It's easy to show that $\mathbb{Q}$ and $\mathbb{Q}\times\mathbb{Q}$ are not ring isomorphic as the first one has no zero divisors where as the second one has zero divisors. But I can't find any solution in case of group isomorphism.
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What are the operations? – Aniruddha Deshmukh Mar 17 '19 at 13:10
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2Possible duplicate of Are the groups $\mathbb{C}$ and $\mathbb{R}$ isomorphic? The additional question there about $\mathbb Q$ and $\mathbb Q[i]$ answers this question. – GEdgar Mar 17 '19 at 13:16
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More generally there's the notion of $\mathbf{Q}$-rank of an abelian group $G$, which is obviously an isomorphism invariant. Namely, this is the maximal $k$ such that $G$ has a subgroup isomorphic to $\mathbf{Z}^k$ ($\infty$ if there's no max). Exercise: what's the $\mathbf{Q}$-rank of $\mathbf{Q}^n$? – YCor Mar 17 '19 at 14:37
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In $\mathbb{Q}$, as a group under addition, any two elements are commensurable; that is, for any $p,q$, there is some $r$ such that both $p$ and $q$ are multiples of $r$. Consequently, every finitely generated subgroup is cyclic.
The same is not true of $\mathbb{Q}\times\mathbb{Q}$; $(1,0)$ and $(0,1)$ are incommensurable.
jmerry
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Since $\Bbb Z\hookrightarrow\Bbb Q $ is a ring epimorphism, the scalar restriction functor $\operatorname{Mod}_{\Bbb Q}\to\operatorname{Mod}_{\Bbb Z} $ is fully faithful, hence every isomorphism of abelian groups $\Bbb Q\to\Bbb Q^2$ lifts to a $\Bbb Q $-vector space isomorphism. Since they have different dimension as vector space, they cannot be isomorphic as abelian groups.
Fabio Lucchini
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