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Prove a subgroup of G/H X G/K is isomorphic to G/(H intersect K)

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose H and K are normal subgroups of G. Prove that G/H x G/K has a subgroup isomorphic to G/(H[itex]\cap[/itex]K)



    2. Relevant equations



    3. The attempt at a solution

    I was trying to find a homomorphism from G to G/H x G/K where G/(H[itex]\cap[/itex]K) is the kernal. Maybe something like if g is in H it getts mapped to (Hg, e), but nothing like that worked. I'm really stuck on this one.
     
  2. jcsd
  3. Nov 6, 2012 #2

    jbunniii

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    That doesn't make any sense, because [itex]G/(H \cap K)[/itex] is not a subgroup of [itex]G[/itex]. What you need is a homomorphism with kernel [itex]H \cap K[/itex].

    Assuming you're working with right cosets, the natural homomorphism from G to G/H is [itex]g \mapsto Hg[/itex], and the natural homomorphism from G to G/K is [itex]g \mapsto Kg[/itex]. Can you use these to construct a homomorphism from G to G/H x G/K?
     
  4. Nov 7, 2012 #3
    Right. Thanks, that is what I meant. So g↦ (Hg, Kg) is a homomorphism and that's easy to prove. Also H∩K will be the kernal. If I'm not mistaken, I just need to show that {Hg x Kg} is a subgroup of G/H x G/K and then the homomorphism will be onto that subgroup by the way I defined it. Then G/(H∩K) is isomorphic to {Hg x Kg}

    Thanks for your help.
     
  5. Nov 7, 2012 #4

    jbunniii

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    No need to do that - surely you have encountered the theorem that the image of a homomorphism is always a subgroup. If not, it's easy to prove - easier than trying to show it for a specific case like this one.

    By the way, I understand what you mean by your notation, {Hg x Kg}, but I don't think it's very good notation. I would suggest writing something like [itex]\{(Hg, Kg) : g \in G\}[/itex].
     
  6. Nov 7, 2012 #5

    micromass

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    Another question: have you seen the isomorphism theorems yet?? They might come in handy.
     
  7. Nov 8, 2012 #6

    jbunniii

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    I have been implicitly assuming that he has the first isomorphism theorem available. Avatarjoe, is that a valid assumption?
     
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