Characters and conjugacy classes This comes up in reading David Speyer's answer to this question. Given a finite group $G$ and two non-conjugate elements $x, y,$ how does one construct a unitary representation $\rho$ of $G$ such that $\rho(x)$ and $\rho(y)$ have different traces? (The same question makes sense for infinite groups, but it is far from clear that this is always possible in the infinite setting, even if you drop "unitary").
 A: Actually, the deleted post by George McNinch was on the right track, at least if one knows the central idempotents of the group algebra $\mathbb{C}G.$ If take the regular module $\mathbb{C}G$ (which affords a unitary representation with resepect to the standard basis of group elements), then the central idempotent $e_{\chi},$ associated to the irreducible character $\chi,$ is represented by a diagonal idempotent matrix of trace $\chi(1)^{2}.$
The right $\mathbb{C}G$ module $e_{\chi}\mathbb{C}$ affords a unitary representation with character $\chi(1) \chi.$ For some choice of $\chi,$ we have $\chi(x) \neq \chi(y),$ so the representation afforded by $e_{\chi} \mathbb{C}G$ will do.
A: I believe you can always do this by taking a rep of the cyclic group generated by $x$, and inducing this up to the whole group. (EDIT: I should mention that this was suggested in comments.) If $y$ is not conjugate to a power of $x$, the trivial will suffice.  If $y$ is a power of $x$, then some non-trivial 1-dimensional representation of $\langle x\rangle$ will work.  
Let $n$ be the order of $x$.  The action of the normalizer of $\langle x\rangle$ by conjugation on this subgroup gives a map to $(\mathbb{Z}/n\mathbb{Z})^\times$ with kernel its centralizer.  Let $U$ be the image of this map.  We just have to choose a representation where the character values of $x$ and $y$ (which are $n$th roots of unity) aren't conjugate under $U$ acting as a subgroup of the Galois group of $\mathbb{Q}[e^{2\pi i/n}]$.  If they are always conjugate, that precisely shows that $x$ and $y$ are conjugate as elements of the group.
