Restriction of multiplicative characters of $\mathbb{F}_{q^n}$

Question

I'm trying to prove the following:

Let $\eta$ be a multiplicative character of $\mathbb{F}_{q^n}$ of order $m$. Then, the restriction $\eta^\ast$ of $\eta$ to $\mathbb{F}_{q}$ is a character of order $\dfrac{m}{gcd(m,(q^n-1)/(q-1))}$. The farther I can get is that $ord(\eta^\ast)$ divides $gcd(m,q-1)$ (since its order must divide $m$), can anyone give me any hint to solve this?

Answer 1

Hint:

• If $g$ is a generator of the multiplicative group $\Bbb{F}_{q^n}^*$ and $N:=(q^n-1)/(q-1)$, then $\gamma:=g^N$ is a generator of the subgroup $\Bbb{F}_q^*$.
• Can you show that the order of a multiplicative character $\eta$ of $\Bbb{F}_{q^n}$ (resp. of $\Bbb{F}_q$) is equal to the order of $\eta(g)$ (resp. of $\eta(\gamma)$)?