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"The multiplicative group $(\mathbb{Z}/p\mathbb{Z})^{\times}$of reduced residue classes modulo an odd prime p is a cyclic group of (even) order p − 1. Thus it has a unique character of order 2."

Why we only have one character of order 2? What has this to do with the first statement above?

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If $\chi:G\to\Bbb{C}^*$, $G$ a finite abelian group, is a character of order two, it follows that $1=\chi(g)^2=\chi(g^2)$ for all $g\in G$. This implies that

  • the kernel of $\chi$ contains all the squares in $G$, and
  • $\chi(g)=\pm1$ for all $g\in G$.

If we furthermore know that $G$ is cyclic of an even order $n=2m$, then

  • the squares form a subgroup $Q\le G$ of order $m$ and index two, and therefore
  • the coset $gQ=G\setminus Q$ for any non-square element $g$. Implying that the product of two non-squares is a square, and that the product of a square and a non-square is a non-square.

So if $\chi$ has order two, then

  1. $\chi(q)=1$ for all $q\in Q$,
  2. $\chi(x)\neq1$ for some $x\in G$. By the above observations $x\notin Q$, and $\chi(x)=-1$.
  3. If $y\notin Q$, then $yQ=xQ$ implying that $y=xq$ for some square $q\in Q$.
  4. So $\chi(y)=\chi(xq)=\chi(x)\chi(q)=(-1)\cdot1=-1$.
  5. Items 1 and 4 imply that $\chi$ is the unique function taking the value $+1$ in $Q$, and the value $-1$ elsewhere.
  6. It easily follows that the function defined in the previous bullet is actually a character of $G$.

As Hempelicious pointed out, the character group of an abelian group is isomorphic to the group itself. But, the isomorphism is not natural. In other words, it relies on our ability to write a finite abelian group as a direct product of cyclic groups. Such a decomposition is not unique, and this leads to complications similar to the well known result that the dual of a f.d. vector space is isomorphic to the vector space itself, but to write down the isomorphism you need to specify a basis. In sharp contrast to the simpler fact that the double dual is naturally isomorphic to the vector space we started with.

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  • $\begingroup$ I guess another way of saying this is that in the abelian case, the character is actually a homomorphism. Since there is only one subgroup of order $2$ in $\mathbb{C}^\ast$, there's still most one such character. $\endgroup$ – Hempelicious Dec 13 '18 at 0:32
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    $\begingroup$ @Hempelicious Cyclicity is also essential. If $G=C_2\times C_2$ then it has three characters of order two. The generators of the two factors can both be mapped to $\mapsto\pm1$. We get a character of order two unless we make the trivial selection. Similarly, that group has three subgroups of index two, and each can serve as the kernel of a character of order two. $\endgroup$ – Jyrki Lahtonen Dec 13 '18 at 4:32
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    $\begingroup$ Yes, sorry, I left that part out, but I meant once we know they're homomorphisms, they can be counted by the index 2 subgroups. This wouldn't be true if the image had more than one element of order 2. $\endgroup$ – Hempelicious Dec 13 '18 at 8:02
  • $\begingroup$ Ok, @Hempelicious. We seem to agree :-) $\endgroup$ – Jyrki Lahtonen Dec 13 '18 at 8:06
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Characters of finite abelian groups are in 1-1 correspondence with the groups themselves. See this question for more details.

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  • $\begingroup$ Honestly: Never heard of that fact. I mean with that my question is obviously answered but its not really obvious that your assumption is correct. $\endgroup$ – user625682 Dec 12 '18 at 22:41
  • $\begingroup$ @FabianSchneider: not an assumption, it's proved at the link. Please read the answers there. $\endgroup$ – Hempelicious Dec 12 '18 at 23:49

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