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Larson 4.4.13

  1. Jan 8, 2008 #1
    [SOLVED] Larson 4.4.13

    1. The problem statement, all variables and given/known data
    A is a subset of a finite group G, and A contains more than one-half of the elements of G. Prove that each element of G is the product of two elements of A.


    2. Relevant equations



    3. The attempt at a solution
    Is that even true? What if G is just the union of the cyclic group with 20 elements and the cyclic group with 21 elements. Let A = C_21. ord(G) = 20+21-1=40. A has more than half of the elements of G but you cannot get any elements of the C_20 subgroup except the identity with a product of elements of the C_12 subgroup.
     
  2. jcsd
  3. Jan 8, 2008 #2

    morphism

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    The union of two groups? What binary operation are you giving it, i.e. how is C_20[itex]\cup[/itex]C_21 a group?
     
  4. Jan 9, 2008 #3
    Oh. I forgot that the binary operation has to be defined between elements of C_20 and C_21, not just within C_20 and within C_21.

    Then it probably is true. Let me think about it.
     
  5. Jan 9, 2008 #4

    Dick

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    Do that. This is not the most difficult question you've posted by a long shot.
     
  6. Jan 9, 2008 #5
    I assume that the two elements in the problem statement are not necessarily distinct. Otherwise, Z_11 and A = {0,1,2,3,4,5} is a counterexample because you cannot add any two distinct elements of A to get 10.

    Let S be the set of all ordered 2-element subsets of A. S has |A|^2-|A| elements. |A|^2 > |G|^2/4-|G|/2. I want to show that at least |G| of the products of the of the two-element sets in A are distinct. This approach does not seem like it will work...

    Maybe I should consider cases. If the e is in A, then any element a in A equal a*e. If e is not in A, I'm not really sure what to do....
     
  7. Jan 9, 2008 #6

    morphism

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    Here's a hint: inverses are key.
     
  8. Jan 9, 2008 #7
    Very nice hint. Short but helpful.

    Let g be any element of G. Then the left coset g*A^{-1} has order greater than |G|/2, so it must intersect A, since A also has the same order. That means there must be elements in A, a_1 and a_2, such that g*a_1^{-1}=a_2. That is g=a_1*a_2.

    Is that right?
     
  9. Jan 9, 2008 #8

    morphism

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    Looks good to me (although I believe the word "coset" is reserved for subgroups)!
     
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