Orthogonality relations for Dirichlet characters modulo q

By [L1], every irreducible complex character of $G$ has degree $1$, and then [L4] shows that their number is $\mathrm{\mid }G\mathrm{\mid }=\phi (q)$ because $\mathrm{\mid }G\mathrm{\mid }={\sum }_i{1}^2$. Thus the irreducible complex characters of $G$ are exactly the Dirichlet characters modulo $q$. Since $G$ is abelian, every conjugacy class is a singleton and every centralizer is all of $G$.

Applying [L2] to the character group of $G$ gives $\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{a\in G}\chi (a)\overline{\psi (a)}={\delta }_{\chi \psi }$, which is exactly the second displayed formula because $\mathrm{\mid }G\mathrm{\mid }=\phi (q)$. Applying [L3] to the same irreducible character list and using step 1.1 turns the centralizer size into $\mathrm{\mid }G\mathrm{\mid }=\phi (q)$ and conjugacy into literal equality of elements, which yields the first displayed formula.