Results 1 to 2 of 2

Math Help - please check for me

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    86

    please check for me

    I have a theorem states that:
    Let G=H\times K, and let H_{1}\triangleleft H and K_{1}\triangleleft K. Then H_{1}\times K_{1}\triangleleft G and G/(H_{1}\times K_{1})\cong(H/H_{1})\times(K/K_{1}).

    Then I have this corollary:
    If G=H\times K, then G/(H\times\{{1}\})\cong K.

    Please check for me if my proof for the corollary is correct.

    Proof:
    We consider H_{1}=H, i.e. H has no proper normal subgroup.
    Also,
    K_{1}=\{1\}.
    By last theorem, G/(H_{1}\times K_{1})\cong (H/H_{1})\times(K/K_{1}).
    We have G/(H\times\{1\})\cong (H/H)\times(K/\{1\}).
    Since H has no proper normal subgroup, so quotient group of H is not defined. Also, (K/\{1\})=K.
    Hence, G/(H\times\{1\})\cong K
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by deniselim17 View Post
    Proof:
    We consider H_{1}=H, i.e. H has no proper normal subgroup.
    Also,
    K_{1}=\{1\}.
    By last theorem, G/(H_{1}\times K_{1})\cong (H/H_{1})\times(K/K_{1}).
    We have G/(H\times\{1\})\cong (H/H)\times(K/\{1\}).
    Since H has no proper normal subgroup, so quotient group of H is not defined. Also, (K/\{1\})=K.
    Hence, G/(H\times\{1\})\cong K
    It is correct except for a phrase you use "H has no proper normal subgroup", why not?
    It does not matter. Just choose H_1 = H. We really do not care if it has proper normal subgroup (by the way it does, it has \{ 1 \}) and then apply the theorem and get your result. But then you say something that is again not necessary "so the quotient group of H is not defined", yes it is! Certainly H\triangleleft H with H/H \simeq \mathbb{Z}_1. It is defined, it is just not a very interesting group.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: March 12th 2011, 04:05 AM
  2. Check this pls
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 31st 2010, 01:23 PM
  3. can some one check this for me?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 9th 2009, 09:24 AM
  4. Somebody check if this is ok!
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 16th 2008, 09:02 AM
  5. Check IVP
    Posted in the Calculus Forum
    Replies: 14
    Last Post: September 12th 2007, 11:33 AM

Search Tags


/mathhelpforum @mathhelpforum