# Character table and conjugacy class of cyclic subgroups

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.

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.