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This question already has an answer here:

I have to show that $\mathbb{C}/\mathbb{Z}$ is isomorphic to the multiplicative group $\mathbb{C} \setminus \{0\}$.

Proof. Let $f:\mathbb{C} \setminus \{0\} \rightarrow \mathbb{C}/\mathbb{Z}$ be the map $$ f(\alpha) = \alpha \mathbb{Z}.$$ This map has inverse $f^{-1}(\alpha \mathbb{Z}) = \alpha$ so it is bijective. Furthermore, $$f(\alpha \beta) = (\alpha \beta)\mathbb{Z} = \alpha \mathbb{Z} \beta \mathbb{Z} =f(\alpha) f(\beta)$$ and $f(1) = \mathbb{Z}$, so the map $f$ is also a homomorphism.

Question.

Is this correct? It feels a bit like I am cheating.

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marked as duplicate by Dietrich Burde, Alex Provost, 6005, Zain Patel, Chill2Macht Jul 25 '16 at 0:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ How is $\mathbb Z/\{0\}$ a multiplicative group? $\endgroup$ – Tim Raczkowski Jul 24 '16 at 16:04
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    $\begingroup$ Ahem... Which one is a difference of sets and which one is a quotient? $\endgroup$ – user228113 Jul 24 '16 at 16:07
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    $\begingroup$ and I believe you mean $\mathbb C\setminus\{0\}$. $\endgroup$ – Tim Raczkowski Jul 24 '16 at 16:07
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    $\begingroup$ the map $f$ is clearly not bijective since $f(\mathbb Z)=\{1\}$. $\endgroup$ – Tim Raczkowski Jul 24 '16 at 16:09
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    $\begingroup$ Also note that $\mathbb{C}/\mathbb{Z}$ is an additive group. $\endgroup$ – Alex Provost Jul 24 '16 at 16:11
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The proof, as written, is definitely wrong -- the inverse map you define does not vanish on $\mathbb{Z}$, so it cannot be a map out of $\mathbb{C}/\mathbb{Z}$.

I think the idea is to use the complex exponential function. Notice that $\alpha \mapsto \exp(2\pi i \alpha)$ defines a group homomorphism from $\mathbb{C}$ to $\mathbb{C} \setminus \{0\}$ that vanishes on $\mathbb{Z}$. Now prove it's an isomorphism (I suggest doing this geometrically -- you can get anything of norm $1$ by taking $\alpha \in \mathbb{R}$, and then scale)...

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You can show that $\mathbb C/\mathbb Z\cong (\mathbb C\setminus\{0\},\times)$ via the homomorphism $z \to e^{2\pi iz}$.

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You seem to be mainly confused about the nature of the additive group $\mathbb{C}/\mathbb{Z}$. By definition of the quotient, two vectors $z,w \in \mathbb{C}$ are equivalent in $\mathbb{C}/\mathbb{Z}$ if and only if $z - w \in \mathbb{Z}$. Geometrically, in the complex plane, this means that $z$ is to be identified with $w$ if and only if $w$ is at the same vertical level as $z$ and shifted an integral number of units to the left or to the right. This means that a fundamental polygon for the quotient, or a particularly nice choice of coset representatives, is given by the band $[0,1] \times \mathbb{R} \subset \mathbb{C}$ modulo the relation $(0,y) \sim (1,y)$. In other words, you take an infinitely long band and glue together the two sides to form a cylinder. Now try think of a way to interpret the punctured complex plane as an infinitely long cylinder.

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Since $\Bbb C\simeq\Bbb R^2$, we have $\Bbb C/\Bbb Z\simeq\Bbb R^2/\Bbb Z\simeq\Bbb R\times(\Bbb R/\Bbb Z)$.

On the other hand, the polar expression $z=\rho e^{i\theta}$ shows that there is an isomorphism $\Bbb C^\times\simeq\Bbb R^{>0}\times\Bbb C^1$ where $\Bbb C^1$ denotes the (sub)group of complex numbers of norm $1$.

To conclude observe that:

1) The map $r\mapsto e^{2\pi i r}$ sets an isomorphism $\Bbb R/\Bbb Z\simeq\Bbb C^1$.

2) The logarithm defines an isomorphism $\Bbb R^{>0}\simeq\Bbb R$.

Just put everything together.

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