I've been working through Artin's Algebra on my own time, and I'm stuck on one of the questions, namely 10.5.3:
Suppose that a group G has exactly two irreducible characters of dimension 1, and let X denote the nontrivial one-dimensional character. Prove that for all g in G, X(g) = +/-1.
I've thought about this for a while and here's what I think so far: We know that all the X(g)'s are going to be roots of unity, because X is a one-dimensional character. For small numbers of conjugacy classes, I can prove this. For two classes, for instance, we need the orthogonality between characters, so the X(g) for the non-identity conjugacy class needs to be real. Therefore, it must be +/-1 for that conjugacy class. I'm having a hard time generalizing this, though.
I also don't think this approach will work in the end because it doesn't use the "exactly two" part of the question. The statement isn't true if there are more than two one-dimensional characters.
So my question is: How do you use the "exactly two" part of the question? What is it about having exactly two irreducible one-dimensional characters that lets you solve this problem?
I'm new to the site, so if I've put this question in the wrong place or if I'm not specific enough, please tell me. Thanks!