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Math Help - Proof involving Cosets

  1. #1
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    Post Proof involving Cosets

    Hey everyone I have a homework problem me and a few friends have been working on for over a week and cant find a solution we all like. Any input would be greatly appreciated. Heres the Problem:

    Let H be a subgroup of G with aha(inverse)∈H for all a∈G. Prove that
    aH = Ha for all a∈G.
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  2. #2
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    Let H be a subgroup of G with aha(inverse)∈H for all a∈G.
    You probably mean "for all h\in H" in addition to "for all a\in G".

    This is easy. To show aH\subseteq Ha for all a\in G you need to show that for every a\in G and every h\in H there exists h'\in H such that ah=h'a. An equivalent equality can be obtained by multiplying both sides on the right by a^{-1}. Does such h'\in H actually exist?

    For the opposite inclusion, again write the statement to be proved and multiply the equality by a^{-1} on the left. Also, a^{-1} can be renamed into some a_1\in G.
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  3. #3
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    thank you so much i seem to always make these 10x harder than they really are. I really appreciate it
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