TITLE: number of characters needed to separate conjugacy classes
QUESTION [11 upvotes]: It is very well-known that, if $G$ is a finite group, then the (complex) characters of $G$ separate the conjugacy classes; that is, if $g, h \in G$ are not conjugate, then there exists a character $\chi$ such that $\chi(g) \ne \chi(h)$. However, I have been wondering about the minimal number of characters needed to separate all the classes.
Consider the (very basic) example $G=C_n$, a cyclic group of order $n$. Then it is possible to find a single character $\chi$ which is injective on $G$, so the answer is $1$ in this case.
Looking at a few character tables using GAP, it seems the answer is often 1.
Is there any literature on this? Any reason why this number could be much easier to determine than I think? Is it always 1 ? I may very well have overlooked something simple.
Note that the answer would be different if you only allowed irreducible characters (then for $C_2\times C_2$, it is 2); it's also different if you allow virtual characters, I guess.
Thanks!
Pierre

REPLY [2 votes]: Obviously Peter Müller's argument is perfectly fine. Here is a different way of doing it.
This algorithm yields a character that separates the conjugacy classes. It is terrible to compute with.
Index a set of representatives of the conjugacy classes $\{g_i\}$. Associate to every character $\chi$, the set of ordered pairs 
$$D(\chi) = \left\{ (g_i,g_j): g_i \neq g_j, { \mathrm {but } }\  \chi(g_i) = \chi(g_j)\right\}.$$
Now we show that if $D(\chi)$ is nonempty, there exists an irreducible character $\chi'$ and a positive integer $N$ (depending on $\chi$ and $\chi'$) such that $D(N\chi + \chi')$ is strictly contained in $D(\chi)$. 
To see this, set $s $ to be the infimum of $|\chi(g_i) - \chi(g_j)|$ over all pairs of distinct representatives $(g_i,g_j)$ that are not in $D(\chi)$ 
(this is an infimum over a finite set of nonzero numbers). Pick $(g,h) \in D(\chi)$; there exists an irreducible $\chi'$ such that $\chi'(g) \neq \chi'(h)$. Select $N$ a positive integer so large that $Ns > 2\chi'(1)$, and set $\psi = N\chi + \chi'$, a character.
If $\chi(g) \neq \chi(g')$, then $\psi (g) -\psi(g') = \left( N(\chi(g) - \chi(g')\right) + (\chi' (g) - \chi(g'))$; the first parenthesized term has absolute value at least $Ns$, while the second one has absolute value at most $2 \chi'(1)$. Hence $D(\psi)$ is contained in $D(\chi)$.
On the other hand, $(g,h)$ is not in $D(\psi)$, since $\psi(g) - \psi (h) = 
\chi'(g) - \chi'(h)$.
This process (begin with any nontrivial character, and continue) must terminate, resulting in a character $\rho$ with empty $D(\rho)$, which is exactly what we want.