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Prove that $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of root of unity in $\mathbb{C}^\times$.

I tried to construct a homomorphism from $\mathbb{Q}$ to $\mathbb{C}^\times$ with kernel $\mathbb{Z}$, but I don't think it worked.

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2 Answers 2

up vote 7 down vote accepted

Correction: Define a map $\mathbb{R} \to \mathbb{C}^\times$ by sending $$r \mapsto e^{2\pi i r}.$$ Now this is clearly a well-defined homomorphism (check this!). It is surjective onto the unit circle (check this!) with kernel $\mathbb{Z}$ (check this!). Thus, $\mathbb{R}/\mathbb{Z}$ is isomorphic to the circle. Recall that the roots of unity comprise the torsion subgroup of the circle (check this!); so they must be isomorphic to the torsion subgroup of $\mathbb{R}/\mathbb{Z}$, which is exactly $\mathbb{Q}/\mathbb{Z}$ (show this!).

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Thanks @ArturoMagidin. I should really get more sleep... –  Rick Jan 30 '12 at 22:04

Note that you are viewing $\mathbb{Q}/\mathbb{Z}$ as an additive group; so your morphism should map sums of rationals into products of complex numbers of norm $1$.

Now, there is a rather famous family of functions that send sums into products; this family is also well-known for giving a very nice description of every element of the complex numbers that has complex norm $1$. Perhaps that can be used somehow?

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