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Apr 29, 2021 at 15:37 comment added user147820 @DanielMiller now I'm being a bit pedantic, for which I apologize, but I think it should be $2^{\aleph_0}$, unless we're assuming the continuum hypothesis. anyway +1, I agree this isomorphism is a very bizarre function indeed!
Dec 12, 2020 at 18:59 comment added Daniel Miller You're right. The subscript should be $\aleph_1$.
Dec 11, 2020 at 2:48 comment added user147820 @DanielMiller perhaps I'm being dense, but surely $S^1$ is not isomorphic to $\mathbb{Z}/\mathbb{Q}\times\bigoplus_\omega\mathbb{Q}$? the former is uncountable, but the latter is a cartesian product of two countable sets and is hence countable. ($\bigoplus_\omega\mathbb{Q}$ is the set of functions from $\omega$ to $\mathbb{Q}$ with finite support)
Jul 17, 2012 at 11:56 comment added Daniel Miller The proof of their existence comes from the structure theorem for divisible abelian groups. Both are isomorphic to $\mathbb{Z}/\mathbb{Q}\times \bigoplus_\omega \mathbb{Q}$.
Jul 17, 2012 at 7:31 comment added Bernikov Could you please give an example of such an isomorphism, or at least an argument to prove that these groups are isomorphic? I find it quite surprising.
Aug 21, 2010 at 18:31 history answered Daniel Miller CC BY-SA 2.5