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 $G\cong ̸H$. 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 $G^a\cong G^b⟺a\equiv b(\mathrm{mod}r)$.

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 $r^2\equiv r(\mathrm{mod}r)$, we obtain that $H^r\cong G^r$. However, since $r\equiv ̸1(\mathrm{mod}r)$, $H\cong ̸G$. This gives the required example.