# Isomorphic nth powers not implies isomorphic

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.
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 $G^a \cong G^b \iff a \equiv b \pmod r$.
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.