Isomorphic nth powers not implies isomorphic

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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.


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.