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Recall that two elements $h_1,h_2$ of a finite group $G$ are called conjugate when $h_2 = gh_1 g^{-1}$ for some $g \in G$, and algebraic-conjugate when $h_2 = gh_1^a g^{-1}$ for some $a \in (\mathbb{Z}/\mathrm{order}(g))^\times$; equivalently when the cyclic subgroups $\langle h_1\rangle$ and $\langle h_2 \rangle$ are conjugate. In general, algebraic-conjugacy is a much coarser relationship than conjugacy. But sometimes they are equivalent: in the symmetric group, for example, both conjugacy and algebraic-conjugacy are simply the cycle structure of the element.

How common is it for algebraic-conjugacy to equal conjugacy?

One reason why you might ask this is the following. Suppose you have a representation $V$ of $G$. Then its character is, of course, a class function: $\mathrm{tr}_V(g)$ depends only on the conjugacy class of $g \in G$. But if $V$ happens to be defined over $\mathbb{Q}$, then $\mathrm{tr}_V(g)$ in fact depends only on the algebraic conjugacy class of $g$. In particular, if every representation of $G$ is defined over $\mathbb{Q}$, then by character theory, algebraic-conjugate elements are automatically conjugate.

An intermediate relation between conjugacy and algebraic conjugacy is what I'll call, based on its character-theory interpretation, real conjugacy: I'll say that $h_1$ and $h_2$ are real-conjugate if $h_2 = g h_1^{\pm 1}g^{-1}$ for some $g$ and some sign.

How common is it for algebraic-conjugacy to equal real-conjugacy?

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    $\begingroup$ I think you got the title backwards - conjugacy always implies algebraic conjugacy, the question is about the converse. $\endgroup$
    – Wojowu
    Commented Jun 23, 2023 at 19:47
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    $\begingroup$ I don't know who coined "algebraic conjugacy" but I find the terminology not much intuitive. Anyway, this holds iff any two elements generating the same cyclic subgroup are conjugate. And this is known to be equivalent to the condition that all reps are defined over $\mathbf{Q}$. For instance, this holds for all Weyl groups. I remember there's a whole Springer LNM on this topic, but didn't immediately find it. $\endgroup$
    – YCor
    Commented Jun 23, 2023 at 21:02
  • $\begingroup$ Ah, this book is mentioned at this question: Denis Kletzing, Structure and Representations of Q-Groups, Springer Lecture Notes in Mathematics 1084, 1984. Available here, apparently currently without restriction. $\endgroup$
    – YCor
    Commented Jun 23, 2023 at 21:20
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    $\begingroup$ @YCor, I think this condition just means all characters are rational valued. Can't their be Schur index issues to define the representation? $\endgroup$ Commented Jun 23, 2023 at 22:00
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    $\begingroup$ @BenjaminSteinberg Oops, you're right. Just to summary and fix the previous comment: in a finite group, the condition that all generators of conjugate cyclic subgroups are conjugate means that any two elements generating the same cyclic subgroup are conjugate, and mans that characters are rational-valued. In Kletzing's book, it is called Q-group. A stronger condition is that all reps are definable over $\mathbf{Q}$, and is satisfied by all Weyl groups (let's say "strong Q-group"). The quaternion group of order 8 is a Q-group but not a strong Q-group. $\endgroup$
    – YCor
    Commented Jun 24, 2023 at 3:15

1 Answer 1

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It is reasonably standard to call a finite group $G$ a rational group if all its complex irreducible characters are rational-valued ( equivalently, if $g \in G,$ then $g$ is conjugate within $G$ to all generators of $\langle g \rangle )$. Thanks to work of Feit-Seitz, and of J.G. Thompson,( see Feit-Seitz "On rational finite groups and related topics", Illinois Journal of Mathematics, 33,1 Spring 1988 and J.G. Thompson "Composition factors of rational finite groups", Journal of Algebra, 319,2,Jan 2008,558-594 ( Feit memorial issue)) it is known that every Abelian composition factor of a rational finite group has order at most $11$, and that there are only five possibilities for isomorphism types of non-Abelian composition factors for $G$ (other than alternating composition factors). The five possible non-alternating non-Abelian composition factors are : ${\rm PSp}(4,3),{\rm Sp}(6,2),{\rm O}^{+}_{8}(2)′, {\rm PSL}(3,4)$ and ${\rm PSU}(4,3)$.

Also, R. Gow proved that solvable finite rational groups are $\{2,3,5 \}$-groups, and P.Hegedus refined this somewhat. In this sense, the rational finite groups are somewhat rare.

On the other hand, if we call an individual element $g$ of a finite group $G$ rational if $g$ is conjugate (within $g$) to all generators of $\langle g \rangle,$ then there may be many rational elements within a non-rational group $G$. For example, all unipotent elements in ${\rm GL}(n,q)$ are rational, so determination of which elements of ${\rm GL}(n,q)$ reduces to determining which semisimple elements are rational, and there are many other instances of similar behaviour among classical groups and other finite groups of Lie type.

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  • $\begingroup$ Thanks. Which five nonabelian groups can arise? Or are there specific works of Feit-Seitz and Thompson I should look at? $\endgroup$ Commented Jun 26, 2023 at 14:13
  • $\begingroup$ @TheoJohnson-Freyd : Now answered in body of question. $\endgroup$ Commented Jun 27, 2023 at 10:10

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