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Subgroups and LaGrange's Theorem

  1. Mar 16, 2009 #1
    Let H and K be finite subgroups of a group G whose orders are relatively prime. Show H and K have only the identity element in common.

    By LaGrange's theorem I know that the orders of H and K must divide the order of G. I have attempted a proof by contradiction but have had no luck arriving at a contradiction. Any ideas?

    Thanks.
     
  2. jcsd
  3. Mar 16, 2009 #2

    matt grime

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    You are correct, and you do need Lagrange. But you're asked what the intersection of H and K is. Firstly, is HnK a group?
     
  4. Mar 16, 2009 #3
    Well, I suppose it would be a group if it contained the identity, g and the inverse of g. This is what I was trying to use to find a contradiction but I'm not sure how this leads to the fact that H and K are relatively prime.
     
  5. Mar 16, 2009 #4

    Dick

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    You should probably answer Matt's question without the 'if'. If the intersection of H and K is a group, then it's a subgroup of H and K. That tells you something about divisors of the order of H and K.
     
  6. Mar 16, 2009 #5

    matt grime

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    You are attempting to deduce something from the fact that they are relatively prime.

    H and K have coprime orders. So, by Lagrange, if L is a subgroup of H and K, then what do you know?
     
  7. Mar 16, 2009 #6
    Oh, okay. I see it now. Thanks very much.
     
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