# Determining whether the unit circle group is a cyclic group

1. Jul 27, 2013

### mahler1

1. The problem statement, all variables and given/known data

Let S be the set of complex numbers z such that |z|=1. Is S a cyclic group?

3. The attempt at a solution

I think this group isn't cyclic but I don't know how to prove it. My only idea is:
If G is a cyclic group, then there is an element x in G such that for all g in G, g can be written as g=x^j (j integer) and x is called a generator of the group G. Now, in this case, if S is not cyclic, I could prove it by contradiction, that means: I assume S is cyclic, then I use the definition of cyclic, that is, there is an element x in S such that if g is in S then g is of the form g=x ^j for some integer j. If I want to show that this cannot be always true, I must find a (generic) complex number z whose absolute value is 1 but z is different from x^j for all integer j. Ok, this may sound stupid but I don't know how to actually do this last part and I got stuck here.

I edit the post to add a new idea: I "know" that the set of complex numbers z with |z|=1 is an uncountable set. Now, if S was a cyclic group then it would have a generator, lets call it x. By definition, for all element s in S, s=x^j for some integer j. Now, the set of integers is countable and so is the set X={x^j} with j integer. As I can't have a surjective function from a countable set to an uncountable one, then, I could find an element s in S that can't be related to any element of the set X.
Is this idea correct? If it is, then I have some doubts on the following points: 1) How can I show that the unit circle is an uncountable set? 2) I automatically assumed that I can't have a surjective function from a countable set to an uncountable set, this can seem obvious but I don't know how to prove it.

Sorry if I've made any grammar or spelling mistakes, english is not my native language

Last edited: Jul 27, 2013
2. Jul 27, 2013

### micromass

The idea to use countable is very good.

Find a bijection with $\mathbb{R}$ or some other set you know to be uncountable.

Well, what's your definition of countable? Try to argue by contradiction. Assume you have a surjective function,...

3. Jul 27, 2013

### mahler1

Well, I still don't know how to do the first part, I mean, showing that the unit circle is uncountable. I thought of a bijective function between this set and another known uncountable set (the real numbers or the interval [0 1]) but I don't know how to explicitly construct a function between these sets. If I come up with any ideas, I'll post and ask again.

4. Jul 27, 2013

### micromass

Do something with sine and cosine functions to parametrize the circle.

5. Jul 27, 2013

### mahler1

Ok, with all the things you've said I came up with the following "proof":

Suppose S is a cyclic group. By definition, there is an element in S, call it x, such that for all s in S, s can be written as s=x^​j for some integer j. Let X={x^​j} be all the elements in S written in that way. Consider the map between X and a subset J of the integers simply by letting f(x^​j)=j. The set Z of all integers is a countable set, then, in particular, J is a countable set and this means that S is also countable.

Every complex number z is of the form z=a+bi for a,b real numbers. So S can be defined as S={a,b in R such that a^​2+b^​2=1}. Now, consider the functions a=cos(t) and b=sin(t) with t in [0 2π); this is a bijective parameterization (parametrization?) of the unit circle. So, there exists a bijective map from [0 2π) to the set S. It follows that S has the same cardinality as the set [o 2π), which is just the interval [0 1) multiplied by 2π. But [0 1) is an uncountable set, then S must be uncountable. But this is absurd since S is countable. The contradiction comes from assuming that there is an element x in S such that for all s in S, s=x^​j with j integer. In conclusion, S is not a cyclic group.

Is something wrong with the proof or is it ok?

6. Jul 27, 2013

### micromass

One small detail that we can't ignore. You defined the map $f(x^j) = j$. But is this well-defined? That is, what if $x^2 = x^4$ (for example). Then $2 = f(x^2)= f(x^4) = 4$. This is clearly false.

So, how do we know that $x^n = x^m$ can't happen for distinct $m$ and $n$?

7. Jul 27, 2013

### pasmith

I would state and prove, as a separate proposition, that if a group is cyclic then it is countable. It is more natural, and avoids micromass's objection, to define a map from the natural numbers to a cyclic group rather than the reverse. Recall that to show that a non-empty set is countable, it is sufficient to show that there exists a surjection from the natural numbers to that set.

It then follows directly from that proposition that if S is uncountable then S cannot be cyclic, and you will show that S is uncountable.

This is really the heart of the proof, so it's not enough to simply state the fact that $t \mapsto \cos (t) + \mathrm{i}\sin (t)$ is a bijection from $[0, 2\pi)$ to $S$. You have actually to prove it.

(Actually I'd save a step and find a bijection directly from $[0,1)$ to S.)

8. Jul 27, 2013

### mahler1

You're right on this point, what I can do is to separate in the only two possible cases: 1) S is a finite cyclic group. 2) S is an infinite cyclic group

Case 1) Suppose S is a finite cyclic group of order n, I want to show that if <x> generates S, then the powers of the set {1, x, ..., x^(n-1)} are all distinct. As S is finite cyclic of order n, then there is an element x in S such that x^k is different from 1 for all 0<k<n. Consider the set {1, x, ..., x^(n-1)} Lets assume x^j=x^i where, without loss of generality,0<= i<j<=n-1. Then, multiplying both sides of the equation by x^i (I can do this because all the elements in a group have inverses)yields x^(j-i)=1. But this is absurd since 0<j-i<n. So, all the powers of that set must be distinct. From the preceding observations we can also see that <x> generates S, as the set {1, x, ..., x^(n-1)} has n different elements, all of which belong to a group of order n.

Case 2) Suppose S is infinite cyclic and S is generated by <x>. Lets show that if j is different from i, where j and i are integers, then x^j is different from x^i. Suppose, without loss of generality, that j>i and that x^j=x^i. As S is a group, I can multiply both sides of the equation by x^i inverse to get
x^(j-i)=1, but this means that S is finite, which is absurd. Then, if j is different from i, it can't happen that x^j=x^i.

Sorry for not writing in latex, I'll try to learn how to use it as soon as I can.

Last edited: Jul 27, 2013
9. Jul 27, 2013

### mahler1

To pasmith: Yes, what you propose seems to be simpler, but I cannot think immediately how to prove all of that, I'll try to follow your suggestions and write the proof. Thanks

10. Jul 27, 2013

### micromass

OK, but how does this help? Isn't just saying that such a group is finite enough? After all, if the circle is uncountable, it cannot be finite.

11. Jul 27, 2013

### mahler1

Sorry, I don't get what you're saying. I assumed S was cyclic and invented that bijective map. If it is well defined, it cannot happen, as you said, that x^n=x^m for n different to m. So, I proved that whenever n is different to m, x^n can't be equal to x^m. To prove this, I assumed that S is cyclic, and if S is cyclic, it can be finite or infinite. In the last part of the proof, I invented another mapping between S and another set to arrive to the conclusion that S is uncountable, which is absurd since from the first mapping it follows that S is a countable set or a finite set.

Last edited: Jul 27, 2013
12. Jul 27, 2013

### micromass

You basically have it. Either the group is finite or it is countably infinite. Either way, it can't be uncountable.

13. Jul 27, 2013

### mahler1

micromass: I edited my last reply because then I reread your answer and realized that i didn't get what I was doing wrong. But, putting all the ideas together, I think I can rewrite and change my original "proof", I'll do it.

14. Jul 28, 2013

### Boorglar

I can suggest another proof that avoids using the uncountability of the unit circle, since this property is not necessary (there are in fact countable subsets of the unit circle that also can't be cyclic groups).

Using the polar coordinate representation of complex numbers, the unit circle is the set of all numbers of the form $cis(2\pi t), t \in [0,1]$. Suppose the unit circle were cyclic. Then let $cis(2\pi t_{0})$ be a generator.

The unit circle can therefore be described as $\left\{cis(2n\pi t_0), n \in Z \right\}$.
Now there are two cases to consider:

1)
If $t_0$ is rational, then $t_0 = \frac{a}{b}$ for some $a, b \in Z, b ≠ 0$. Then the unit circle consists of $\left\{cis(2\pi \frac{na}{b} ), n \in Z \right\}$. But we can easily find a number in the unit circle that is not in the set (which one?) and thus we reach a contradiction.

2)
If $t_0$ is irrational, then $cis(\pi) = -1$ is not in the set (why?). Again, we reached a contradiction.

Therefore, the unit circle is not a cyclic group.
Keep in mind that the set of all solutions to the equation $x^n = 1, n \in N$ is a countable subset of the unit circle, and yet it is not cyclic either.

15. Aug 1, 2013

### mahler1

Thanks for your suggestion and for the counterexample of the countable subset, it's much more simpler to prove the statement using your ideas.
In 1), if we choose t=1/2, then it's easy to see that cis(1/2) is not in the set because in that case we would have 2πna/b=1/ iff π=b/(4na) which is absurd since π is irrational and clearly b/(4na) is rational.
In 2), cis(π) is not in the set, as you said, because if this is true, then we would have that π=2nπt0 iff 1/2n = t0 which is absurd since 1/2n is a rational number and t0 is irrational.