I was reading this theorem from An introduction to ergodic theory by Peter Walters but could not figure out why the statement given in bold is true.
Theorem. Let $T$, given by $T(g)=a g$, be an ergodic rotation of a compact abelian group $G$. Then $T$ has discrete spectrum. Every eigenfunction of $T$ is a constant multiple of a character, and the eigenvalues of $T$ are $\{\gamma(a): \gamma \in \widehat{G}\}$.
Proof. Let $\gamma \in \hat{G}$, the character group of $G$. Then $$ \gamma(T g)=\gamma(a g)=\gamma(a) \gamma(g) $$ Therefore each character is an eigenfunction and so $T$ has discrete spectrum since the characters are an orthonormal basis of $L^{2}(m)$. If there is another eigenvalue besides the members of $\{\gamma(a): \gamma \in \hat{G}\}$ then the corresponding eigenfunction would be orthogonal to all members of $\hat{G}$, and so is zero. Hence $\{\gamma(a): \gamma \in \hat{G}\}$ is the group of all eigenvalues and the only eigenfunctionsare constant multiples of characters.
Why is the statement in bold true? Could you please give some hints? I know, I have to use the fact that eigenvalues of $T$ form a subgroup of the unit circle.
If $U$ is the operator induced by an invertible measure preserving transformation then $U$ is a unitary operator: $U^{*}=U^{-1}$. For such an operator eigen vectors corresponding to distinct eigen values are orthogonal:
Let $Uf=\lambda f, Ug=\mu g$ with $\lambda \neq \mu, f\neq 0$ and $g \neq 0$. Since $U$ preserves norms we see that $|\lambda|=|\mu|=1$. Now $ \langle Uf, g \rangle =\langle f, U^{*}g \rangle =\langle f, U^{-1}g \rangle=\langle f, \frac 1 \mu g\rangle$. This gives $\lambda \langle f, g \rangle =\mu \langle f, g \rangle $ because $\frac 1 {\overline {\mu}}=\mu$. It follows that $ \langle f, g \rangle =0$,