I’ve read that the Legendre symbol $$\left (\frac{\cdot}{p} \right ): x \mapsto \left (\frac{x}{p} \right ) $$ is a character of the group $G=(\mathbb{Z}/p\mathbb{Z})^\times$.
However, that means the character group $\hat G$ is the abelian group formed by the group homomorphisms $\chi : G \rightarrow \mathbb{C}^{\times}$ under multiplication, right? I’m trying to understand how this group is isomorphic to $G=(\mathbb{Z}/p\mathbb{Z})^\times$. In particular, why does it have $p-1$ elements?
Edit: to clarify, I already know that this isomorphism is a general fact for all finite abelian groups, what I don’t understand is why there are $p-1$ elements in the character group in this particular example. I know there must be $p-1$ of them, but I have trouble “picturing” them.
The character group $\hat{G}$ of a group $G$ is defined as $$\hat{G}=\lbrace \chi:G\to \mathbb{C}^\times \mid \chi\text{ homomorphism}\rbrace$$
as Sean Eberhard wrote in the comment. (note: some require that the image lies in $\mathbb{C}^1$, the unit circle, but for finite groups this does not matter). This allows for the image to be more than just $\lbrace \pm 1\rbrace$. Indeed, e.g. for $G=\mathbb{Z}/4\mathbb{Z}$, there is a character defined by $1\mapsto i$.
As Dietrich Burde mentions, it is a general fact that $\hat{G}\cong G$ for all finite abelian groups. A proof can be found in Keith Conrad's notes, https://kconrad.math.uconn.edu/blurbs/grouptheory/charthy.pdf, Theorem 3.13. In your case, as $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic, actually Theorem 3.11 suffices.