Let $G$ be a non-abelian finite group. Let $C_1$ and $C_2$ be distinct conjugacy classes in $G$ with following conditions:
1) $C_1$ and $C_2$ contain of elements of same prime order $p$.
2) For every complex irreducible character of $G$, its values on $C_1$ and $C_2$ are complex conjugates (in case they are real, they should be same).
Can we conclude that the cyclic subgroup generated by any element of $C_1$ and one such by any element of $C_2$ are conjugate in $G$? (If yes, how? and if not, what is example?)
In the book on Finite Simple Groups (First proceeding) an article mentions such fact concerning $G$ to be a Janko group and $C_1$ and $C_2$ certain conjugacy classes of elements of same prime order.
Best Answer
Let $C_1'$ be the conjugacy class $\{ g^{-1} : g \in C_1\}$. Then $\chi(g^{-1})$ is the complex conjugate of $\chi(g)$ for all characters $\chi$ and $g \in G$, so the hypothesis says that all irreducible characters of $G$ take the same values on $C_1'$ and on $C_2$. So $C_1'=C_2$ and the answer to the question is yes.