Results 1 to 7 of 7

Math Help - Cardinality of right and left cosets.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Cardinality of right and left cosets.

    Suppose that G is a group, H is a subgroup of G, and g_1,g_2 \in G. Show that |g_1 H|=|Hg_2|.

    (Do not assume that G is finite).
    I wanted to try and find a bijection between them but I wasn't sure quite how to do it.

    Alternatively, is there a way to prove it without using bijections?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Showcase_22 View Post
    I wanted to try and find a bijection between them but I wasn't sure quite how to do it.

    Alternatively, is there a way to prove it without using bijections?

    Define: \phi: g_1 H \to Hg_2 as \phi(g_1 h) = hg_2

    Show the map is a bijection.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Right, i'm not really sure how to show that it's a bijection but here's what i'll try:

    \phi:g_1H \rightarrow Hg_2 as \phi(g_1h)=hg_2.

    So I need to show that an inverse exists ie. \phi^{-1}(hg_2)=g_1 h

    Is this the best way of doing it or would proving that g_1H\subset Hg_2 and Hg_2 \subset g_1H be better?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Showcase_22 View Post
    Right, i'm not really sure how to show that it's a bijection but here's what i'll try:

    \phi:g_1H \rightarrow Hg_2 as \phi(g_1h)=hg_2.

    So I need to show that an inverse exists ie. \phi^{-1}(hg_2)=g_1 h
    Not really... You have to show that the map is both one-one and onto.

    This is equivalent to doing the following steps:

    One-One: Show that \phi(x) = \phi(y) \Rightarrow x = y

    Onto: Show that \forall y \in Hg_2, \exists x \in g_1H : \phi(x) = y

    Quote Originally Posted by Showcase_22 View Post
    would proving that g_1H\subset Hg_2 and Hg_2 \subset g_1H be better?
    This is wrong, be careful.

    The question is asking you to prove they have the same number of elements , its not asking you to prove the sets are equal.

    The argument you are using is claiming the sets g_1H and Hg_2 are the same. But that's not necessarily true here.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Define, \phi_1 : g_1H \to H by g_1h\mapsto h.
    Define, \phi_2 : H \to Hg_2 by h\mapsto hg_2.
    Show that these are bijections (also that the first one is well-defined).
    Therefore, |g_1H| = |H| and |H|=|Hg_2|, hence, |g_1H| = |Hg_2|.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Show |g_1 H|=|Hg_2|

    Injective: \forall g_1h_1, g_1h_2 \in g_1H, \phi (g_1h_1)=\phi (g_1h_2) \Rightarrow g_1h_1=g_1h_2

    Suppose \phi(g_1h_1)=\phi(g_1h_2) \Rightarrow g_1h_1=g_1h_2. (Need To Show: h_1=h_2).

    Since g_1h_1,g_1h_2 \in g_1H using the property of groups that they have inverses means:

    g_1h_1=g_1h_2
    g_1^{-1}g_1h_1=g_1^{-1}g_1h_2
    eh_1=eh_2
    h_1=h_2

    Therefore the function is injective.

    Surjective: \forall y \in Hg_2 \exists x \in g_1H: \phi(x)=y

    I'm not really sure how to show it is surjective.

    I Need To Show: \phi(g_1h)=hg_2 holds for all elements in Hg_2.

    Can I define \phi to be \phi(g_1h_i)=h_i g_2 where h_i \in H? Since all the h_i's are different this would be a surjection (i'm wondering whether this is a reasonable assumption).

    Combining this result with the proof that it is injective would prove \phi is bijective.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6

    Thumbs up

    You are right

    Surjectivity is not an assumption.

    For the sake of clarity, let me explain the details:

    Every element of Hg_2 is of the form hg_2 for some h \in H. Now \phi maps g_1 h \in g_1H to this hg_2. Thus every element in the range is mapped onto by some element in the domain and hence the map is onto.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Left and Right cosets.
    Posted in the Advanced Algebra Forum
    Replies: 12
    Last Post: February 2nd 2010, 02:35 PM
  2. Left cosets
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 11th 2010, 10:43 AM
  3. left and right cosets
    Posted in the Advanced Algebra Forum
    Replies: 10
    Last Post: November 4th 2009, 09:42 PM
  4. Left and Right Cosets
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 23rd 2009, 12:06 AM
  5. right and left cosets
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 15th 2008, 07:08 PM

Search Tags


/mathhelpforum @mathhelpforum