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Least common multiple of the orders of the elements in a finite Abelian group

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data
    (From an exercise-section in a chapter on Lagrange's theorem:) Let G be a finite abelian group and let m be the least common multiple of the order of its elements. Prove that G contains an element of order m.

    3. The attempt at a solution
    We have x^m = e for all x in G. By Lagrange's theorem we have that every order of every element in G is a divisor of |G|. Thus if we could show that m = |G| we would arrive at the desired conclusion. This is however impossible because there are examples of finite Abelian groups for which this property doesn't hold. We do know that |G| is a multiple of all orders of all elements of G, and hence that m <= |G| and m divides |G|.

    A second approach is letting O be the maximal order of all the elements of G, and then prove that the order of every element of G divides O.

    A third approach could be to form some kind of a minimal 'basis' for the group of elements b1, ..., bk. Then, every element of G can be written as a product of powers of the elements in the basis. However, in general, the representation nor the basis seems to be unique.

    None of the three approaches has brought me any progress so far. Is there one I am missing? I also took a glance at the chapter on finitely generated abelian groups. It looks as if it could be of some use with this problem, but it's ten chapters ahead, so I don't think it is necessary to use it.

    Any hint in the right direction will be appreciated.
     
  2. jcsd
  3. Sep 27, 2008 #2
    "least common multiple of the order of its elements"

    or

    "least common multiple of its elements" ?

    I don't know if I can answer your question I'm just curious what the difference is?
     
  4. Sep 27, 2008 #3

    Dick

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    Write m=p1^k1*p2^k2*...*pn^kn for distinct primes pi. Can you prove that for every i there is an element of the group ai such that the order of the element ai is pi^ki? Hint: the power of p appearing the LCM(a,b,c,d...) is the largest power of p appearing in any one of the a,b,c,d...
     
  5. Sep 27, 2008 #4
    Hence for every pi^ki there exists an ai in G such that pi^ki divides the order of ai. Somehow I must show that there is also such an ai for which the order has no prime factors other than pi, but I don't see how to do that.
     
  6. Sep 27, 2008 #5

    Dick

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    The group generated by ai is a CYCLIC group (it's isomorphic to Z(n) for some n). For any number k that divides the order of a cyclic group, there is an element of order k.
     
    Last edited: Sep 27, 2008
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