Let $F=\mathbb{F}_{p}$. Let $E=\mathbb{F}_{p^{2}}$ be the degree 2 extension over F. Let $\tau$ be the nontrivial Galois automorphism of E. Let $\phi$ be a degree one representation of $E^{*}$. We know that there are $p^{2}-1$ such degree one representations. Now we define $\phi^{\tau} : G \to \mathbb{C}^{*}$ as $\phi^{\tau}(x)=\phi(\tau (x))$ $\forall$ $x$ in $E^{*}$. Clearly $\phi^{\tau}$ is a degree one representation of $E^{*}$ too. Now what I want to know is the following:
How many such representation $\phi$ are there such that $\phi=\phi^{\tau}$?
In this context, I must say that I am reading the representations of $GL_{2}(\mathbb{F}_{p})$. This question appears when we start constructing the cuspidal representations. The text says that the number of $\phi$ such that $\phi \neq \phi^{\tau}$ is $p(p-1)$ which suggests that the answer to my question $p-1$. But I can't find the logic. May be I am overlooking something very simple.
Thanks in advance!!!
Each $\phi$ is defined by its value on some primitive element $\xi$ of $E$, that is an element generating the group $E^\times$. Now $\phi^\tau(\xi)= \phi(\xi^\tau)=\phi(\xi^p)=\phi(\xi)^p$. Then $\phi^\tau=\phi$ iff $\phi(\xi)^p=\phi(\xi)$. This means that $\phi(\xi)$ is a $(p-1)$-th root of unity. Therefore there are $p-1$ $\phi$ with $\phi^\tau=\phi$.