Is it true that group of real numbers under addition isomorphic to group of complex numbers under addition

Question

Is it true that set of all real numbers under addition isomorphic to set of all complex numbers under addition

Answer 1

It’s true, but the only argument I know involves Axiom of Choice. Both $\mathbb C$ and $\mathbb R$ are vector spaces over $\mathbb Q$, and they have bases, both of continuum cardinality. Since the bases may be put in 1-1 correspondence, this correspondence extends to an isomorphism of $\mathbb Q$-vector spaces, so $\mathbb C$ and $\mathbb R$ are isomorphic as rational vector spaces, and certainly as abelian groups.

ADDITION: In response to your request, let me explain. The point of a basis $\mathfrak B$ is that every element of the vector space $V$ over a field $k$ may be written uniquely as a linear combination $\sum_i\lambda_ib_i$ where the $\lambda$’s are “scalars” in $k$ and the $b$’s are elements of $\mathfrak B$. So if you have vector spaces $V$ and $W$ over $K$, with bases $\mathfrak B$ and $\mathfrak C$ respectively, and if you have a one-to-one correspondence $\varphi\colon\mathfrak B\to\mathfrak C$, (onto as well as one-to-one), then you define your $k$-isomorphism by extending by linearity. If $v\in V$, then it has an expansion $v=\sum_i\lambda_ib_i$. You send it to $w\in W$, by $w=\sum_i\lambda_i\varphi(b_i)$.

You may ask about what happens when the bases are infinite, as in your specific question. The neat way to handle this is to say that the sum is over all elements of the base, but with only finitely many of the scalars $\lambda_i$ being nonzero.