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Modern Algebra unified subgroup question

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    If H and K are subgroups of G, show HUK is a subgroup of G if and only if H < K or K < H ( the < meaning that all the elements of H are in K or all the elements of K are in H).

    2. Relevant equations
    None

    3. The attempt at a solution
    I believe the problem here is HUK might not be a closed group. Certainly all the elements of HUK are also in G.

    If all the elements of H are in K, then hk is an element of HUK for all h,k.

    If the intersection of H and K does not equal H or K, that means that hk may not be in HUK as it is not closed.

    These are my thoughts so far. Am I on the right track here? How do I start turning this into a proof?
     
  2. jcsd
  3. Sep 30, 2014 #2

    Dick

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    Yes, you are on the right track. Suppose h is not an element of K and k is not an element of H. Can you show hk is not an element of H or K? Use proof by contradiction.
     
  4. Oct 1, 2014 #3

    WWGD

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    There is a nice, helpful result that a subset H of a group G is a subgroup of G if for any a,b in H, ## ab^{-1}## is in H.
    Now if a,b are either both in A or both in B, no problem, but consider what happens when a is in H and b is in K-H ( of course then we show we must have H subset K ).
     
  5. Oct 4, 2014 #4
    h is not an element of K and k is not an element of H. Then hk being an element of HUK is a contradiction because.. i'm lost >.<
     
  6. Oct 4, 2014 #5

    Dick

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    Suppose hk is an element of K. Say hk=k'. Solve for h and think about it.
     
  7. Oct 4, 2014 #6

    WWGD

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    Well, notice that subgroups are closed under the group operation. Let a in H, let b be in K-H and

    ## ab:=c \in H ## . Then ##b= a^{-1} c ## , so that b is the product of elements of H, and the product of

    elements in H must be in H.
     
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