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Prove that the roots of unity is a cyclic group

  • #1
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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##.

Homework Equations




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?
 

Answers and Replies

  • #2
Math_QED
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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##.

Homework Equations




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.
 
  • #3
andrewkirk
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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}##.
 
  • #4
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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?
 
  • #5
andrewkirk
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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##.
 
  • #6
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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##?
 
  • #7
andrewkirk
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Yes, that is correct.
 
  • #8
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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.
 
  • #9
andrewkirk
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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
$$\prod_{j=1}^k O_j$$
 
  • #10
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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##.
 
  • #11
andrewkirk
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Yes that works for me.
 

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