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1. Homework Statement

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

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

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