The character ring of a direct product is the tensor product of the factor character rings

Let ${\pi }_G:G\times H\to G$ and ${\pi }_H:G\times H\to H$ be the coordinate projections. Pulling $V$ and $W$ back along these homomorphisms and taking their tensor product gives the stated $G\times H$ action, because the two factor actions multiply componentwise. The tensor-product construction is well defined on $V{\otimes }_{\mathbb{C}}W$.

Let $X$ be an irreducible $G\times H$-module and restrict it to $N:=G\times \{1_H\}$. By [F5] this nonzero restriction has an irreducible constituent $S$ with character $\chi \in \mathrm{Irr}(G)$. Conjugation by $(g,h)$ acts on $N$ by conjugation by $g$, so $\chi$ is fixed because characters are constant on conjugacy classes. Hence the inertia group of $\chi$ in $G\times H$ is all of $G\times H$.

For irreducible characters $\chi ,{\chi }^{\mathrm{\prime }}$ of $G$ and $\psi ,{\psi }^{\mathrm{\prime }}$ of $H$, the finite sum over $G\times H$ factors as $⟨\chi ⊠\psi ,{\chi }^{\mathrm{\prime }}⊠{\psi }^{\mathrm{\prime }}{⟩}_{G\times H}=\frac{1}{\mathrm{\mid }G\mathrm{\mid }\mathrm{\mid }H\mathrm{\mid }}{\sum }_{g\in G,h\in H}\chi (g)\psi (h)\overline{{\chi }^{\mathrm{\prime }}(g)}\overline{{\psi }^{\mathrm{\prime }}(h)}=⟨\chi ,{\chi }^{\mathrm{\prime }}{⟩}_G⟨\psi ,{\psi }^{\mathrm{\prime }}{⟩}_H$. Thus the external products of irreducible characters are orthonormal by [F4].

By [F3], the character of this representation at $(g,h)$ is ${\chi }_V(g){\chi }_W(h)$. Maschke decomposes representations affording honest characters into irreducibles; distributivity of tensor products over direct sums then shows that this formula agrees with the bilinear extension from the irreducible-character bases.

The representation $\tilde{S}(g,h):=S(g)$ extends $S$ to $G\times H$, and the quotient by $N$ is canonically $H$. Gallagher's theorem therefore says that the irreducibles above $\chi$ are exactly $S⊠W$ for $W$ irreducible over $H$, without repetition. Since $X$ lies above its chosen constituent $\chi$, this proves that every irreducible of $G\times H$ occurs exactly once in the family of external products. The argument includes a trivial factor, for which the relevant irreducible-character set is a singleton.

The trivial character is the unit in each character ring, and its external product is the trivial character of $G\times H$. Thus $\mathrm{\Phi }(1\otimes 1)=1$.

Steps 1.3 and 2.2 show that the external products are precisely the irreducible characters of $G\times H$, each once, and are orthonormal. They are therefore an orthonormal $\mathbb{Z}$-basis of $R(G\times H)$ by [F1, F4].

For irreducible basis elements, the pointwise product formula in step 2.1 gives $\mathrm{\Phi }((\chi \otimes \psi )({\chi }^{\mathrm{\prime }}\otimes {\psi }^{\mathrm{\prime }}))=\mathrm{\Phi }(\chi {\chi }^{\mathrm{\prime }}\otimes \psi {\psi }^{\mathrm{\prime }})=(\chi ⊠\psi )({\chi }^{\mathrm{\prime }}⊠{\psi }^{\mathrm{\prime }})$. Bilinearity extends this identity to all virtual characters, including zero and negative combinations. Thus $\mathrm{\Phi }$ is multiplicative under the tensor-product algebra structure [F8].

By Maschke's theorem, any honest character of $G\times H$ is a nonnegative integral sum of its irreducible characters. Step 3.1 identifies each such character as one external product, proving the positivity assertion.

The irreducible characters form free $\mathbb{Z}$-bases of $R(G)$ and $R(H)$, so [F7] gives the basis $(\chi \otimes \psi )$ of $R(G){\otimes }_{\mathbb{Z}}R(H)$. Step 3.1 shows that the bilinear map sends this basis bijectively to the basis of the target; hence it is a $\mathbb{Z}$-module isomorphism.

Steps 2.3, 3.2, and 4.2 prove the unital ring isomorphism; steps 3.1 and 4.1 prove the orthonormal-basis and positivity claims.