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Basic group theory prove xH=yH

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Let H be a subgroup of G
    Prove xH=yH ⇔ x-1.y[itex]\in[/itex]H


    2. Relevant equations



    3. The attempt at a solution
    If x.H = y.H then x,y[itex]\in[/itex]H
    since H is a subgroup x-1,y-1[itex]\in[/itex]H
    and the closure of H means x-1.y[itex]\in[/itex]H

    Proving the reverse is my problem despite the fact that I'm sure is very easy but i just can't see it.

    What I want to do is show that x-1.y[itex]\in[/itex]H implies x,y[itex]\in[/itex]H
    In which case x.H=H=y.H.

    How is the best way to show this?
     
  2. jcsd
  3. Oct 7, 2012 #2
    If [itex]xH = yH[/itex] then [itex]x,y \in H[/itex]. This is not true. Consider the integers. And conside the sub group [itex]5Z[/itex]. Now, [itex]3+5Z = 8+5Z[/itex] but neither 8 nor 3 is in [itex]5Z[/itex]. So, if [itex]xH=yH[/itex] then for every [itex]h\in H[/itex] there is a [itex]h' \in H[/itex] with [itex]xh=yh'[/itex]. Now, use that fact to prove this direction.

    Try to do something similar for the other direction. You know that there is some [itex]h \in H[/itex] with [itex]y = xh[/itex], see what you can make of that.
     
  4. Oct 11, 2012 #3
    Thanks.

    So we know that x=y.h for some h[itex]\in[/itex]H
    therefore e=x-1.y.h
    e.h-1= x-1.y
    Therefore x-1.y[itex]\in[/itex]H since h-1[itex]\in[/itex]H

    So the reverse could be

    x-1.y=h[itex]\in[/itex]H
    therefore y=x.h and this tells us xH=yH by the theorem that xH=yH ⇔ x=y.h for some h[itex]\in[/itex]H
     
  5. Oct 11, 2012 #4
    yes, that seems correct to me.
     
  6. Oct 12, 2012 #5
    Thanks for the help
     
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