[proofplan] We produce $kr$ characters of $G \times H$ from the irreducibles of $G$ and $H$, then show they exhaust the irreducible characters by orthonormality plus a degree-sum count. The construction is the external tensor product: given representations $\rho_i: G \to \operatorname{GL}(V_i)$ and $\rho'_j: H \to \operatorname{GL}(W_j)$, the map $(g, h) \mapsto \rho_i(g) \otimes \rho'_j(h)$ is a representation of $G \times H$ on $V_i \otimes W_j$ whose character at $(g, h)$ is $\chi_i(g)\,\psi_j(h)$. Computing the $G \times H$ inner product, the double sum factors as a product of $G$ and $H$ inner products, giving $\langle \chi_i \psi_j, \chi_r \psi_s \rangle_{G \times H} = \delta_{ir}\delta_{js}$ — so the $kr$ products are orthonormal and hence pairwise distinct irreducible characters. Finally, the degree-sum identity $\sum_{i,j} (\chi_i \psi_j)(1)^2 = |G||H| = |G \times H|$ matches the order of $G \times H$, which by [Number of Irreducible Characters](/theorems/2428) means we have all irreducibles. [/proofplan] [step:Construct the external tensor product representation $(\rho_i \otimes \rho'_j)$ on $V_i \otimes W_j$] Fix irreducible representations $\rho_i: G \to \operatorname{GL}(V_i)$ affording $\chi_i$ and $\rho'_j: H \to \operatorname{GL}(W_j)$ affording $\psi_j$. Define \begin{align*} \rho_i \boxtimes \rho'_j: G \times H &\to \operatorname{GL}(V_i \otimes W_j) \\ (g, h) &\mapsto \rho_i(g) \otimes \rho'_j(h), \end{align*} where $\rho_i(g) \otimes \rho'_j(h)$ denotes the linear map on $V_i \otimes W_j$ determined on pure tensors by $(\rho_i(g) \otimes \rho'_j(h))(v \otimes w) = \rho_i(g)v \otimes \rho'_j(h)w$ and extended linearly via the universal property of the tensor product (well-definedness of this extension is the bilinearity argument used in [Tensor Product of Representations](/theorems/2441), Step 1). To check that $\rho_i \boxtimes \rho'_j$ is a homomorphism, fix $(g_1, h_1), (g_2, h_2) \in G \times H$. On a pure tensor, \begin{align*} (\rho_i \boxtimes \rho'_j)(g_1, h_1)\,(\rho_i \boxtimes \rho'_j)(g_2, h_2)(v \otimes w) &= (\rho_i(g_1) \otimes \rho'_j(h_1))(\rho_i(g_2)v \otimes \rho'_j(h_2)w) \\ &= \rho_i(g_1)\rho_i(g_2)v \otimes \rho'_j(h_1)\rho'_j(h_2)w \\ &= \rho_i(g_1 g_2)v \otimes \rho'_j(h_1 h_2)w \\ &= (\rho_i \boxtimes \rho'_j)(g_1 g_2, h_1 h_2)(v \otimes w). \end{align*} The first equality applies the definition of $(\rho_i \boxtimes \rho'_j)(g_2, h_2)$, then the definition of $(\rho_i \boxtimes \rho'_j)(g_1, h_1)$ to the pure tensor produced. The second uses bilinearity of $\otimes$. The third uses that $\rho_i, \rho'_j$ are individually homomorphisms. The fourth recognises the result as the value of $\rho_i \boxtimes \rho'_j$ at $(g_1, h_1)(g_2, h_2) = (g_1 g_2, h_1 h_2)$, the product in $G \times H$. Pure tensors span $V_i \otimes W_j$, so this identity extends linearly to all of $V_i \otimes W_j$. The identity element $(1_G, 1_H)$ acts as $I_{V_i} \otimes I_{W_j} = I_{V_i \otimes W_j}$. Hence $\rho_i \boxtimes \rho'_j$ is a representation of $G \times H$. [/step] [step:Compute the character $(g, h) \mapsto \chi_i(g)\,\psi_j(h)$] Fix $(g, h) \in G \times H$. Both $\rho_i(g)$ and $\rho'_j(h)$ have finite order (as $G$ and $H$ are finite), hence are diagonalisable over $\mathbb{C}$. Choose bases $\{v_a\}_{a=1}^{m}$ of $V_i$ and $\{w_b\}_{b=1}^{n}$ of $W_j$ with $\rho_i(g)v_a = \lambda_a v_a$ and $\rho'_j(h)w_b = \mu_b w_b$. Then $\{v_a \otimes w_b\}_{a,b}$ is a basis of $V_i \otimes W_j$, and \begin{align*} (\rho_i(g) \otimes \rho'_j(h))(v_a \otimes w_b) = \lambda_a \mu_b\, (v_a \otimes w_b). \end{align*} Summing eigenvalues, \begin{align*} \chi_{\rho_i \boxtimes \rho'_j}(g, h) = \sum_{a, b} \lambda_a \mu_b = \biggl(\sum_a \lambda_a\biggr)\biggl(\sum_b \mu_b\biggr) = \chi_i(g)\,\psi_j(h). \end{align*} Write $\chi_i \psi_j: G \times H \to \mathbb{C}$, $(g, h) \mapsto \chi_i(g)\,\psi_j(h)$. So $\chi_i \psi_j$ is the character of the representation $\rho_i \boxtimes \rho'_j$ of $G \times H$. [/step] [step:Show $\{\chi_i \psi_j\}$ are orthonormal in $\mathcal{C}(G \times H)$ via factoring the double sum] Fix indices $i, r \in \{1, \ldots, k\}$ and $j, s \in \{1, \ldots, r\}$. The $G \times H$ inner product of class functions is \begin{align*} \langle \chi_i \psi_j, \chi_r \psi_s \rangle_{G \times H} = \frac{1}{|G \times H|} \sum_{(g, h) \in G \times H} \overline{\chi_i(g)\psi_j(h)}\, \chi_r(g)\psi_s(h). \end{align*} Using $|G \times H| = |G|\,|H|$ and that the conjugate of a product is the product of conjugates, and that the sum over $G \times H$ factors as iterated sums over $G$ and $H$, \begin{align*} \langle \chi_i \psi_j, \chi_r \psi_s \rangle_{G \times H} &= \frac{1}{|G|\,|H|} \sum_{g \in G} \sum_{h \in H} \overline{\chi_i(g)}\,\chi_r(g)\, \overline{\psi_j(h)}\,\psi_s(h) \\ &= \biggl(\frac{1}{|G|} \sum_{g \in G} \overline{\chi_i(g)}\,\chi_r(g)\biggr) \biggl(\frac{1}{|H|} \sum_{h \in H} \overline{\psi_j(h)}\,\psi_s(h)\biggr) \\ &= \langle \chi_i, \chi_r \rangle_G\, \langle \psi_j, \psi_s \rangle_H. \end{align*} The factorisation in the second line uses that the summand is the product of a function of $g$ and a function of $h$, so the double sum splits as a product of single sums. By orthonormality of the irreducible characters of $G$ ([Row Orthogonality](/theorems/2430)), $\langle \chi_i, \chi_r \rangle_G = \delta_{ir}$, and similarly $\langle \psi_j, \psi_s \rangle_H = \delta_{js}$. Therefore \begin{align*} \langle \chi_i \psi_j, \chi_r \psi_s \rangle_{G \times H} = \delta_{ir}\,\delta_{js}. \end{align*} [guided] The $G \times H$ inner product factors because the inner product is a sum over the group, and $G \times H$ is a Cartesian product of groups. Concretely, \begin{align*} \langle \chi_i \psi_j, \chi_r \psi_s \rangle_{G \times H} = \frac{1}{|G \times H|} \sum_{(g, h) \in G \times H} \overline{\chi_i(g)\psi_j(h)}\, \chi_r(g)\psi_s(h). \end{align*} The summand $\overline{\chi_i(g)}\,\chi_r(g) \cdot \overline{\psi_j(h)}\,\psi_s(h)$ is the product of a function of $g$ alone and a function of $h$ alone — a separable summand. The general principle is: $\sum_{(a, b) \in A \times B} f(a) g(b) = (\sum_a f(a))(\sum_b g(b))$. Applying this with $|G \times H| = |G|\,|H|$: \begin{align*} \langle \chi_i \psi_j, \chi_r \psi_s \rangle_{G \times H} = \biggl(\frac{1}{|G|} \sum_{g \in G} \overline{\chi_i(g)}\,\chi_r(g)\biggr) \biggl(\frac{1}{|H|} \sum_{h \in H} \overline{\psi_j(h)}\,\psi_s(h)\biggr) = \langle \chi_i, \chi_r \rangle_G\, \langle \psi_j, \psi_s \rangle_H. \end{align*} By [Row Orthogonality](/theorems/2430), the irreducible characters of any finite group are orthonormal in the standard inner product. Since $\chi_1, \ldots, \chi_k$ are the distinct irreducible characters of $G$ by hypothesis, $\langle \chi_i, \chi_r \rangle_G = \delta_{ir}$. Similarly $\langle \psi_j, \psi_s \rangle_H = \delta_{js}$. Multiplying gives $\delta_{ir}\delta_{js}$. In particular $\langle \chi_i \psi_j, \chi_i \psi_j \rangle_{G \times H} = 1$, so each $\chi_i \psi_j$ is the character of an irreducible representation of $G \times H$ by [Irreducibility Criterion](/theorems/2426). And distinct pairs $(i, j) \neq (r, s)$ give orthogonal characters, hence pairwise distinct: there are exactly $kr$ distinct irreducible characters in our list. [/guided] [/step] [step:Confirm $\langle \chi_i \psi_j, \chi_i \psi_j \rangle_{G \times H} = 1$ implies each is irreducible] The diagonal case $i = r$, $j = s$ of Step 3 gives $\langle \chi_i \psi_j, \chi_i \psi_j \rangle_{G \times H} = 1$. By Step 2, $\chi_i \psi_j$ is the character of the representation $\rho_i \boxtimes \rho'_j$, with $(\chi_i \psi_j)(1, 1) = \chi_i(1)\psi_j(1) > 0$ since irreducible characters of finite groups have strictly positive degree. By the [Irreducibility Criterion](/theorems/2426) applied to $\chi_i \psi_j$ as a character of $G \times H$, this character is irreducible. The off-diagonal case of Step 3 gives $\langle \chi_i \psi_j, \chi_r \psi_s \rangle_{G \times H} = 0$ for $(i, j) \neq (r, s)$, so the $kr$ products are pairwise distinct irreducible characters of $G \times H$. [/step] [step:Match the squared-degree sum to $|G \times H|$ and conclude completeness] The number of irreducible characters of $G \times H$ equals the sum of squares of their degrees being $|G \times H|$ — formally, by [Number of Irreducible Characters](/theorems/2428) applied to $G \times H$, the squared degrees of irreducible characters sum to $|G \times H|$. We verify the $kr$ characters $\chi_i \psi_j$ already exhaust this sum. Computing the sum over our list: \begin{align*} \sum_{i = 1}^{k} \sum_{j = 1}^{r} (\chi_i \psi_j)(1, 1)^2 = \sum_{i, j} \chi_i(1)^2\, \psi_j(1)^2 = \biggl(\sum_{i=1}^{k} \chi_i(1)^2\biggr) \biggl(\sum_{j=1}^{r} \psi_j(1)^2\biggr) = |G|\,|H| = |G \times H|, \end{align*} using $(\chi_i \psi_j)(1, 1) = \chi_i(1)\psi_j(1)$ from the construction, separability of the squared summand, and applying [Number of Irreducible Characters](/theorems/2428) once to $G$ (giving $\sum_i \chi_i(1)^2 = |G|$) and once to $H$ (giving $\sum_j \psi_j(1)^2 = |H|$). Suppose for contradiction there were an irreducible character $\zeta$ of $G \times H$ not in the list $\{\chi_i \psi_j\}$. Then by orthonormality, $\zeta(1, 1)^2$ would add a strictly positive contribution to the total sum of squared degrees over irreducibles, exceeding $|G \times H|$ — contradicting [Number of Irreducible Characters](/theorems/2428) applied to $G \times H$. Hence every irreducible character of $G \times H$ is some $\chi_i \psi_j$, and the irreducible characters of $G \times H$ are exactly the $kr$ functions $\{\chi_i \psi_j\}$. [guided] We have produced $kr$ pairwise distinct irreducible characters of $G \times H$. To conclude these are **all** of them, we use a counting argument based on the squared-degree identity. By [Number of Irreducible Characters](/theorems/2428) applied to $G \times H$, the irreducible characters $\zeta$ of $G \times H$ satisfy \begin{align*} \sum_{\zeta \in \widehat{G \times H}} \zeta(1, 1)^2 = |G \times H|. \end{align*} Apply the same identity to $G$ alone and $H$ alone: \begin{align*} \sum_i \chi_i(1)^2 = |G|, \qquad \sum_j \psi_j(1)^2 = |H|. \end{align*} The squared degrees of our $kr$ characters sum as \begin{align*} \sum_{i, j} (\chi_i \psi_j)(1, 1)^2 = \sum_{i, j} \chi_i(1)^2\, \psi_j(1)^2 = |G|\,|H| = |G \times H|. \end{align*} This already saturates the total $|G \times H|$. Any additional irreducible character $\zeta$ not in our list would contribute $\zeta(1, 1)^2 \geq 1$ to the same sum, exceeding $|G \times H|$ — a contradiction. So the $kr$ characters $\{\chi_i \psi_j\}$ form the complete list of irreducible characters of $G \times H$. The structural insight: the irreducible characters of a direct product factor as products of irreducibles of the factors. This is the character-theoretic shadow of the fact that $\mathbb{C}[G \times H] \cong \mathbb{C}[G] \otimes_\mathbb{C} \mathbb{C}[H]$ as $\mathbb{C}$-algebras. [/guided] [/step]