Is $\mathbb{C}^{\times}/\mathbb{R}^{\times}$ isomorphic to $\mathbb{C}^{\times}/\{\pm 1\}$

Question

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?

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 $\mathbb{C}^\times/\mathbb{R}^\times\cong S^1/\{\pm 1\}\cong \mathbb{C}^\times/\{\pm 1\}$.