Let $R(q)$ be the number of real characters mod $q$. A character $\chi \mod q$ is called real if $\chi(a)\in\mathbb{R}$ for every $a\in \mathbb{Z}$, which means $\chi(a)\in\{-1,1\}$ for every $a\in\mathbb{Z}$ with gcd$(a,q)=1$.

I want to show that this $R$ is multiplicative, so $R(ab)=R(a)R(b)$ for $a,b\in\mathbb{Z}$ with gcd$(a,b)=1$. I'm trying to prove this with induced characters, but I'm not getting it completely. Can someone help me to understand it? Thanks!

  • $\begingroup$ Do you mean number of irreducible real characters? Also, how would induced characters help? $\endgroup$ – A. Pongrácz Dec 30 '18 at 20:37
  • $\begingroup$ @A.Pongrácz not specifically irreducible.. Also, I thought, characters mod $ab$ might be induced by characters mod $a$ and mod $b$? Not sure if this could help $\endgroup$ – jbuser430 Dec 30 '18 at 20:43
  • $\begingroup$ There are infinitely many real characters $\pmod q$ for every $q$. E.g., any positive integer multiple of the trivial character is a real character. (Check out the definition of a character again.) I believe you want to talk about irreducible characters. $\endgroup$ – A. Pongrácz Dec 30 '18 at 20:45
  • $\begingroup$ @A.Pongrácz I think we only consider irreducible characters then in my course! :) $\endgroup$ – jbuser430 Dec 30 '18 at 20:47
  • $\begingroup$ It seems to me that you are also using the word "induced character" in a loose sense. Mind that it has a meaning. How would you "induce" a(n irreducible!) character of $ab$ from a(n irreducible!) character of $a$ and one of $b$? $\endgroup$ – A. Pongrácz Dec 30 '18 at 20:47

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