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