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.



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



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> How common is it for algebraic-conjugacy to equal real-conjugacy?
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How often does algebraic-conjugacy imply conjugacy?
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.


