1. Not finding help here? Sign up for a free 30min 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!

Modern Algebra: Basic problem dealing with Cosets

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    If H is a subgroup of G and Ha = bH for elements a and b in G, show that aH = Hb.

    2. Relevant equations
    None needed

    3. The attempt at a solution
    I've basically just been fiddling around by right and left side multiplication of inverses and what not and can't seem to get it in the right form.. anyone want to guide me in the right direction?

    My attempt:
    Ha = bH --> (b^-1)Ha(a^-1) = (b^-1)bH(a^-1) = (b^-1)H = H(a^-1)
     
  2. jcsd
  3. Sep 30, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    Involving the inverses of [itex]a[/itex] and [itex]b[/itex] seems unproductive, since you know nothing about them beyond that they exist.

    Let [itex]h \in H[/itex]. If [itex]Ha = bH[/itex], what can you say about the element [itex]ha \in bH[/itex]?
     
  4. Sep 30, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I wouldn't try to do it that way. I would, instead, use individual elements of H. And then use inverse of the elements of H, not a and b. "Ha= bH" means that, for any x in H, there exist y in H such that xa= by. Multiply on both sides, on the right, by [itex]y^{-1}[/itex] to get [itex]xay^{-1}= b[/itex]. Now multiply both sides, on the left, by [itex]x^{-1}[/itex]: [itex]ay^{-1}= x^{-1}b[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Modern Algebra: Basic problem dealing with Cosets
  1. Modern Algebra problem (Replies: 3)

Loading...