Infinite groups: Does a^k=b^k imply a=b? Counter-examples? Question in the end, but some background for the question: I'm reading Pinter's Book of Abstract Algebra (for fun, so not a course exercise) and Chapter 12 on equivalence classes exercise D.4 really raised a question I cannot answer:
Let $G$ be a group. In each of the following, a relation on $G$ is defined. Prove it is an equivalence relation. Then describe the equivalance class of $e$.
4. let $a \sim b$ iff there is an integer $k$ such that $a^k = b^k$
Now for a group of finite order you'll have $a^i=e$ for some $i$ and essentially all items are equivalent to $e$. I'll skip the details, it's a simpler case of the infinite one...
For infinite group the reflexivity is trivial with $a^1 = a^1$ and symmetry as well as $a^k = b^k$ works also backwards. Transitive takes only noticing that if $a^i = b^i$ and $b^j = c^j$, then $a^{ij} = c^{ij}$. Now the equivalence class of $e$ is obviously $e$ alone, as $a^k = e^k = e$ pretty much implies $a = e$ (update: I thought that this would follow from the Book's "every $a^k$ is unique proof earlier, but that only applies to elements of infinite order, so this is not true, but not strictly relevant to main question).
Now, what I'm really curious to know, that will every item in a group of infinite order be in a equivalence class of its own with this relation?
For an abelian group I've managed to deduce, that as $a^k=b^k$ we'll have $a^{-k}a^k = e = a^{-k}b^k$ and as things commute, you can rearrange it to $(a^{-1}b)^k=e$, which will again imply that $a^{-1}b=e$, so we get $a=b$.
But how-about non-abelian infinite groups? Does $a^k=b^k$ always imply $a=b$? Is there some math magic that will let us prove that without commutation, or is there an example of a group where that equality does not hold and you can arrive to same item $c$ from two places with equal power?
 A: A simple counterexample with an abelian infinite group : consider $\mathbb{S}^1 = \lbrace z \in \mathbb{C}, |z|=1 \rbrace$, with the multiplication. For any $z_0 \in \mathbb{S}^1$, there are exactly $k$ solutions in $\mathbb{S}^1$ to the equation $z^k = z_0^k$.
A: No. Look at groups of matrices, $G = PSL_n(A)$, with for example $A = \bf Q$ or $A = \bf R$ if you are interested in continuous groups.
The equation $A^k = B^k$ has many solutions in general, even if the group  $G$ is simple (assuming $A$ infinite).
A: As previously mentioned, every non-torsion-free group is a trivial counterexample (if $x^k=1$ with $k\ge 1$ and $x\neq 1$, we have $x^k=1^k$).
For a torsion-free abelian group conversely $x^k=y^k$ ($k\ge 1$) implies $x=y$. This more generally holds in torsion-free nilpotent groups (this is less trivial, and follows, e.g., from the fact that every finitely generated torsion-free nilpotent group embeds into the group of upper unipotent matrices over $\mathbf{Q}$ of size $n$ for some $n$, while the unique root property is not hard to prove there).
However there are torsion-free groups where this fails too. The simplest example is maybe the semidirect product $\mathbf{Z}\rtimes\mathbf{Z}$ with the only nontrivial action. This group has presentation $\langle x,t\mid txt^{-1}=x^{-1}$ and, as a semidirect product of torsion-free groups, is torsion-free. (In addition, it is metabelian, i.e. 2-step-solvable.)
Indeed $(tx^m)^2=t^2$ for every $m\in\mathbf{Z}$, so $t^2$ has infinitely many "square roots".
This group is actually the fundamental group of the Klein bottle.
It also has the presentation $\langle t,u\mid t^2=u^2\rangle$, where $u=tx$. Derek Holt mentioned in a comment that more generally the groups $\langle t,u\mid t^k=u^k\rangle$ ($k\ge 2$) are also examples (that such groups are torsion free and satisfy $t\neq u$ follows from the general theory of amalgams).
