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How do we show that $\mathbb{C}^{\times}$ and $S^{1}$ are isomorphic as groups?

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    The key to my confusion here is that you are asking about a *group* isomorphism and not a *topological* isomorphism (aka homeomorphism).2013-06-18

2 Answers 2

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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. $

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    @wnoise:what are you referring to by "this"?2011-06-10
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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.