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Let $H = \langle i \rangle =\{ i, -1, -i, 1 \}\le \mathbb{C}^\times$. Then is $\mathbb{C}^\times/H$ isomorphic to $\mathbb{C}^\times$?

I don't think there exists an isomorphism $\varphi : \mathbb{C}^\times \to \mathbb{C}^\times/H$ because for any $z = re^{i\theta} \in \mathbb{C}^\times$, we can consider $z' = re^{i(\theta+\pi/2)}$ which gives $zH = z'H$. So this makes me think an isomorphism property can be broken somehow but I can't seem to make it work out. Or perhaps the quotient group is isomorphic to the multiplicative group of complex numbers.

user330531
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    Consider the map $z\mapsto z^4$. – Angina Seng Oct 22 '17 at 19:09
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    @LordSharktheUnknown Oh that's wonderfully obvious, which I didn't see, thanks. So that map is a surjective homomorphism with kernel being $H$. Using the first isomorphism theorem, $\mathbb{C}^\times \cong \mathbb{C}^\times / H$. You can make that an if you want answer and I'll check it off. – user330531 Oct 22 '17 at 19:17
  • @LordSharktheUnknown Please give me your thoughts. Do you agree with the following explicit construction of an isomorphism by Brian Bi linked here ? –  Oct 15 '18 at 11:04

1 Answers1

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Let $n$ be any positive integer, and $\mu_n$ the group of complex $n$-th roots of unity. The map $\Bbb C^\times\to\Bbb C^\times$ given by $z\mapsto z^n$ is a surjective group homomorphism with kernel $\mu_n$. By the First Isomorphism Theorem, $\Bbb C^\times\cong\Bbb C^\times/\mu_n$.

This is the $n=4$ case.

Angina Seng
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