Results 1 to 6 of 6

Math Help - abstract algebra

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    166

    abstract algebra

    Prove that if H is a normal subgroup of G of prime index p then for all subgroups K of G either
    (i) K is a subgroup of H, or
    (ii) G = HK and |K : K intersect H| = p

    I tried to do this in 2 cases. The first case is to let K be a subest of H, then since K is a subgroup of G, it is a subgroup of H. For case 2, I let K not contained in H, then try to show G = HK and |K : K intersect H| = p, but don't know how... please help, thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Let \pi:G\to G/H be the quotient map and consider \pi(K). (It is a subgroup of the simple group G/H.)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2008
    Posts
    166
    Quote Originally Posted by Opalg View Post
    Let \pi:G\to G/H be the quotient map and consider \pi(K). (It is a subgroup of the simple group G/H.)
    I don't get it...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    I don't get it...
    What Opalg means is this:

    The group G/H has prime order, therefore, it only has no proper non-trivial subgroup. The natural projection \pi : G\to G/H defined by \pi (a) = aH is a homomorphism. Thus, by the property of homomorphism, if K is a subgroup of G then \pi (K) is a subgroup of G/H. This means that \pi (K) = G/H or \pi (K) = \{ H \}. We have two cases, (i) \pi (K) = G/H (ii) \pi (K) = \{ H\}. Now consider each case.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2008
    Posts
    166
    Quote Originally Posted by ThePerfectHacker View Post
    if K is a subgroup of G then \pi (K) is a subgroup of G/H. This means that \pi (K) = G/H or \pi (K) = \{ H \}. We have two cases, (i) \pi (K) = G/H (ii) \pi (K) = \{ H\}. Now consider each case.
    How do you know \pi (K) = G/H or \pi (K) = \{ H \}?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    How do you know \pi (K) = G/H or \pi (K) = \{ H \}?
    Because G/H has only non-proper and trivial subgroups. That is G/H - the full group itself. And \{ H \} - the set containing the identity element.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 6th 2010, 03:03 PM
  2. Replies: 0
    Last Post: April 23rd 2010, 11:37 PM
  3. Abstract Algebra
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 22nd 2008, 05:21 PM
  4. abstract algebra
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 4th 2008, 06:00 PM
  5. Abstract Algebra help
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 12th 2008, 01:59 AM

Search Tags


/mathhelpforum @mathhelpforum