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Let $G$ be a topological group. We define an equivalence relation on $G$ as follows:

For $a,b\in G$ we set $a\sim b$ if the following two maps are topologically conjugate:

$$x\mapsto ax,\qquad x\mapsto bx$$

By topological conjugacy of two maps $f, g$ on a topological space $G$ we mean, as usual, existence of a homeomorphism $H:G\to G$ with $Hf=gH$.

Obviously the usual group theoretical conjugate elements are equivalent in this new sense. But possibly the equivalence classes are larger than algebraic conjugacy class.

In the finite (discrete ) group case the equivalence relation is the following: two elements are equivalent if the corresponding permutations are conjugate in $P(G)$, the permutation group of $G$.

My first question:

In the finite group case, the cardinality of the conjugacy class of an element $a$ is $\frac{|G|}{|C(a)|}$ where $C(a)$ is the centralizer of $a$. Now what is a formula for the cardinality of an equivalence class containing $a$. I mean a formula in terms of $G$ itself not in terms of $P(G)$. More precisely can we canonically associate a subgroup $\tilde{C}(a)\subset G$ to every element $a\in G$ such that the equivalence class $[a]$ is in bijection correspondance to $\tilde{C}(a)$ in a canonical way. Another question: From undergraduate group theory I remember that various equivalence relations on finite groups enable us to obtain some useful counting theorems. Even Sylow's theorems were direct or undirect consequence of some equivalence relations. So I wonder is this equivalence relation useful to produce some new counting results?

My second question:

For the obvious case $G=\mathbb{R}$ we get triviality. There are only two equivalence classes. On the other hand for circle case every equivalence class is the (singleton) conjugacy class since the rotation number is a topological invariant. Are there examples for which the measure of an equivalence class approach to zero or approach to 1 (w.r.t. the Haar measure)? That means: can one produce a sequence of example such that an equivalent class has measure $1/n$ or $1-1/n$? This possibly prevent us to give finite group example.

However one should first answer to: Is every equivalence class necessarily measurable?

Improve this question
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    $\begingroup$ For a finite group two elements are equivalent if and only if they have the same order. This is because if g has order n then it acts as a disjoint union of |G|/n n-cycles corresponding to the cosets $\endgroup$
    – Benjamin Steinberg
    Commented Sep 22, 2023 at 12:26
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    $\begingroup$ The number of elements of order dividing n divides the order of the group but not the number of elements of order exactly n. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 22, 2023 at 12:30
  • $\begingroup$ @BenjaminSteinberg Thank you for your interesting comment. Since "order" plays a crucial role so your comment is a motivation to consider a concept of order for elements of a topological group. The order is not a number but is an equivalent class. $\endgroup$
    – Ali Taghavi
    Commented Sep 22, 2023 at 12:57
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    $\begingroup$ It is an equivalence relation and an equivalence class. $\endgroup$
    – Moishe Kohan
    Commented Sep 22, 2023 at 22:49
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    $\begingroup$ In a discrete group $G$, $g,h$ are equivalent iff $g$, $h$ have the same order and the (possibly infinite) index of $\langle g\rangle$ and $\langle h\rangle$ is the same. (If $G$ is not virtually infinite cyclic, the latter condition can be dropped.) $\endgroup$
    – YCor
    Commented Sep 25, 2023 at 8:45

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