# Prove that the roots of unity is a cyclic group

• Mr Davis 97
In summary: If this element has finite order, it generates a finite subgroup. If it has infinite order, it generates the whole group.
Mr Davis 97

## Homework Statement

Let ##\mu=\{z\in \mathbb{C} \setminus \{0\} \mid z^n = 1 \text{ for some integer }n \geq 1\}##. Show that ##\mu = \langle z \rangle## for some ##z \in \mu##.

## The Attempt at a Solution

My thought would be just to write out all of the elements of ##\mu## in exponential form and show that ##e^{\frac{2 \pi i}{n}}## generates all of the other elements. Would this be the best way to do this?

Delta2
Mr Davis 97 said:

## Homework Statement

Let ##\mu=\{z\in \mathbb{C} \setminus \{0\} \mid z^n = 1 \text{ for some integer }n \geq 1\}##. Show that ##\mu = \langle z \rangle## for some ##z \in \mu##.

## The Attempt at a Solution

My thought would be just to write out all of the elements of ##\mu## in exponential form and show that ##e^{\frac{2 \pi i}{n}}## generates all of the other elements. Would this be the best way to do this?

Yes.

Delta2
The claim is actually false as given. The usual interpretation of
$$\{z\in \mathbb{C} \setminus \{0\} \mid z^n = 1 \text{ for some integer }n \geq 1\}$$
is
$$\{z\in \mathbb{C} \setminus \{0\} \mid \exists n\in \mathbb N \smallsetminus \{0\}\ :\ z^n = 1\}$$
which is
$$\{e^{\frac{2\pi k i }n} \mid n,k\in \mathbb Z \wedge 0\le k < n\}$$
which I am pretty sure is not cyclic, as the group generated by ##e^{\frac{2 \pi k i}n}## does not contain ##e^{\frac{2 \pi i}{2n}}##.

I suspect what they meant to say was :

"For each ##n\in \mathbb N\smallsetminus \{0\}##, define ##\mu_n =
\{z\in \mathbb{C} \setminus \{0\} \mid z^n = 1\}##. Show that for every ##n##, ##\mu_n## is cyclic."

To prove the second one, you don't need to write out all the elements. Instead, just show that all elements are of the form
##e^{\frac{2\pi k i }n}## where ##0\le k < n##. Then show that each of those elements can be expressed as a power of ##e^{\frac{2\pi i }n}##.

Delta2
andrewkirk said:
The claim is actually false as given. The usual interpretation of
$$\{z\in \mathbb{C} \setminus \{0\} \mid z^n = 1 \text{ for some integer }n \geq 1\}$$
is
$$\{z\in \mathbb{C} \setminus \{0\} \mid \exists n\in \mathbb N \smallsetminus \{0\}\ :\ z^n = 1\}$$
which is
$$\{e^{\frac{2\pi k i }n} \mid n,k\in \mathbb Z \wedge 0\le k < n\}$$
which I am pretty sure is not cyclic, as the group generated by ##e^{\frac{2 \pi k i}n}## does not contain ##e^{\frac{2 \pi i}{2n}}##.

I suspect what they meant to say was :

"For each ##n\in \mathbb N\smallsetminus \{0\}##, define ##\mu_n =
\{z\in \mathbb{C} \setminus \{0\} \mid z^n = 1\}##. Show that for every ##n##, ##\mu_n## is cyclic."

To prove the second one, you don't need to write out all the elements. Instead, just show that all elements are of the form
##e^{\frac{2\pi k i }n}## where ##0\le k < n##. Then show that each of those elements can be expressed as a power of ##e^{\frac{2\pi i }n}##.
Your interpretation of the problem is correct. It is not a cyclic group. Actually what I need to show is this:
Prove that for each integer ##m \geq 1##, there is a unique subgroup ##H_m\leq \mu## with ##|H_m|=m## and that ##H_m## is cyclic.

It seems rather easy to show there exists one. Let ##\mu_m =\{z\in \mathbb{C} \setminus \{0\} \mid z^m = 1\}##. We can easily show that this is a cyclic subgroup of order ##m##. My problem is with the uniqueness requirement. What does it mean to show that ##\mu_m## is unique?

You have to show that if ##H_m## is a subgroup of order m then ##H_m=\mu_m##.

Start by showing that any ##g\in H_m## must satisfy ##g^m=1##.

andrewkirk said:
You have to show that if ##H_m## is a subgroup of order m then ##H_m=\mu_m##.

Start by showing that any ##g\in H_m## must satisfy ##g^m=1##.
Well, since ##|H_m|=m##, if ##g \in H_m## then ##g^m=1## since m is a multiple of the order of ##g##. So ##g \in \mu_m##. Does this show that ##H_m \subseteq \mu_m##? And since ##|H_m|=m=\mu_m##, we have ##H_m = \mu_m##?

Yes, that is correct.

Mr Davis 97
andrewkirk said:
Yes, that is correct.
I think I have only one more question. I want to show that if ##H \le \mu## is finitely generated, then ##H## is finite. Here is my idea. Let ##H## be an arbitrary subgroup of ##\mu## that is finitely generated. So ##H=\langle z_1,z_2, \dots, z_k\rangle##. We can only multiply elements in the generating set and powers of those elements in a finite number of ways, since each ##z## in the generating set has a finite order. But whenever we multiply any combination of elements, we get another element of finite order. Hence the size of ##H## is finite.

Mr Davis 97 said:
We can only multiply elements in the generating set and powers of those elements in a finite number of ways, since each ##z## in the generating set has a finite order.
This is correct, but a little vague, and doesn't lead directly to the result.

To put it more compactly, we can observe that, since the group is abelian and each generator has finite order, every element of $H$ can be expressed as
$$\prod_{j=1}^k z_j^{r_j}$$
for some ##k##-tuple ##r_1,...,r_k## such that ##\forall j:\ 0\le r_j <O_j##, where ##O_j## is the order of ##z_j##.
Then an upper bound for the number of elements is
$$\prod_{j=1}^k O_j$$

Mr Davis 97
andrewkirk said:
This is correct, but a little vague, and doesn't lead directly to the result.

To put it more compactly, we can observe that, since the group is abelian and each generator has finite order, every element of $H$ can be expressed as
$$\prod_{j=1}^k z_j^{r_j}$$
for some ##k##-tuple ##r_1,...,r_k## such that ##\forall j:\ 0\le r_j <O_j##, where ##O_j## is the order of ##z_j##.
Then an upper bound for the number of elements is
$$\prod_{j=1}^k O_j$$
Would this same argument show that ##\mu## is not finitely generated? For suppose it were. Then by the same argument there is an upper bound for the number of elements, so it is finite. But ##\mu## is not finite, since there are infinitely many elements in ##\mu##. since ##\forall n \in \mathbb{N}##, ##e^{\frac{2 \pi}{n}i} \in \mu##.

Yes that works for me.

Mr Davis 97

## 1. What is the definition of a cyclic group?

A cyclic group is a mathematical group that is generated by a single element, also known as a generator. This means that all elements in the group can be expressed as powers of the generator.

## 2. How can we prove that the roots of unity form a cyclic group?

To prove that the roots of unity form a cyclic group, we need to show that they satisfy the properties of a cyclic group. This includes closure, associativity, identity element, inverse element, and commutativity. Once we have shown that these properties hold, we can conclude that the roots of unity form a cyclic group.

## 3. What are the roots of unity?

The roots of unity are the complex numbers that, when raised to a certain power, result in 1. In other words, they are the solutions to the equation x^n = 1, where n is a positive integer. The roots of unity can be represented on a complex plane as points evenly spaced around the unit circle.

## 4. Why are the roots of unity important in mathematics?

The roots of unity have numerous applications in mathematics, including number theory, group theory, and geometry. They also play a crucial role in the study of polynomials and their roots. Additionally, the properties of cyclic groups, such as the roots of unity, have implications in cryptography and coding theory.

## 5. Can you provide an example of how the roots of unity form a cyclic group?

Yes, an example of the roots of unity forming a cyclic group is the set of 3rd roots of unity, which are 1, -1/2 + √3i/2, and -1/2 - √3i/2. These three complex numbers, when raised to the power of 3, all equal 1. They also satisfy the properties of a cyclic group, making them a valid example of the roots of unity forming a cyclic group.

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