Proof The row orthogonality relations have already been proved. They are recorded here merely for comparison with the column relations.
For $1 \leqslant s \leqslant k$, let $\psi_{s}$ be the class function which satisfies
$$
\psi_{s}\left(g_{r}\right)=\delta_{r s} \quad(1 \leqslant r \leqslant k) .
$$
By Corollary $15.4, \psi_{s}$ is a linear combination of $\chi_{1}, \ldots, \chi_{k}$, say
$$
\psi_{s}=\sum^{k} \lambda_{i} \chi_{i} \quad\left(\lambda_{i} \in \mathbb{C}\right) .
$$
We know that $\left\langle\chi_{i}, \chi_{j}\right\rangle=\delta_{i j}$, so
$$
\lambda_{i}=\left\langle\psi_{s}, \chi_{i}\right\rangle=\frac{1}{|G|} \sum_{g \in G} \psi_{s}(g) \overline{\chi_{i}(g)}
$$
Now $\psi_{s}(g)=1$ if $g$ is conjugate to $g_{s}$, and $\psi_{s}(g)=0$ otherwise; also there are $|G| /\left|C_{G}\left(g_{s}\right)\right|$ elements of $G$ which are conjugate to $g_{s}$, by
MATH0073 COURSE NOTES ：
character $\bar{\chi}$. The function $\rho: G \rightarrow \mathrm{GL}(n, \mathbb{C})$ which is given by the composition
$$
g \rightarrow N g \rightarrow(N g) \tilde{\rho} \quad(g \in G)
$$
is a homomorphism from $G$ to $\mathrm{GL}(n, \mathbb{C})$. Thus $\rho$ is a representation of G. The character $\chi$ of $\rho$ satisfies
$$
\chi(g)=\operatorname{tr}(g \rho)=\operatorname{tr}((N g) \bar{\rho})=\tilde{\chi}(N g)
$$
for all $g \in G$. Moreover, $\chi(1)=\tilde{\chi}(N)$, so $\chi$ and $\tilde{\chi}$ have the same degree.