It's known that $\mathbb{C}^{\times}/\mathbb{R}^{\times}$ is isomorphic to $S^1/\{\pm1\}$ (and the isomorphism is given by $\varphi(z)=z/|z|\{\pm 1\})$. Also, $\mathbb{C}^{\times}$ is isomorphic to $S^1$. So a natural question came to me (witohut no natural answer): is true that $\mathbb{C}^{\times}/\mathbb{R}^{\times}$ isomorphic to $\mathbb{C}^{\times}/\{\pm 1\}$? If the answer is yes, how is the isomorphism given (I tried by First Isomorphism Theorem and by writing explicitly without any success), and if not, how to prove it?

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    $\begingroup$ $\mathbb{C}^\times/\mathbb{R}^\times\cong S^1/\{\pm 1\}\cong S^1$ is not isomorphic to $\mathbb{C}^\times/\{\pm1\}\cong\mathbb{C}^\times$. $\mathbb{C}^\times$ is not isomorphic to $S^1$. $\endgroup$ Mar 28, 2016 at 21:29
  • $\begingroup$ @Batominovski: Why not? $\endgroup$ Mar 28, 2016 at 21:37
  • $\begingroup$ The easiest way to see is that both are (real) Lie groups with different dimensions. $\endgroup$ Mar 28, 2016 at 21:38
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    $\begingroup$ Sure they can. $\mathbb{R}^2\cong\mathbb{R}$ as groups, for instance. $\endgroup$ Mar 28, 2016 at 21:39
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    $\begingroup$ My claim doesn't come from that false observation (which I don't know how you use it anyway). If $\phi:\mathbb{R}\to\mathbb{R}^2$ is an isomorphism, then $1\in\mathbb{Z}$ is mapped to $\phi(1)\in\mathbb{R}^2$. Then, consider $\mathbb{R}^2=\mathbb{R}\phi(1)\times \mathbb{R}x$ for some $x\notin \mathbb{R}\phi(1)$. Then, we have that isomorphism as $\mathbb{R}/\mathbb{Z}\cong \phi(\mathbb{R})/\phi(\mathbb{Z})\cong\mathbb{R}^2/\big(\mathbb{Z}\times\{0\} \big)$. $\endgroup$ Mar 28, 2016 at 21:55

1 Answer 1


Yes, this is true. Note that the map $z\mapsto z^2$ is a surjective homomorphism $\mathbb{C}^\times\to\mathbb{C}^\times$ with kernel $\{\pm1\}$, so it gives an isomorphism $\mathbb{C}^\times\cong \mathbb{C}^\times/\{\pm 1\}$. Similarly, the squaring map also gives an isomorphism $S^1\cong S^1/\{\pm 1\}$. Thus $\mathbb{C}^\times/\mathbb{R}^\times\cong S^1/\{\pm1\}\cong S^1\cong \mathbb{C}^\times\cong\mathbb{C}^\times/\{\pm1\}$.

Alternatively, you could observe that $\{\pm 1\}$ is the unique subgroup of order $2$ in both $\mathbb{C}^\times$ and in $S^1$, so any isomorphism $\mathbb{C}^\times\to S^1$ must map $\{\pm1\}$ to itself. It follows that such an isomorphism induces an isomorphism $\mathbb{C}^\times/\{\pm 1\}\to S^1/\{\pm1\}$, so $\mathbb{C}^\times/\mathbb{R}^\times\cong S^1/\{\pm 1\}\cong \mathbb{C}^\times/\{\pm 1\}$.

  • $\begingroup$ Brilliant and elegant! Now this post provide an example of normal subgroups for whoses quocients are isomorphic but they're not (there's a trivial example considering \mathbb{Z}_2\times\mathbb{Z}_4$). $\endgroup$ Mar 28, 2016 at 21:55

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