If $a^{k+1}b^{k-1}=b^{k-1}a^{k+1}$, where $a,b$ belong to a group $G$. Can I conclude that $ab=ba$, if $k\ne1$ and $k\ne -1$? I encountered this while solving a problem. This is not a separate problem.
If $a^{k+1}b^{k-1}=b^{k-1}a^{k+1}$, where $a,b$ belong to a group $G$. Can I conclude that $ab=ba$, if $k\ne1$ and $k\ne -1$?
 A: No, also if $k \neq \pm 1$. Just a counterexample, for $k = 2$: $BS(3,3) = \mathbb{Z}^{*2}/<a^3 b = ba^3>$ is not abelian, by the theorem 1.3 here.
A: Easier and finite counterexample with $k=3$ or $k=5$, take the quaternion group $Q=\{\pm1, \pm i, \pm j, \pm k\}$.
A: In the free product of $C_{k+1}=\langle a'\rangle$ and $C_{k-1}=\langle b'\rangle$, $a'$ and $b'$ commute as soon as $\{k+1,k-1\}\cap\{-1,1\}$ is empty, that is, if $k\notin\{-2,0,2\}$. (Also $a'$ and $b'$ don't commute in some suitable finite quotient of this free product.)
If $k=2$, an example was given by dcolazin. More generally, if we take elements $a$ of order 3 and $b$ of order 2 in the non-abelian group of order 6, we have $[a^{k+1},b^{k-1}]=1$ for all $k\in 3\mathbf{Z}+2$, while $a,b$ do not commute. The case $k=-2$ follows by changing $(a,b)$ to $(b^{-1},a^{-1})$.

Hence, for every $k\in\mathbf{Z}\smallsetminus\{0\}$, there exists a group $G$ (which can be chosen to be finite) and elements $a,b$ such that $[a,b]\neq 1$ and $[a^{k+1},b^{k-1}]=1$.

Note: the smallest counterexample has size unbounded with respect to $k$. Indeed, when $G$ has order $\le n$ and $k=n!$, then the implication holds, so the smallest counterexample for $k=n!$ has order $>n$.

The result can be interpreted as follows: for $k\in\mathbf{Z}$ consider the group $G_k$ with presentation
$$\langle a,b\mid [a^{k+1},b^{k-1}]=1\rangle.$$
Then for $k\neq 0$, the commutator $[a,b]$ is nontrivial. One can certainly say more about this group; for instance it's torsion-free. For $k\neq 0$, the group $G_k$ is large (=virtually surjects homomorphically onto a nonabelian free group), because it surjects onto both the free products $C_{k+1}\ast\mathbf{Z}$ and $C_{k-1}\ast\mathbf{Z}$ (as we see by killing either $a^{k+1}$ or $b^{k-1}$).
