Isomorphic nth powers not implies isomorphic

Definition

Suppose $n>1$. Then, we can find groups $G,H$ such that $G^{n}\cong H^{n}$ (i..e, the $n^{th}$ direct powers are isomorphic groups) but $G\not \cong H$. In fact, we can choose both $G$ and $H$ to be countable torsion-free (i.e., aperiodic) abelian groups.

Facts used

1. There exist abelian groups whose isomorphism classes of direct powers have any given period: This states that for any $r$, we can find a (countable torsion-free) abelian group $G$ such that $G^{a}\cong G^{b}\iff a\equiv b{\pmod {r}}$.

Proof

Pick $r=n$ from fact (1), let $G$ be the group $G$ as arising from fact (1), and let $H=G^{r}$. Then, $H^{r}=(G^{r})^{r}=G^{r^{2}}$ Since $r^{2}\equiv r{\pmod {r}}$, we obtain that $H^{r}\cong G^{r}$. However, since $r\not \equiv 1{\pmod {r}}$, $H\not \cong G$. This gives the required example.