1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...