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COSETS are equal for finite groups

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove that if H is a subgroup of a finite group G, then the number of right cosets of H in G equals the number of left cosets of H in G

    2. Relevant equations

    Lagrange's theorem: for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G.

    3. The attempt at a solution

    I know how to find subgroups of groups, and how to get the cosets from there. But i just dont understand how to show that right and left cosets will be equal because the group is finite..

    Im stuck :( thanks for any help!!
  2. jcsd
  3. Dec 3, 2007 #2


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    Try reading the proof of Lagrange's theorem.
  4. Dec 3, 2007 #3
    well, i know that since cosets from the subgroup of H form partitions of the group G, and G is finite, then G is completely separated into a finite number of cosets. I know that since each coset has |H| elements, then |G|=|H|*n so therefore |H| divides |G|, proving the theorem. SO, what does this have to say about left and right cosets being equal?
  5. Dec 3, 2007 #4


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    What is n?
  6. Dec 3, 2007 #5
    n is the number of cosets of G
  7. Dec 3, 2007 #6
    And then there was light :).
  8. Dec 3, 2007 #7
    i still dont get it.. haha sorry :(
  9. Dec 3, 2007 #8
    Good. Do the right cosets have the same properties? Think about how you might obtain a right coset given a left coset.
  10. Dec 3, 2007 #9


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    Technically you should re-word the proof to specify which cosets you're working with, i.e. left or right. So in fact n = number of left cosets (if you used left cosets), or n = number of right cosets (if you used right cosets).

    Evidently it doesn't matter, so there must be as many left as right cosets.
  11. Dec 3, 2007 #10
    If there are as many left and right cosets, then it does not matter. However, your remark is begging the question.
  12. Dec 4, 2007 #11


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    Have you already proven, or can you prove, that all left cosets have the same number of members and that all right cosets have the same number of members? That's pretty straight forward. Have you proved that both left cosets and right cosets "partition G" (every member of G is in exactly one coset). It's obvious that if x is a member of H, xH and Hx are just equal to order of H. If every left coset has |H| members and G has |G| members, and left cosets partition G, how many left cosets are there? Same thing for right cosets.
  13. Feb 28, 2010 #12
    I am working on the same problem for my modern algebra class, and I think I have the right answer. However, one of my partners in my group has gone a different way in proving this which I'm not sure if it correct or not. She goes through Lagrange's theorem exactly, and then defines a mapping, alpha, from H to Ha and shows that it is onto and one-to-one.

    Is that a proper way to prove this problem?
  14. Feb 28, 2010 #13


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    That's a great way to solve to problem.
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