Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Ivo Terek, Ali Caglayan, Tim Raczkowski, Micah
If this question can be reworded to fit the rules in the help center, please edit the question.

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

1 Answer 1

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

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.