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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.

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    $\begingroup$ Not quite, it is the group of homomorphisms $\chi : G \to \mathbb C^\times$. $\endgroup$
    – Sean Eberhard
    May 4 at 11:50
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    $\begingroup$ We have $\widehat{G}\cong G$ for finite abelian groups - see for example here. $\endgroup$
    – Dietrich Burde
    May 4 at 12:49
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    $\begingroup$ It is a character, yes. There are other characters, not just the Legendre symbol. So $\hat G$ is bigger. $\endgroup$
    – David A. Craven
    May 4 at 12:52
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    $\begingroup$ If $p=5$ then there is a character taking $2$ to $i$. Why do you think everything goes to $\pm1$? $\endgroup$
    – Gerry Myerson
    May 4 at 13:08
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    $\begingroup$ 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. $\endgroup$
    – Sean Eberhard
    May 4 at 13:25

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Gokimo
    May 4 at 14:42
  • $\begingroup$ But I’m starting to see where my confusion originated. I will just think a bit more about it until it’s clear to me and accept your answer if I don’t have any other doubts. $\endgroup$
    – Gokimo
    May 4 at 14:48
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    $\begingroup$ @Gokimo, have you looked at the proof of the theorem? It tells you how to construct the isomorphism! $\endgroup$
    – Mariano Suárez-Álvarez
    May 4 at 14:49
  • $\begingroup$ @MarianoSuárez-Álvarez Thank you, it is all clear now! $\endgroup$
    – Gokimo
    May 5 at 16:45

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