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Normal and Simple groups

  1. Nov 28, 2009 #1
    H,K are normal subgroups of a (finite) group G, and K is also normal in H. If G/K and G/H are simple, does it follow that H=K?
    I'm almost convinced it does, but I'm having trouble proving it. I mean, the cosets of H partition G and the cosets of K partition G in the same way and on top of that partition H, right? I'm not sure when to bring in normality and the fact that the quotients are simple.
     
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  3. Nov 28, 2009 #2

    morphism

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    What if G=H?
     
  4. Nov 29, 2009 #3
    no, sorry, I should have included that. H and K are proper normal subgroups in G.
     
  5. Nov 29, 2009 #4

    morphism

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    Okay. Now, can you describe the relationship(s) between G/H, G/K and H/K?
     
  6. Nov 29, 2009 #5
    This is actually where I had trouble.
    [tex] G/H = \left\{ H, g_{1}H, g_{2}H,... \right\} [/tex] and [tex] G/K = \left\{ K, g_{1}K, ... \right\} = \left\{ K, g_{1}K, ... \right\} \cup \left\{ K, h_{1}K, ... \right\} [/tex] for [tex] g_{i} \in G-H[/tex] and [tex]h_{j} \in H [/tex]
    That last on is [tex] G/K = \left\{ K, g_{1}K, ... \right\} \cup H/K [/tex] for [tex] g_{i} \in G-H[/tex]. Then, can you say that [tex] G/H = \left\{ K, g_{1}K, ... \right\} [/tex] for [tex] g_{i} \in G-H[/tex], since the intersection of H/K and [tex]\left\{ K, g_{1}K, ... \right\}[/tex] is just the identity element, K. Then the cosets of H in G must coincide with the cosets of K in G. Thus H=K?
    Or am I totally missing it. I must be, since I haven't used either normality and simple quotients...
     
  7. Nov 29, 2009 #6

    morphism

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    I have to run, so I'll leave you with a quick exercise (or reminder, in case you've seen this result previously): (G/K)/(H/K) = G/H. This isomorphism pretty much solves your problem. One way to prove it is to write down an obvious map G/K -> G/H and look at its kernel.
     
  8. Nov 29, 2009 #7
    oh alright. So the proof of the isomorphism uses normality of K in G and H in G (specifically for the map to be a group homomorphism) and the First Isomorphism Theorem. Now, the kernel of the map is normal in its domain (ie G/K), so ker = H/K is normal in G/K. But G/K is simple, so either H/K = {1} or H/K = G/K. If H/K = G/K then (G/K)/(H/K) = {1} and so G/H = {1}. But this implies G=H, which is not the case. Therefore H/K = {1}, and so H=K. Great, thanks a lot!

    But one more thing. I didn't use the fact the G/H is simple. Does it mean you can relax that condition to just exclude it?
     
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