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Set into Group

  1. Aug 29, 2007 #1
    I have seen this problem a long time ago. It is really supprising, maybe you shall like it as well.

    Given any non-empty set we can define a binary operation on this set to turn it into a group.
     
  2. jcsd
  3. Aug 30, 2007 #2

    morphism

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    If the set is countable, then this is no problem, because we can turn it into the appropriate cyclic group. I forgot the actual solution, but for uncountable sets I think we can use the well-ordering theorem to get an equivalent ordinal, and then we give the set of finite subsets of this ordinal the group operation of symmetric difference. There is a bijection between this latter set and our original one, so we simply copy the new-found group structure. I believe this even gives us an abelian group.

    Is this the solution you saw?
     
  4. Aug 30, 2007 #3

    mathwonk

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    if you want one less exotic, or non commutative, try the vector space over Q with basis your given set, or even the free group generated by your set.

    again you can use the "same cardinality" trick of morphism, to replace the elements of your group by the original set elements.
     
  5. Sep 1, 2007 #4

    Chris Hillman

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    Vector space over Q with basis given X

    Just thought I'd mention a keyword: Hamel basis.
     
  6. Sep 2, 2007 #5

    mathwonk

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    well, given a set S, i just meant the vector space which is the set of all functions on the set S, into Q, with value zero except at a finite number of elements of S.

    so this space has an obvious basis, namely the functions with exactly one non zero value.

    the phrase hamel basis to me usually means you have a vector space first, and then invoke the zorn lemma to prove a basis exists, in cases where an obvious one is not at hand.

    so it is much easier and more constructive, to go from a basis to a vector space spanned by that basis, than the other way around.
     
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