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Roots of unity form a cyclic group

  1. Jun 1, 2012 #1
    In a lot of places, I can read that the roots of unity form a cyclic group, however I can find no proofs. Is the reasoning as follows:

    Let's work in a field of characteristic zero (I think that's necessary). Let's look at the nth roots of unity, i.e. the solutions of [itex]x^n - 1[/itex]. There are n different roots, since the derivative is [itex]nx^{n-1}[/itex], which is not zero since the characteristic is zero. Now suppose the group of roots is not cyclic, then the exponent of that group is [itex]m < n [/itex]. In that case the group is also the set of solutions of [itex]x^m-1[/itex], however this can only have m solutions. Contradiction.
  2. jcsd
  3. Jun 1, 2012 #2
    Isn't this simpler than you think?

    The nth roots of unity are 1, e^(i 2pi/n), e^(i 2pi 2/n), e^(i 2pi 3/n), ..., e^[i 2pi (n-1)/n].
    Just keep multiplying the second member by itself to generate the others.
  4. Jun 1, 2012 #3
    Do you have to know the form in complex plane? I mean, is there a way to prove without that?
  5. Jun 1, 2012 #4


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    Yes, this works if you're working in [itex]\mathbb{C}[/itex] or it subfields. But the OP wants to prove it for all fields of characteristic 0.
  6. Jun 1, 2012 #5
    This is what I get for entering the math boards .. and now I make my exit.
    Good Day, sirs.
  7. Jun 1, 2012 #6
    I was curious about this point too. The OP seems to be assuming that the field is algebraically closed, but that's not necessarily true. [itex]\mathbb{Q}[/itex] is a field of characteristic 0. What is the group of the 5-th roots of unity in [itex]\mathbb{Q}[/itex]? It's just {1}. Trivially cyclic, but this case needs to be considered. Likewise, the 6-th roots of unity are {-1, 1}.

    Does "roots of unity" imply that the field must be algebraically closed? Or would {1} or {-1, 1} be regarded as the group of n-th roots of unity depending on n being odd or even?

    Must any algebraically closed field of characteristic zero contain a copy of the complex numbers? If so, that might save sam_bell's proof.
  8. Jun 1, 2012 #7


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    This is certainly not true. However, I do think that every algebraically closed field of characteristic zero contains a copy of the algebraic numbers.
  9. Jun 2, 2012 #8
    I think we assume we've adjoined the roots of unity to any starting base field (of characteristic zero?).

    All roots alpha_i satisfy the equation (alpha_i)^n=1. So doesn't that mean they form a group. Then is it it a fact from group theory that one of the alpha_i generate the group? If so, does this not work in finite fields?
    Last edited: Jun 2, 2012
  10. Jun 2, 2012 #9
    To add to steve and miscromass's remarks.

    The complex numbers contain transcendentals like pi. I don't think pi is in the algebraic closure of Q. On the other hand, C is algebraically closed (and topologically?).
  11. Jun 2, 2012 #10
    You could always extend the field to be complete. You would then obtain n roots. The original group is a subgroup and subgroups of cyclic fields are always cyclic, so it suffices to prove this for a complete field.

    You can do this by induction. Assume it is true for all k less than or equal to n-1. If n has more than one distinct prime factor, then it can be decomposed into a direct sum of two groups whose orders have no common factors. Each of these orders is at most n-1, so the induction hypothesis says they are cyclic. Since their orders are relatively prime, their direct sum is also cyclic. So it remains to prove the case when n is a power of a single prime number... If such a group were not cyclic it would have more than p elements of order p contradicting the fact that there cannot be more than p pth roots of unity.

    [EDIT]: The OP had the right answer all along. Yes, if a finite abelian group is not cyclic then its exponent is a proper divisor of the order, so your logic is exactly right. But I think you do need the first step I wrote showing that it is sufficient to prove the case in which the field is algebraically closed.
    Last edited: Jun 2, 2012
  12. Jun 2, 2012 #11
    Somewhere I found the following
    Apparently for them the being cyclic follows immediately, however I don't really get the step. Can somebody enlighten me?
  13. Jun 2, 2012 #12
    I think this is about as general as the proof can be. But I have some details missing, but I think this is a good outline.

    1. Define the polynomials F[x]
    2. Mod out by the ideal (x^n-1), that is, E=F[x]/(x^n-1)
    3. E contains F, and it is a vector space over F
    4. I forget how, but we should have n elements which satisfy x^n-1=0 (at least one of them is in F).
    5. We can show they form a group (abelian because we are in a field)
    6. So why then is it cyclic? If we have 4 elements, why can't it be Z_2 x Z_2? Then every element has order 2, which I think is somehow going to contradict something. Someone, help please. :)

    EDIT: Found a source for the cyclic part in general http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf, it applies to my Z_2 x Z_2 like so:

    If every element of G has order two, then every element of G satisfies x^2-1=0. Then there are no more than 2 roots, which means G has [STRIKE]less than[/STRIKE] at most two roots, contradiction.

    Okay, all the steps are not clear to me, but there you have the claim and a source in more generality.
    Last edited: Jun 2, 2012
  14. Jun 2, 2012 #13


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    We have the following theorem

    A finite subgroup of the multiplicative group [itex]K^\times[/itex] of a field K is cyclic.

    Let G be a finite subgroup of [itex]K^\times[/itex]. Let p be a prime divisor of |G|. The elements of order p of G are zeroes of the polynomial [itex]X^p-1[/itex]. But there are at most p zeroes of this polynomial, so G has at most p-1 elements of order p. It follows that the elements of order p form a cyclic subgroup of G. The result now follows from the fundamental theorem of finite abelian groups.
  15. Jun 2, 2012 #14
    Thank you
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