# Show that $\chi(\cdot)$ is a non-trivial character on $(\mathbb{Z}/p\mathbb{Z})^{\times}$.

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 $$p-1$$: $$(1\ w\ w^2\ \dots\ w^{p-2})$$.