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Homework Help: Proof: Cosets equal or disjoint

  1. Mar 14, 2012 #1


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    1. The problem statement, all variables and given/known data

    Two left cosets aH, bH of H in G are equal if and only
    if a^{−1}b ∈ H. This is also equivalent to the statement b ∈ aH.
    Suppose that aH = bH. Then e ∈ H. So, b = be ∈ bH. If
    aH = bH then b ∈ aH. So, b = ah for some h ∈ H. But, solving for
    h, we get h = a
    b ∈ H

    2. Relevant equations
    Let G be a group, H a subgroup. a,b ∈ G.
    Def: Left coset aH={ah |h ∈ H}

    I never managed to understand this simple concept of cosets, but maybe understanding this will help. I do not see why when saying that aH = bH implies that b ∈ aH, which implies b = ah for some h ∈ H. That is how {ah |h ∈ H} = {bh |h ∈ H} implies b ∈ aH and b=ah, and not bh ∈ aH and bh =ah ? The definition of the coset only gives information on h, not on b...
  2. jcsd
  3. Mar 14, 2012 #2


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    Science Advisor

    It is the same as you have in the first proof above. H is a subgroup so the identity, e, is in H and therefore bH contains b. Since we are given that aH= bH, b is in aH. Any member of aH is of the form ah so we have b= ah for some h in H.

    That is how {ah |h ∈ H} = {bh |h ∈ H} implies b ∈ aH and b=ah, and not bh ∈aH[/quote]
    What h are you talking about? Certainly it is true that bh∈ aH for some h in H- specifically h= e.

    If that were true then we could multiply both sides of the equation, on the right, by h-1 and get b= a. Saying that bh= a or b= ah', possibly with h= h' or [itex]h\ne h'[/itex], does not lead to ah= bh. Any member of

    Yes, of course, b is "given" and does not change. You don't need to know any thing more about b. The only information on h you need (or have) is that it is a member of H.
  4. Mar 20, 2012 #3


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    Thanks for the reply!
    I think I figured it out just after I posted the message, I think my problem was that I missed the "Suppose that aH = bH" part, which was crucial for the proof to work.
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