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This question is asking to prove that the quotient group $\mathbb{R}/\mathbb{Z}$ is isomorphic to the group of complex numbers with modulus 1 (under multiplication). It's hard for me to visualize the structure of $\mathbb{R}/\mathbb{Z}$, so I can't think of an isomorphism. I feel like the first isomorphism theorem can help? Help is appreciated.

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Start by writing down what it means for $x$ to be a member of $\mathbb{R}/\mathbb{Z}$. – Alex R. Jan 10 '13 at 0:04
Your feeling is correct. If you want to show that $G/H$ is isomorphic to $K$, what do you need to apply the first isomorphism theorem? – Jonas Meyer Jan 10 '13 at 0:06
In terms of visualizing: you start with the real numbers and mod out by the integers. In other words, you associate any two real numbers that differ by an integer. Thinking carefully, one can see that everything in $[0, 1)$ is "different," but that $0$ and $1$ are the same (since they differ by an integer, namely, $1$). In this way, $[0,1]$ is like a circle (the group of complex numbers with modulus 1). Similarly, the point $1/2$ is the same as $3/2, 5/2, 7/2,$ etc., since they differ by an integer, so really you just have this one circle. Perhaps this gives you a clearer mental picture... – Benjamin Dickman Jan 10 '13 at 0:18
Don't forget to accept one answer – Alan Simonin Jan 10 '13 at 1:07

The map $t\mapsto e^{2\pi it}$ is a homomorphism from $\mathbb{R}$ onto the circle group. Its kernel is $\mathbb{Z}$. Invoke the first isomorphism theorem.

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Thanks, I've been thinking of the map $x \to e^{ix}$, but this doesn't give the right kernel. I guess putting in the $2\pi$ is what I needed. – user56744 Jan 10 '13 at 0:09
$\mathbb{R}/\mathbb{Z}$ is isomorphic to $\mathbb{R}/2\pi \mathbb{Z}.$ – ncmathsadist Jan 10 '13 at 0:17

Hint: Try to look at the obvious homomorphism (from addition in $\mathbb{R}$ to multiplication in $\mathbb{C}$) defined by $\phi: \mathbb{R} \to \mathbb{C}$ with $$\phi(x) = e^{2\pi ix}.$$

What is the image? What is the kernel? Then you should apply the first isomorphism theorem.

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You can see the problem this way : $\mathbb{R}/\mathbb{Z}=[0,1)$. Then, at each point in this interval correspond an unique point in the circle $\{ e^{2\pi i \theta} : \theta \in [0,1) \}$. About the group operations, when you add two numbers in $[0,1)$, you multiply them in the set representing the circle (adding the argument).

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@Serkan Yes you are right... Making the changes – Alan Simonin Jan 10 '13 at 0:43

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