How do the roots of unity form a group with respect to multiplication?

Question

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

Answer 1

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 $T = \{e^{ik\pi} | k\in \mathbb{Q}\}$.