Let $G$ be a finite group. Two elements $x$ and $y$ of $G$ are said to be rationally conjugate, written $x \sim_{r} y$, if and only if $\langle x\rangle$ and $\langle y\rangle$ are conjugate subgroups of $G$. Let $A$ be the set of all character values of $G$, and let $H$ be the Galois group of the field extension $\mathbb{Q}(A)/\mathbb{Q}$. Is it true that the orbits of $H$ on conjugacy classes of $G$ are the same as the equivalence classes of $\sim_{r}$? Is there any proof or reference for this result?
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4$\begingroup$ The answer is yes, and I think this is clear, once you have shown that there is a well-defined action of $H$ on the conjugacy classes: Let $\varepsilon$ be a primitive $|G|$-th root of unity. Then the Galois group $\Gamma$ of $\mathbb{Q}(\varepsilon)/\mathbb{Q}$ acts on the conjugacy classes, and $H$ is a factor group of $\Gamma $. It is well known and not difficult to show that the orbits of $\Gamma$ are the rational conjugacy classes. (c.f. Serre, Linear Representations of Finite Groups, §12.4.) $\endgroup$– Frieder LadischCommented Nov 19, 2014 at 12:47
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1$\begingroup$ An earlier question which may be related to this one? mathoverflow.net/questions/186581/… $\endgroup$– Yemon ChoiCommented Nov 20, 2014 at 0:02
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1$\begingroup$ @FriederLadisch, make this an answer! $\endgroup$– Nick GillCommented Nov 20, 2014 at 16:37
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$\begingroup$ In fact, this seems to be a duplicate of the question linked by @YemonChoi, only that the question here is formulated a little better. I have made my comment an answer to the old question, which already had an answer. I think this question can be closed as duplicate. $\endgroup$– Frieder LadischCommented Nov 21, 2014 at 20:05
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