Are $(\mathbb{R},+)$ and $(\mathbb{C},+)$ isomorphic as additive groups?

I know that there is a bijection between $\mathbb{R}$ and $\mathbb{C}$, and this question asks whether they are isomorphic as abelian groups, are they referring to the additive abelian group? If so is there any simple isomorphism I can find? I know nothing about Hamel basis. Thanks.

  • 1
    $\begingroup$ NB: There is no continuous isomorphism. This illustrates the relevance of topological groups (as compared to abstract groups) in analytical contexts. $\endgroup$ – Martin Brandenburg Feb 13 '13 at 22:59

Assuming the axiom of choice, yes.

Observe that both these abelian groups are actually $\mathbb Q$-vector spaces, and they have the same dimension, so they must be isomorphic as vector spaces, and such isomorphism is also a group isomorphism. This is in fact a stronger requirement than just group isomorphism, but nevermind that.

It is consistent with the failure of the axiom of choice that these two are not isomorphic, though. So one cannot give an explicit isomorphism between them.

  • $\begingroup$ From what you saying, do you imply that there is a way to construct $\mathbb{R}$ without accepting AC? $\endgroup$ – mez Feb 15 '13 at 8:39
  • 1
    $\begingroup$ @mezhang: Sorry about the previous comment. The real numbers are constructed without choice, yes. $\endgroup$ – Asaf Karagila Feb 15 '13 at 8:52
  • $\begingroup$ ok, what about this: With the failure of AC, existence of real numbers provide no contradiction? $\endgroup$ – mez Feb 15 '13 at 13:05
  • $\begingroup$ @mezhang: We can construct the real numbers in ZF. Their existence is not only consistent, but provable. $\endgroup$ – Asaf Karagila Feb 15 '13 at 15:28
  • $\begingroup$ And in this manner $\mathbb{R}\simeq\mathbb{R}/\mathbb{Q}$. $\endgroup$ – Ash GX Nov 12 '13 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.