Irreducible representations of product groups and the relationship between characters Let $G_1$ and $G_2$ be finite groups. Let $V$ be a finite-dimensional complex vector space. Let $\rho: G_1\times G_2\to\text{GL}(V)$ be an irreducible representation with character $\chi$. Let $\rho_i: G_i\to\text{GL}(V)$ be the obvious representations realized as restrictions of $\rho$ with $\chi_i$ its character, for $i = 1, 2$.
I'm trying to show that each $\rho_i$ is irreducible, but have been stuck for a while.
I first want to claim that $\chi(t, s) = \chi_1(t)\cdot \chi_2(s)$, but I'm unconvinced that this must hold since $\rho$ isn't necessarily a homomorphism. Because of this, I can't claim that
$\chi(g, h) = \chi(g, e)\cdot\chi(e, h) = \chi_1(g)\cdot\chi_2(h)$. However, if this were to hold, I think I'd have a route to proving what I want. If $\{C_1, \ldots, C_n\}$ and $\{K_1, \ldots, K_m\}$ are the conjugacy classes of $G_1$ and $G_2$, respectively, then it's clear that $\{(C_i, K_j) : 1\leq i\leq n \text{ and }1\leq j\leq m\}$ form the conjugacy classes of $G_1\times G_2$ and that ${\mid (C_i, K_j) \mid} = {\mid C_i\mid}\cdot{\mid K_j\mid}$.
Let $c_i$ denote a representative of $C_i$, and similarly define $k_i$. Since $\rho$ is irreducible, we get the following
\begin{align}
1 &=\langle \rho, \rho\rangle \\
&=\frac{1}{{\mid G_1\mid}{\mid G_2\mid}}\sum_{\substack{i\leq n\\j\leq m}}{\mid{(C_i, K_j)\mid}{\mid\rho(c_i, k_j)\mid^2}}\\
&=\frac{1}{{\mid G_1\mid}{\mid G_2\mid}}\sum_{\substack{i\leq n\\j\leq m}}{\mid{C_i\mid}{\mid K_j\mid}{\mid\rho_1(c_i)\mid^2}{\mid\rho_2(k_j)\mid^2}}\\
&=\left(\frac{1}{{\mid G_1\mid}}\sum_{\substack{i\leq n}}{\mid{C_i\mid}{\mid\rho_1(c_i)\mid^2}}\right)\left(\frac{1}{{\mid G_2\mid}}\sum_{\substack{j\leq m}}{\mid{K_j\mid}{\mid\rho_2(k_j)\mid^2}}\right)\\
&= \langle\rho_1, \rho_1\rangle\langle\rho_2, \rho_2\rangle.
\end{align}
Since $\langle \rho_i, \rho_i\rangle$ is necessarily a positive integer for each $i$, it then follows that $\langle \rho_i, \rho_i\rangle = 1$ for both $i$, showing that $\rho_1$ and $\rho_2$ are irreducible. Therefore, if I can show that $\rho(t, s) = \rho_1(t)\cdot\rho_2(s)$, then I'm done. I'm just stuck here.
 A: This is false. Presumably this is something you're trying to do as part of a longer exercise the goal of which is to eventually show that $\rho$ is the external tensor product $V_1 \boxtimes V_2$ of an irreducible representation of $G_1$ and an irreducible representation of $G_2$. What this means is that the restriction $\rho_1$ ends up being a direct sum of $\dim V_2$ copies of $V_1$, while the restriction $\rho_2$ ends up being a direct sum of $\dim V_1$ copies of $V_2$. So in general these restrictions are not irreducible.
It is true that $\rho(g, h) = \rho_1(g) \rho_2(h)$, because $\rho$ is by definition a homomorphism. The statement you need to make your argument work, which is not true, is $\chi(g, h) \stackrel{?}{=} \chi_1(g) \chi_2(h)$ (and you can check that this is not true by substituting $g = h = e$, where you will get $\dim V$ on the LHS but $(\dim V)^2$ on the RHS); I guess you conflated $\rho$ and $\chi$ since it is $\chi$ which is not a homomorphism. From $\rho(t, s) = \rho_1(t) \rho_2(s)$ we can't immediately conclude anything about $\chi(t, s)$.
A correct although somewhat unsatisfying proof that irreducible representations of $G_1 \times G_2$ are external tensor products is the following: if $\chi_1, \chi_2$ are irreducible characters of $G_1, G_2$ respectively it's easy to check that $\langle \chi_1 \chi_2, \chi_1 \chi_2 \rangle = \langle \chi_1, \chi_1 \rangle \langle \chi_2, \chi_2 \rangle$ (which you've done) so $\chi_1 \chi_2$ (the character of the external tensor product of the representations with characters $\chi_1, \chi_2$) is an irreducible character of $G_1 \times G_2$. We know that the number of irreducible characters of $G_1 \times G_2$ is the number of conjugacy classes, which is $c(G_1 \times G_2) = c(G_1) \times c(G_2)$. So this construction must have produced all irreducible characters.
