Two representations of a finite group have no irreps in common if and only if their characters are orthogonal. We want to prove that two representations of a finite group $\Gamma_{1}$ and $\Gamma_{2}$ have no irreducible representation in common if and only if their characters are orthogonal, i.e.,
$\sum_{k=1}^{c} n_k \chi_{1k}\chi_{2k}^{*}=0$,
where $\chi_{1k}$ and $\chi_{2k}$ are the characters of the $k-$th class in $\Gamma_{1}$ and $\Gamma_{2}$ respectively.
We can say that for common irreps the characters are the same and therefore the above relation doesn't equal to zero. But it is trivial. Could you please help?
 A: Let
$$
\Gamma_1 \sim \bigoplus_{\alpha\in\widehat{G}}m_\alpha^{(1)}\Gamma_\alpha, \qquad \Gamma_2 \sim \bigoplus_{\alpha\in\widehat{G}}m_\alpha^{(2)}\Gamma_\alpha
$$
be the decompositions of $\Gamma_1$ and $\Gamma_2$ into irreducible representations. Here $\widehat{G}$ denotes the equivalence classes of irreducible reps of $G$, $\Gamma_\alpha$ denotes a specific representation in the class $\alpha$, and $m_\alpha^{(i)}$ denotes the multiplicity of the $\alpha$ in $\Gamma_i$. Taking the trace of both sides gives
$$
\chi_{\Gamma_i} = \sum_{\alpha\in\widehat{G}}m_\alpha^{(i)}\chi_\alpha
$$
for $i=1,2$. Defining the inner product
$$
\langle a, b \rangle := \frac{1}{|G|}\sum_{g\in G}\overline{a(g)}b(g)
$$
and using orthonormality of the irreducible characters $\chi_\alpha$ gives
$$
\langle \chi_{\Gamma_1}, \chi_{\Gamma_2}\rangle = \sum_{\alpha\in\widehat{G}}m_\alpha^{(1)}m_\alpha^{(2)}.
$$
Since the $m_\alpha^{(i)}$ are non-negative integers, this equals zero iff $m_\alpha^{(1)}m_\alpha^{(2)} = 0$ for all $\alpha\in\widehat{G}$. This is true iff for each $\alpha\in\widehat{G}$, either $m_\alpha^{(1)} = 0$ or $m_\alpha^{(2)} = 0$, i.e. $\Gamma_1$ and $\Gamma_2$ have no irreducible representations in common.
