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the circle group is the multiplicative group of all complex numbers of absolute value 1. How can i show that this group is isomorphic with $\mathbb R/ \mathbb Z$. Any hints for the right map is great.

Basil R
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2 Answers2

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Hint: The complex exponent map.

Asaf Karagila
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The map $\phi\colon\Bbb R\to T$ given by $\phi(t)=e^{2\pi it}$ is a group morphism. Hence we can apply first isomorphism theorem.

Davide Giraudo
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