Let $G = \mathbb{Z}/p\mathbb{Z}$ with $p$ an odd prime.
If $p \nmid a$ then multiplication by $a$ on the elements of G is bijective and therfore this is an permutation on G.
Define $\chi(a)$ as the signum of this permutation.
Show that $\chi(\cdot)$ is a non-trivial character on $(\mathbb{Z}/p\mathbb{Z})^{\times}$.
The character-part i got. I struggle with the non-trivial.
So I need to find a permutation with signum $-1$. Thought about using that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic, so that it has an generator $w$. But why defines multiplication with $w$ a (p-1)-cycle. (then it would have signum $-1$...).
Well, just write down the cycle that $w$ forms starting from $1$: it maps $1$ to $w$, $w$ to $w^2$, $w^2$ to $w^3$, and so on. Since no positive power of $w$ is $1$ before $w^{p-1}$, this forms a cycle of length $p-1$: $(1\ w\ w^2\ \dots\ w^{p-2})$.