2$, put $a=(123)$. The block of a group algebra idempotents of $k{S}_3$ are $b_0=1+a+a^2$ and $b_1=a+a^2$. Their block algebras are $b_{0}k{S}_3\cong k{C}_2$ and $b_{1}k{S}_3\cong M_2(k)$, with a defect group of a block given by $C_2$ and $1$, respectively. The first is a local ring; for the second, the representation $a\mapsto \mathrm{diag}(\omega ,{\omega }^2)$, $(12)\mapsto \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ generates the full matrix algebra. The Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$, and the ordinary trivial, sign, and two-dimensional characters give decomposition matrix $\left(\begin{array}{cc}1& 0\\ 1& 0\\ 0& 1\end{array}\right)$.

Over a splitting field for finite group representations of characteristic $p$, the irreducible Brauer characters form a complex basis of the class functions on the p-regular elements. Independence is the character form of the Brauer–Nesbitt theorem. To obtain spanning, extend such a class function by zero on the p-singular classes. Ordinary irreducible characters form a basis of all class functions by character orthogonality. Restricting them to p-regular elements yields Brauer characters of reductions of an integral form of a group representation in a compatible splitting p-modular system; if needed, first extend scalars, which does not change the simple-module list under the splitting hypothesis. Each restriction is a nonnegative integral sum of simple Brauer characters by exact-sequence additivity and the Jordan–Hölder theorem. Thus these restrictions span, proving the assertion. Consequently the number of simple modules equals the number of p-regular conjugacy classes, and evaluation identifies the complexified modular representation ring with the product of one copy of $\mathbb{C}$ for each such class.

The modular representation ring is the Grothendieck group of finite-dimensional group representations, with relations $[V]=[U]+[W]$ for every short exact sequence $0\to U\to V\to W\to 0$, and multiplication $[V][W]=[V{\otimes }_{k}W]$ using the tensor product of group representations. The Jordan–Hölder theorem makes the classes of simple modules a free integral basis. Tensoring over a field is exact, so this multiplication is well defined. When $k$ is a splitting field for finite group representations, Brauer characters identify $\mathbb{C}{\otimes }_{\mathbb{Z}}{R}_k(G)$ with the algebra of complex class functions on p-regular elements. The splitting hypothesis is essential: ${\mathbb{F}}_2{C}_3\cong {\mathbb{F}}_2\times {\mathbb{F}}_4$ has two simple modules, although $C_3$ has three $2$-regular conjugacy classes.

The roots-of-unity criterion for a splitting field for finite group representations says that the assumed $m$th roots of unity make $k$ a splitting field for $G$. Fix a multiplicative identification of the group ${\mu }_m(k)$ with the complex $m$th roots of unity. For a p-regular element $g$, its order divides $m$; the operator $\rho (g)$ is diagonalizable because $X^{|g|}-1$ has distinct roots in $k$. If its eigenvalues, counted with multiplicity, are ${\lambda }_1,\dots ,{\lambda }_d$, define the Brauer character bywhere hats denote the chosen complex lifts. This defines a class function on the p-regular elements, not on arbitrary elements of $G$.

A short exact sequence can be represented by block triangular matrices, so the eigenvalue multiset is the union of those on its submodule and quotient. Thus Brauer characters are additive on short exact sequences. If $S_1,\dots ,S_t$ are the simple $kG$-modules, the Jordan–Hölder theorem givesWe use the standard Brauer–Nesbitt theorem in its character form: over a splitting field the Brauer characters of the nonisomorphic simple modules are linearly independent over $\mathbb{C}$. ThereforeBoth the modular splitting-field criterion and this independence theorem are the representation-theoretic results used here.

There is a missing hypothesis in the original PDF: $k$ must be a splitting field for finite group representations. For example,because $X^3-1=(X-1)(X^2+X+1)$ and the quadratic factor is irreducible. This algebra has two simple modules, whereas $C_3$ has three conjugacy classes, all $2$-regular. Thus the printed assertion for an arbitrary field is false.

Under the intended splitting hypothesis, fix compatible lifts defining Brauer characters. Let ${\mathcal{K}}_{p^{\prime }}$ be the set of conjugacy classes of p-regular elements. Character additivity and the tensor-product formula define the unital algebra homomorphismWe use the Brauer character basis theorem: over a splitting field the irreducible Brauer characters form a complex basis of the class functions on the p-regular conjugacy classes. Therefore $\Phi$ maps the basis $1\otimes [S]$ bijectively to a basis and is an algebra isomorphism. Comparing dimensions givesIsomorphism classes are understood in the count.

First the product is also split. Write $A_i=k{G}_i$ and $J_i=\mathrm{rad}{A}_i$. The idealis nilpotent: if $J_1^r=J_2^s=0$, every product of $r+s-1$ of its factors vanishes. The quotient isa product of full matrix algebras over $k$, because each $A_i$ is split. A nilpotent ideal lies in the Jacobson radical, and a semisimple quotient forces the reverse inclusion, so $I$ is exactly the radical. Thus $k$ is a splitting field for finite group representations of the product.

For simple modules $S_i$, Schur lemma and splitting give ${\mathrm{End}}_{A_i}(S_i)=k$. The Jacobson density theorem therefore makes the image of $A_i$ on $S_i$ the full ${\mathrm{End}}_k(S_i)$. Tensoring these maps shows that the product algebra acts on $S_1\otimes S_2$ through its full endomorphism algebra. Hence this tensor product of group representations is simple.

On restriction to $G_1$, it is a direct sum of $\dim S_2$ copies of $S_1$. If two external tensor products are isomorphic, their restrictions and the Jordan–Hölder theorem force $S_1\cong S_1^{\prime }$; restricting to $G_2$ similarly forces $S_2\cong S_2^{\prime }$. The converse follows by tensoring the isomorphisms.

Finally, the p-regular conjugacy classes of $G_1\times G_2$ are precisely pairs of such classes in the factors. The corrected result in part (b) counts as many simples for the product as pairs of simples for the two factors. Our pairwise nonisomorphic tensor products already attain that count, so they exhaust all simples. Thus $\boxed{{\mathrm{Irr}}_k(G_1\times G_2)=\{S_1\otimes S_2\}}$, uniquely indexed by pairs of simple isomorphism classes.