Let's define this relation over the elements of an infinite group $(G,\cdot,e)$
$$a\asymp_sb :\rightarrow\exists n\in\Bbb Z(as^n=b)$$
where $a^n$ is defined as follow
1)$a^0=e$
2)$a^{n+1}=aa^n$
Qestion 1, is it an equivalence relation?Is my proof correct for all the groups?
1- it is reflexive
$a\asymp_sa$ because $as^0=ae=a$
2- it is symmetric
If $a\asymp_sb$ then $b\asymp_sa$ because if $as^n=b$ then $a=b(s^n)^{-1}=bs^{-n}$
3- it is transitive
If $a\asymp_sb$ and $b\asymp_sc$ then $a\asymp_sc$ because if $as^n=b$ and $bs^m=c$ then $as^ns^m=as^{n+m}=c$
Qestion 2, does $G / \asymp_s$ has a name? What are some basic results about it?
Is very interesting to see that if we take $(\Bbb Z,+,0)$ then
$\Bbb Z /\asymp_ 0$ is isomorphic to $\Bbb Z $ because $a\asymp_0b$ implies $a=b$
$\asymp_ 1$ induces only one Equivalence class because $m\asymp_ 1 n$ for every pair of $m$ and $n$
$\asymp_ 2$ creates two Equivalence classes, the odd numbers and the even ones..
In my opinion is interesting to ask if in a group $G$ exist an element $1_G$ such that $\asymp_ {1_G}$ induces only one equivalence class on $G$ or in other words if every element of $G$ is is the form $b=z1_G^n$ for a fixed $z$.
Another queston could be tha following: when for group $G$ exist an element $u$ such that $|G / \asymp_u|$ is finite? In that case every element of $G$ can be written in the form $a=z_iu^n$ for some elements $z_1,z_2,...,z_{|G / \asymp_u|}$ and $\neg(z_i \asymp_u z_j)$
Qestion 3, Is there an easy and fast way to know if two elements are equivalent via $\asymp_s$
This problem looks, at first, like the conjugacy problem for groups. In fact its structure is similar to the conjugacy problem.
$\asymp_s$ -Problem - given $s\in G$, $a\in G$ and $b\in G$ do exists an $n\in \Bbb Z$ such that $$as^n= b$$?
Conjugacy problem - given $a\in G$ and $b\in G$ do exists an $\gamma\in G$ such that $$a\gamma=\gamma b$$?
The "$\asymp_s$ -Problem" is like asking if $log_s(a^{-1}b)$ has a solution. Because if $log_s(a^{-1}b)$ exists then $a\asymp_s b$:
$$as^{log_s(a^{-1}b)}=b$$
Where the map the map $log_s:\langle s\rangle\rightarrow \Bbb Z$ is the inverse map of the group power defined above $s^{-}:\Bbb Z \rightarrow G$.
Is this problem as hard as the conjugation problem? Is it related to the conjuagation in some way?