Proving without Zorn's Lemma: additive group of the reals is isomorphic to the additive group of the complex numbers

Question

Let $(\mathbb{R},+)$ be the additive group of the reals and $(\mathbb{C},+)$ be the additive group of the complex numbers. Prove that those groups are isomorphic.

I think I got a solution using the fact that the real numbers have a Hamel Basis $B$, then proving that $B\times {0}\cup 0\times B$ is a Hamel basis for the complex numbers and noting that these sets have the same cardinality. However this result relies on Zorn's lemma and is thus not very elementary. Is there a simpler way to get it?

Answer 1

Unfortunately, there is no more elementary argument than going through some form of AC, because the result actually does depend on some amount of choice. As shown by e.g. C.J. Ash (see this 1973 J. Australian Math Society paper), an isomorphism between $(\mathbb{R},+)$ and $(\mathbb{C},+)$ implies the existence of a non-measurable set of reals. The paper has the full argument, but the short version is that (assuming that all sets of reals are measurable) one takes an isomorphism $f:\mathbb{R}\oplus\mathbb{R}\mapsto\mathbb{R}$, defines the sets $S_n=f[\mathbb{R}\oplus[n,n+1)]\cap(0,1)$ (that is, the image of $\mathbb{R}\oplus[n,n+1)$ under $f()$, intersected with the unit interval), and then shows that (a) the $S_n$ partition $(0,1)$ and (b) they all have the same measure. This is enough to contradict countable additivity.

Answer 2

This two groups are vector spaces over $\mathbf{Q}$, and as these two groups have the same cardinal, any basis (over $\mathbf{Q}$) of one of them has the same cardinal than has any basis (over $\mathbf{Q}$) of the other one. This allows you to show that these two $\mathbf{Q}$-vector spaces are isomorphic as $\mathbf{Q}$-vector spaces. As any such isomorphism is also a group isomorphism, you're done.

Take for instance a Hamel basis $(e_i)_{i\in I}$, and a bijection $f : I \to I\coprod I$, then the $\mathbf{Q}$-linear map $F$ defined on basis vectors by $F(e_i)=(e_{f(i)},0)$ or $(0,e_{f(i)})$ claimed that $f(i)$ is in the first copy of $I$ or in the second is an isomorphism from $\mathbf{R}$ to $\mathbf{R}^2$, and there is a obvious group isomophism from $\mathbf{R}^2$ to $\mathbf{C}$, the one sending $(x,y)$ to $x+iy$...