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How do the roots of unity form a group with respect to multiplication (closure, association, identity) ?

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closed as off-topic by Jonas Meyer, Ivo Terek, Ali Caglayan, Tim Raczkowski, Micah Jul 26 '15 at 0:27

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What exactly are you asking? For a proof that the nonzero complex numbers form a group under multiplication? – Mariano Suárez-Alvarez Feb 11 '11 at 23:07
If you mean all of the complex numbers then you should really think about $\mathbb{C} - \{0\}$. For inverse recall the formula $z\overline{z} = |z|^2$. – AnonymousCoward Feb 11 '11 at 23:09
I mean how do the roots of unity form a group with respect to multiplication. – Zo Has Feb 11 '11 at 23:13
Ok I changed my answer. That hint should be enough I think. – AnonymousCoward Feb 11 '11 at 23:19
You should edit your question so it reflects what you meant. – AnonymousCoward Feb 11 '11 at 23:28
up vote 2 down vote accepted

A group will satisfy the four conditions: closure, identity, existence of inverses, and associativity.

$\mathbb{C}$ under multiplication is not a group. Can you see which axiom fails?

On the other hand, $\mathbb{C}-\{0\}$ will be a group, the only axiom that is non-trivial is the existence of a multiplicative inverse. Here the formula $z\overline{z} = |z|^2$ is useful.

If you can prove this, then a nice next exercise is to show that the set $S = \{z\in \mathbb{C} |\,|z| = 1 \}$ is also a group under multiplication. Then show $z\mapsto z^2$ is an automorphism of this group.

Edit: I see you have changed your question. To show that the $n$th roots of unity are a group under multiplication, think about the identity $e^{a+b} = e^ae^b$.

An easy extension of this problem is to prove that the union of all the groups of $n$th roots of unity is a group, and that it is exactly the set $T = \{e^{ik\pi} | k\in \mathbb{Q}\}$.

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thanks for the hint. – Zo Has Feb 11 '11 at 23:25
As a further exercise, take a complex number $z$ such that $|z| = 1$ and $z\not\in T$. Does $\{z^k\,|\, k\in\mathbb{Z}\}$ form a group under multiplication? What is it isomorphic to? – AnonymousCoward Feb 11 '11 at 23:58

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