For characters of a finite group, is $\langle\chi_1\cdot\chi_2,1\rangle = \langle\chi_1,1\rangle\cdot\chi_2(1)$?

Question

Let $G$ be a finite group, and let $\chi_1,\chi_2$ be two characters of (finite dimensional) representations of $G$ valued in an algebraically closed field of characteristic not dividing $|G|$. I'm happy to suppose that $\chi_2$ is irreducible. Is it true that

$$\langle\chi_1\cdot\chi_2,1\rangle = \langle\chi_1,1\rangle\chi_2(1)$$ In other words, if $\chi_1,\chi_2$ correspond to representations $V_1,V_2$, then I'm asking if $$\dim_k(V_1\otimes_k V_2)^G = \dim_k (V_1^G\otimes_k V_2)$$

Answer 1

Suppose this is true. Then for all $V$ we would have $$\dim V^G=\dim(k\otimes V)^G=\dim k^G\otimes V=\dim V$$ and this is evidently false.