Isomorphism between complex numbers minus zero and unit circle How do we show that $\mathbb{C}^{\times}$ and $S^{1}$ are isomorphic as groups?
 A: First, note that the additive groups of $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic, since $\mathbb{R}$ and $\mathbb{R}^2$ have the same dimension as vector spaces over $\mathbb{Q}$.
In particular, there exists a group isomorphism $\varphi\colon \mathbb{R} \to \mathbb{R}^2$ such that $\varphi(1) = (1,0)$.  Then $\varphi(\mathbb{Z}) = \mathbb{Z}\times\{0\}$, so
$$
S^1 \;\cong\; \mathbb{R}/\mathbb{Z} \;\cong\; \mathbb{R}^2/(\mathbb{Z}\times\{0\}) \;\cong\; S^1\times\mathbb{R} \;\cong\; \mathbb{C}^\times.
$$
A: Every divisible abelian group is equal to the direct sum of its torsion part and of a $\mathbb Q$-vector space : $$A=Tors(A) \oplus V$$
In the situation at hand, the torsion part of both groups under study is the denumerable group $\mu_\infty (\mathbb C)$ of roots of unity and we deduce 
$$\mathbb C^\times= \mu_\infty (\mathbb C)\oplus  V   \quad  \quad  S^1=    \mu_\infty (\mathbb C)  \oplus  W        $$
Since for cardinality reasons $V$ and $W$ have continuous dimension , they are isomorphic and so are our groups $\mathbb C^\times$ and $ S^1$ .
Terminology In the multiplicative notation, an element $a\in A$ of an abelian  group is said to be torsion if $a^n=1$ for some positive integer $n$.  
Remark Jim's answer has the charm of being direct and  slick. However some users might like the fact that the  present solution  is a simple application  of the general structure theorem for divisible abelian groups. That  theorem, and much, much more, is to be found in Kaplanski's elegant booklet (90 pages!) Infinite Abelian Groups.
