Results 1 to 3 of 3

Math Help - [SOLVED] Homomorphisms groups and subgroup help

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    43

    [SOLVED] Homomorphisms groups and subgroup help

    Been working for hours on my homework...brain is fried. Can not figure out this last problem. Any help/guidance would be appreciated. Thanks.

    Prove or disprove that if phi and psi are homomorphisms between groups G and G', then the set



    H = { x element of G | psi(x) = phi(x)}

    is a subgroup of G.

    thanks again!
    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 kpizle View Post
    Been working for hours on my homework...brain is fried. Can not figure out this last problem. Any help/guidance would be appreciated. Thanks.

    Prove or disprove that if phi and psi are homomorphisms between groups G and G', then the set



    H = { x element of G | psi(x) = phi(x)}

    is a subgroup of G.


    thanks again
    Problem: Let \Phi:G\mapsto G' and \Psi:G\mapsto G' be homomorpisms. Is the set \mathcal{H}=\left\{g\in G:\Phi(g)=\Psi(g)\right\} a subgroup of G? Prove or disprove your answer.

    Claim: It is.

    Proof: Let us verify the four neccessary conditions to be a subgroup

    1. Clearly \mathcal{H} will inherit G's associativity.

    2. You should know that e\in\text{Ker}\left(\Phi\right),\text{Ker}\left(\P  si\right). Thus \Phi\left(e_G\right)=e_{G'}=\Psi\left(e_G\right)\i  mplies e_G\in\mathcal{H}.

    3. Now suppose that g\in\mathcal{H}, then \Phi\left(g\right)=\Psi\left(g\right), so that \Phi^{-1}(g)\Psi(g)=\Phi^{-1}(g)\Psi(g)=e_{G'}\implies \Phi^{-1}(g)=\Psi\left(g^{-1}\right)\implies\Phi\left(g^{-1}\right)=\Psi\left(g^{-1}\right) which of course implies g^{-1}\in\mathcal{H}.

    4. Suppose g,g'\in\mathcal{H} then \Phi\left(g\right)=\Psi\left(g\right) and \Phi\left(g'\right)=\Psi\left(g'\right). Therefore \Phi\left(gg'\right)=\Phi(g)\Phi\left(g'\right)=\P  si\left(g\right)\Psi\left(g'\right)=\Psi\left(gg'\  right). Therefore gg'\in\mathcal{H}

    This completes the proof.


    Remark: I always find the direct satsifaction of the subgroup axioms to be more instructive. You may (and should) attempt to redo this using the condition that if a,b\in\mathcal{H}\implies ab^{-1}\in\mathcal{H}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    43
    Drexel28:

    Thank you very much.

    After a night's sleep, this seems much clearer.

    I was not sure whether to try to prove with the 3 conditions of a subgroup, or the Subgroup Criterion. After attempting both (after working 8 other of these suckers), I felt like I was doing it wrong. But looking at my attempt and your reply, I was on the right track.

    Thanks again!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Abelian and isomorphic groups, homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 1st 2011, 04:52 AM
  2. Number of homomorphisms between two groups
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 3rd 2011, 05:10 AM
  3. groups...normalizer of subgroup..
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: July 25th 2010, 06:57 PM
  4. A subgroup of symmetry groups
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 9th 2008, 03:10 PM
  5. groups homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: February 3rd 2008, 02:20 PM

Search Tags


/mathhelpforum @mathhelpforum