# Exactly two irreducible characters of dimension 1

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!

• You wrote your question fine. Provided citation, background, your own thoughts, asked clearly. +1. One thing I edited into your question was additional relevant tags. They help users search for questions relevant to their interests much easier. – whacka Nov 21 '15 at 18:11

Square the nontrivial character. The result can't be the nontrivial character again, so (because there are exactly two characters of dimension $1$) it must be the trivial character.

In other words, the characters of dimension $1$ form a group under pointwise multiplication (it can be identified with the group $\text{Hom}(G, \mathbb{C}^{\times})$, which is noncanonically isomorphic to the abelianization $G/[G, G]$ when $G$ is finite), and there's only one group of order $2$.

One-dimensional characters are group homomorphisms $\chi:G\to\mathbb{C}^\times$. If $G$ is finite then the range is comprised of elements of finite order, i.e. roots of unity, so they land in the circle group $\mathbb{S}^1$. In any case, any homomorphism from $G$ into an abelian group $A$ factors uniquely through its abelianization,

$$G\to A \quad\textrm{equals}\quad G\xrightarrow{\textrm{proj}} G^{\textrm{ab}}\longrightarrow A.$$

Note $G^{\textrm{ab}}=G/[G,G]$. This is because $[G,G]$ must be in the kernel if $A$ is abelian, and $G^{\textrm{ab}}\to A$ is uniquely determined by $G\to A$.

Thus, one-dimensional characters of $G$ correspond to characters of $G^{\textrm{ab}}$. And conversely, since any character $G^{\rm ab}\to\mathbb{S}^1$ may be precomposed with $G\xrightarrow{\textrm{proj}}G^{\textrm{ab}}$ yielding a one-dimensional character of the original group $G$. Indeed, these two correspondences are mutually inverse, so one-dimensional characters of $G$ are in bijective correspondence with characters of $G^{\textrm{ab}}$.

If $A$ is any abelian group, its character group has size $|A|$ (indeed it's isomorphic to $A$, but not canonically... unless $A$ is the Klein-four group, but that's another story). If your $G$ has two one-dimensional characters then $G^{\textrm{ab}}$ has size two, so must be $C_2$. What must be the range of the nontrivial group homomorphism $C_2\to\mathbb{S}^1$ be?