32

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

user10
  • 5,688
  • 9
    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 09 '11 at 22:43
  • @Zev Chonoles: This is page 662, corollary 9.29 in Rotman's book (Advanced Modern Algebra). The proof is there (just trying to show it from first principles). – user10 Jun 09 '11 at 22:53
  • 1
    @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 09 '11 at 23:08
  • @Zev Chonoles: thanks though, may I ask, what makes you think they aren't isomorphic? – user10 Jun 09 '11 at 23:10
  • 4
    @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 09 '11 at 23:11
  • 2
    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 09 '11 at 23:16
  • @Jack: That makes sense, thanks for helping me out. user10, sorry for doubting :) – Zev Chonoles Jun 09 '11 at 23:18
  • @user10: I'm out of votes for today, but I'll make sure I come back and give you +1 when it resets. – Zev Chonoles Jun 09 '11 at 23:19
  • 1
    I suppose I can take some amount of solace in the fact that 3 other people had a similar incorrect intuition :) – Zev Chonoles Jun 09 '11 at 23:33
  • There's a proof in the Wikipedia article on the circle group (http://en.wikipedia.org/wiki/Circle_group#Group_structure) using the structure theorem for divisible groups. – joriki Jun 09 '11 at 23:36
  • The key to my confusion here is that you are asking about a group isomorphism and not a topological isomorphism (aka homeomorphism). – robjohn Jun 18 '13 at 13:03

2 Answers2

40

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 $$ S^1 \;\cong\; \mathbb{R}/\mathbb{Z} \;\cong\; \mathbb{R}^2/(\mathbb{Z}\times\{0\}) \;\cong\; S^1\times\mathbb{R} \;\cong\; \mathbb{C}^\times. $$

Jim Belk
  • 49,278
  • 12
    I like the way you nicely compartmentalize the insane part of it into R = R^2. – Jack Schmidt Jun 09 '11 at 23:49
  • @gary that map is not a bijection. – wckronholm Jun 10 '11 at 00:41
  • 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 00:55
  • @wckronholm: where does it fail? The kernel is trivial. Is it not onto, then? – gary Jun 10 '11 at 00:57
  • 1
    @gary, the kernel is not trivial, it is all $z$ such that $z/|z|=1$, that is, all of $\mathbb{R}^+$. – Thomas Andrews Jun 10 '11 at 01:47
  • @Thomas: ouch, you're right. I was only thinking of the kernel of z1.z2, forgetting that we were dividing by the norm. A highschool-level mistake!@ Every line thru (0,0) is collapsed into the point (1,m)/$Sqr(1+m^2)$ , for points in (x,mx). – gary Jun 10 '11 at 02:11
  • And this is yet another reason why some have trouble with non-constructive mathematics. – wnoise Jun 10 '11 at 04:35
  • @wnoise:what are you referring to by "this"? – gary Jun 10 '11 at 14:18
11

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.

Georges Elencwajg
  • 150,790