In a group, is it true that $x=y\Longleftrightarrow x^n=y^n$?

Question

Follow up to this question. I am reading this paper. In the proof of Theorem 2.1, second paragraph, the Author says:

We are in a group, so we can simplify with respect to the group law ”◦”.

and goes ahead by:

$x_n^{\circ m}=x_q^{\circ p}\Leftrightarrow x_n^{\circ mq}=x_q^{\circ pq}$

($n,m,p,q\in\mathbb{N}^+$).

To my understanding, this is: $$ x = y \Longleftrightarrow x^n = y^n $$ for $x,y$ in a group and $n\in\mathbb{N}^+$. Is this really true as it is stated? I tried to prove it and failed, and could not find a proof by googling. Please note that the actual group of interest in the Theorem is, from the claim:

a non-trivial complete totally ordered Abelian group which is dense in itself

Answer 1

It's not true exactly as stated: if the group is not torsion-free (for example, any finite group), then it's false: if $x^n = e$ then $x^n = e^n$.

For torsion-free abelian groups it's true: if $x^n = y^n$ then $(xy^{-1})^n = e$, torsion-free implies $xy^{-1} = e$ and so $x = y$.

Ordered groups are necessary torsion-free: if, for example, $x > e$ then also $x^n = x\cdot x^{n-1} > e \cdot x^{n - 1} = x \cdot x^{n - 2} > \ldots > e$, so $x^n \neq e$, similarly if $x < e$ then $x^n \neq e$.