Results 1 to 4 of 4

Math Help - another group theory question

  1. #1
    Junior Member
    Joined
    Nov 2010
    Posts
    32
    Awards
    1

    another group theory question

    Let G be a group and H be a subgroup of G. For any g_1 , g_2 \in G prove that:
    g_2 \in g_1H \Rightarrow g_1H = g_2H
    Use this to prove that any two left cosets of H are equal or disjoint.

    help would be much appreciated! thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by situation View Post
    Let G be a group and H be a subgroup of G. For any g_1 , g_2 \in G prove that:
    g_2 \in g_1H \Rightarrow g_1H = g_2H
    Use this to prove that any two left cosets of H are equal or disjoint.

    help would be much appreciated! thanks
    I'm sure you're aware that (ab)H=a(bH) and hH=H for all h\in H. So, note that g_2\in g_1 h\implies g_2=g_1h for some h\in H and so g_1h^{-1}=g_2. Thus, g_2H=(g_1h^{-1})H=g_1(h^{-1}H)=g_1H
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2010
    Posts
    32
    Awards
    1
    Quote Originally Posted by Drexel28 View Post
    I'm sure you're aware that (ab)H=a(bH) and hH=H for all h\in H. So, note that g_2\in g_1 h\implies g_2=g_1h for some h\in H and so g_1h^{-1}=g_2. Thus, g_2H=(g_1h^{-1})H=g_1(h^{-1}H)=g_1H

    thanks man, just a quick one about the result hH=H how is this the case? if you wouldnt mind explaining that further it would be much appreciated!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by situation View Post
    thanks man, just a quick one about the result hH=H how is this the case? if you wouldnt mind explaining that further it would be much appreciated!
    No problem! By definition hH=\left\{hh':h'\in H\right\} clearly then since each element is the multiplication of elements in H we have that hH\subseteq H. But, if h'\in H then h^{-1}h'\in H and so h(h^{-1}h')=h'\in hH and so H\subseteq hH. Therefore, H=hH (this is a result usually proven along with Cayley's theorem, that the map \sigma_h:H\to H:h'\mapsto hh' is a permutation)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Question on group theory
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 27th 2010, 09:14 AM
  2. group theory question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 9th 2009, 10:47 AM
  3. Question in group theory.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 3rd 2008, 11:05 PM
  4. A question in group theory.
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 10th 2008, 11:38 AM
  5. Group Theory Question, Dihedral Group
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 4th 2008, 10:36 AM

Search Tags


/mathhelpforum @mathhelpforum