Legendre symbol as a group character

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.

• Not quite, it is the group of homomorphisms $\chi : G \to \mathbb C^\times$. May 4 at 11:50
• We have $\widehat{G}\cong G$ for finite abelian groups - see for example here. May 4 at 12:49
• It is a character, yes. There are other characters, not just the Legendre symbol. So $\hat G$ is bigger. May 4 at 12:52
• If $p=5$ then there is a character taking $2$ to $i$. Why do you think everything goes to $\pm1$? May 4 at 13:08
• Actually the Legendre symbol is the only nontrivial real-valued character. Indeed, such a character satisfies $\chi(x^2) = \chi(x)^2 = 1$ for all $x$. If $y$ is an element such that $\chi(y) = -1$ then $y$ must be a nonsquare and $\chi(x^2y)=-1$ for all $x$, so $\chi$ is the Legendre symbol. May 4 at 13:25

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.
• Thank you for your answer. Perhaps I didn’t choose the right words in my question. I already know that this isomorphism is a general fact for all finite abelian groups, what I don’t see is why there are $p-1$ elements in the character group in this particular example. May 4 at 14:42