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Is this a surjective homomorphism?

  1. Oct 9, 2008 #1
    I'm trying to prove that if M,N are normal in G and MN = G, then

    [tex]G/(M\cap N)\cong G/M \times G/N[/tex]

    In an attempt to use the 1st Isom. Thm, I have a homomorphism from G to G/M x G/N :

    [tex]g \mapsto (gM, gN)[/tex]

    The kernel is [tex]M\cap N[/tex], so I just have to show that the function is onto to get the isomorphism.

    My guess is that (aM, bN) = (abM, abN). I am having a difficult time showing this, or I may be wrong.

    Any help?
     
    Last edited: Oct 9, 2008
  2. jcsd
  3. Oct 9, 2008 #2

    morphism

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    The fact that G=MN=NM is important here. Write a=n1m2 and b=m2n2 and expand.
     
  4. Oct 9, 2008 #3
    I don't understand how if you have a=n_1 m_2, then you can write b=m_2 n_2, if a and b are arbitrary
     
  5. Oct 9, 2008 #4

    morphism

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    Sorry, that was an unfortunate typo. It should've been "a=n1m1" instead.
     
  6. Oct 9, 2008 #5
    So suppose [tex]a=n_1m_1[/tex] and [tex]b=m_2n_2[/tex]

    Then [tex]aM=n_1 M[/tex] and [tex]bN=m_2 N[/tex].

    Also, [tex]abM=n_1m_1m_2n_2M=n_1m_3n_2M=n_1n_2m_4M=n_3M[/tex]

    I've been trying this method for days but I can't connect the two. I must be missing some important fact about cosets and/or normal groups
     
  7. Oct 9, 2008 #6

    morphism

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    Here's another hint: [itex](n_1 M, m_2 N) = (n_1 M, n_1 N) (m_2 M, m_2 N)[/itex].
     
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