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

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They aren't. Do you mean $\mathbb{C}^\times$ and $S^1\times\mathbb{R}_{>0}$ (where $\mathbb{R}_{>0}$ is the positive real numbers under multiplciation)? – Zev Chonoles Jun 9 '11 at 22:43
@user10: I see that you're right, Rotman is making this claim. I know for a fact that $\mathbb{C}^\times \cong S^1\times \mathbb{R}_{>0}$, but I don't think that that alone would rule out the possibility that also $\mathbb{C}^\times\cong S^1$ (it would seem peculiar though). Still, I'm inclined to believe the book over myself on this, I'm too rusty with divisible groups and whatnot. Hopefully someone can come along and clear it up. – Zev Chonoles Jun 9 '11 at 23:08
@Zev: S^1 ≅ S^1 × R. Basically S^1 is Q/Z × V where V is a zillion dimensional vector space over Q. The logarithm takes the positive reals under multiplication to the additive reals. – Jack Schmidt Jun 9 '11 at 23:11
Simply that $\mathbb{C}^\times$ has an "extra dimension", the radius, that $S^1$ doesn't. However, I'm starting to get an idea of how they might be isomorphic. I think that $\mathbb{R}_{>0}$ just adds on $c$ more copies of $\mathbb{Q}$, which doesn't change anything. I doubt that there will be a formula describing the isomorphism $\mathbb{C}^\times\cong S^1$, since it will involve taking a $\mathbb{Q}$-basis for $\mathbb{R}$ or something like that. – Zev Chonoles Jun 9 '11 at 23:16
I suppose I can take some amount of solace in the fact that 3 other people had a similar incorrect intuition :) – Zev Chonoles Jun 9 '11 at 23:33
up vote 31 down vote accepted

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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I like the way you nicely compartmentalize the insane part of it into R = R^2. – Jack Schmidt Jun 9 '11 at 23:49
@gary that map is not a bijection. – wckronholm Jun 10 '11 at 0:41
Wouldn't the map h: z->z/||z|| do it? $h(z_1.z_2)$:= $z_1.z_2/||z1_.z_2||$= $(z_1/||z_1||)(z_2/||z_2||)$ ; kernel is trivial, and map is onto. – gary Jun 10 '11 at 0:55
@wckronholm: where does it fail? The kernel is trivial. Is it not onto, then? – gary Jun 10 '11 at 0:57
@Jim Belk: thanks! – user10 Jun 10 '11 at 1:23

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.

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