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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

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  • $\begingroup$ I don't think the second equivalence relation should be called an action. Perhaps it is better to say that we have two equivalence relations that happen to coincide. $\endgroup$ – S. Carnahan Nov 8 '14 at 22:11
  • $\begingroup$ Perhaps this later question mathoverflow.net/questions/187529/… is related? $\endgroup$ – Yemon Choi Nov 20 '14 at 0:03
  • $\begingroup$ I have edited your question to address the issue raised by S. Carnahan. I hope this is ok with you. $\endgroup$ – Frieder Ladisch Nov 21 '14 at 20:20
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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.

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The answer is yes. For details, see Roger C. Alperin "Rational subsets of finite groups" International Journal of Group Theory Vol. 3 No. 2 (2014), pp. 53-55 (and the references cited there).

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    $\begingroup$ I saw that Alireza is editor of this journal :) $\endgroup$ – user21574 Nov 10 '14 at 3:00
  • $\begingroup$ Unfortunately, I could not find my question in the mentioned paper. $\endgroup$ – Ali Reza Ashrafi Nov 11 '14 at 4:42

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