Suppose $G$ is a finite group and $A$ is the set of all character values of $G$. By character values, I mean entries of the character table of $G$. Let $\Gamma = \operatorname{Gal}({\mathbb{Q}(A)}/{\mathbb{Q}})$. Then $\Gamma$ has an action on the set of all conjugacy classes of $G$. Considering orbits, we get an equivalence relation on $G$.  

There is another equivalence relation on $G$ in which elements $x$ and $y$ are equivalent if and only if the cyclic subgroups generated by $x$ and $y$ are conjugate subgroups of $G$. My question is about the relationship between the classes defined by the two relations. In an exact phrase:


Question: Is it true that both equivalence relations coincide? 


Any suggestions, comments or references to this problem are highly appreciated.


Regards, Alireza



Rational Conjugacy Classes of Finite Groups
The answer is yes, and this is not difficult to see, but first you have to show that you have indeed an action of $\Gamma$ on the conjugacy classes. Let $\varepsilon$ be a primitive $|G|$-th roots of unity and let $\widehat{\Gamma}=\operatorname{Gal}(\mathbb{Q}(\varepsilon)/\mathbb{Q})$. Then $\mathbb{Q}(A)\subseteq \mathbb{Q}(\varepsilon)$ and $\Gamma$ is a homomorphic image of $\widehat{\Gamma}$ with kernel the Galois group of the extension $\mathbb{Q}(\varepsilon)/\mathbb{Q}(A)$. The action of $\Gamma$ on the conjugacy classes of $G$ is induced by the action of $\widehat{\Gamma}$ on the conjugacy classes of $G$. This action is defined as follows: For $\alpha\in \widehat{\Gamma}$, there is $k\in \mathbb{Z}$ such that $\varepsilon^{\alpha} = \varepsilon^k$. Then for $g\in G$, we set $g^{\alpha}=g^k$, which defines an action of $\widehat{\Gamma}$ on $G$ and thus on its conjugacy classes. That the orbits of this action are the same as the equivalence classes of the second relation follows from the well known fact that $\widehat{\Gamma} \cong (\mathbb{Z}/|G|\mathbb{Z})^\*$ naturally (and some elementary considerations: if $x$ and $y$ generate the same cyclic subgroup, then $x=y^k$, where we may assume that $k$ is prime to $|G|$).  

The action of $\widehat{\Gamma}$ on the conjugacy classes factors through $\Gamma$ since $\operatorname{Gal}(\mathbb{Q}(\varepsilon)/\mathbb{Q}(A))$ fixes the conjugacy classes. This follows since the characters separate the conjugacy classes.


