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Question about Cosets and Lagrange's Theorem

  1. Dec 22, 2014 #1
    If a is an element of G and H is a subgroup of H, let Ha be the right coset of H generated by a.

    is Ha a subgroup?

    I have this question because i feel like the answer should be know, yet my textbook notation makes it look like it is.

    Why I think it should only be a set:
    Let G = <a> and H = <a^3>. and |a| = 6,.
    Then H = {1,a^3}, Ha = {a,a^4} Ha^2 = {a^2,a^5}.
    To me it looks like Ha and Ha^2 can not be subgroups because they are not closed under the operation, unless the operation is multiplying the elements by a^3.. I don't know i'm confused.

    But then I think they should be considered subgroups, because it goes on to say that Ha = H iff a is in H. So Ha must be a subgroup if it equals a subgroup... Anyway, anyone have insight here?
     
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  3. Dec 22, 2014 #2

    Stephen Tashi

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    The problem doesn't not say that [itex] a [/itex] must be an element of [itex] H [/itex]. What the textbook goes on to say about the case when [itex] a [/itex] is an element of [itex] H [/itex] is not relevant to the problem.
     
  4. Dec 23, 2014 #3

    jbunniii

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    ##Ha## is a subgroup if and only if ##Ha = H## if and only if ##a \in H##.

    One easy way to see this is that if ##Ha \neq H## then ##Ha## and ##H## are disjoint, so ##Ha## does not even contain the identity element.
     
  5. Dec 24, 2014 #4
    so when you create G out of disjoint cosets, you are joining H (a subgroup of G) with Ha, Hg, Hh, etc etc, which are only disjoint sets of elements of G?
     
  6. Dec 24, 2014 #5

    jbunniii

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    Yes, ##G## is the disjoint union of ##H## and the other cosets of ##H##. Only ##H## is a subgroup. The other cosets are subsets of ##G## with the same size as ##H##, but they aren't subgroups.
     
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