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Order of subgroup of an abelian group

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose H and K are subgroups of an abelian group G (not neccessarily finite). Let the order of H and K be a and b respectively. Prove that there exists a subgroup of order L, where L = lcm(a,b).

    2. Relevant equations

    Product Formula: |HK|/|H| = |K|/|H intersect K|
    lcm(a,b)*gcd(a,b)=ab
    Lagrange theorem
    A finite abelian group G has a subgroup of order d for every divisor d of |G|


    3. The attempt at a solution

    Using the product formula, i obtain: |HK||H intersect K| = ab
    Then using largrange theorem, i have |H intersect K| divides |H| and |H intersect K| divides |K|. Hence, |H intersect K| divides ab. However, i am unable to proceed further. Or is there another method to solve this problem

    Thank You.
     
  2. jcsd
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