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This is an exercise from J.J.Rotman's book:

Prove that the following groups are all isomorphic:

$$G_1=\frac{\mathbb R}{\mathbb Z},G_2=\prod_p{\mathbb Z(p^{\infty})}, G_3=\mathbb R\oplus\big(\frac{\mathbb Q}{\mathbb Z}\big)$$

What I have done is:

Since $tG_1=\frac{\mathbb Q}{\mathbb Z}$, which $t$ means the torsion subgroup; and the fact that $G_1\cong tG_1\oplus\frac{G_1}{tG_1}$ so I should show that $\frac{\mathbb R}{\mathbb Q}\cong\mathbb R$. A theorem tells me that $\frac{\mathbb R}{\mathbb Q}$ is a vector space over $\mathbb Q$ because it is abelian divisible torsion-free. The same is true for $\mathbb R$. I didn't work on a basis for any of these infinite structures good, so I can't go ahead well. :(

For $G_2$ the only first idea to me is $\frac{\mathbb Q}{\mathbb Z}\cong\sum_p{\mathbb Z(p^{\infty})}\leq G_2$.

Any helps or suggestions? Thanks.

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$\sum_p {\mathbb Z}(p^\infty)$ is equal to $t G_2$. I believe it is proved in Rotman's book that any torsion-free divisible group can be given the structure of a vector space over ${\mathbb Q}$, so that applies to $G_2/tG_2$. Now use the result that vector spaces having the same dimensions are isomorphic. – Derek Holt Nov 13 '12 at 14:55
@DerekHolt: So, as $\frac{G_2}{tG_2}$ can be regarded as a number of copies of $\mathbb Q$ then should I work on $c$ (the cardinality of the continuum) in order to show taht this direct sum of $c$ copies of $\mathbb Q$ is isomorphic to $\mathbb R$?? – Babak S. Nov 13 '12 at 15:08
Yes, my guess is that that is the method intended by Rotman. – Derek Holt Nov 13 '12 at 17:01
@DerekHolt: Thanks so much Prof. About my first approach, I have been thinking yet: Can we easily conclude that if $\mathcal{B}$ be a basis for vector space $\mathbb R$ over $\mathbb Q$ then $\mathcal{B}+\mathbb Q$ is a basis for vector space $\frac{\mathbb R}{\mathbb Q}$ over $\mathbb Q$? Thanks. – Babak S. Nov 13 '12 at 18:47

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