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Normal Subgroups proof.

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Let G be and group and let N and H be normal subgroups of G with $N \subset H \subset G. Prove that H/N is a subgroup of G/N, and that it is a normal subgroup. Note that aKa^{-1} = {aKa^{-1} | k in K}.


    2. Relevant equations



    3. The attempt at a solution

    I understand that since H is contained in G and N is contained in H that it would make sense that the factor group H/N is not only a subgroup, but a normal subgroup. However, I am struggling trying to figure out a way to transition from aKa^{-1} to either H or N. We've also learned the 2-step check of closure and inverses for proving subgroups, but i'm not quite sure how to apply that to factor groups. If I could get some suggestions where to start that would be great.
     
    Last edited: Apr 17, 2012
  2. jcsd
  3. Apr 17, 2012 #2
    Have you covered in your class that the elements of [itex]H / N[/itex] and [itex]G / N[/itex] are the cosets [itex]N[/itex] in [itex]H[/itex] and [itex]G[/itex]? I might try this problem by taking an arbitrary element [itex]h' \in H / N[/itex] and [itex]g' \in G / N[/itex] and showing that [itex]g' h' g'^{-1} \in H / N[/itex]. Since the elements chosen were arbitrary, [itex]H / N \triangleleft G / N[/itex].
     
  4. Apr 19, 2012 #3
    let x,y be in aKa^-1.

    thus x = aka^-1, for some k in K, and y = ak'a^-1 for some k' in K.

    then xy^-1 = (aka^-1)(ak'a^-1)^-1 = (aka^-1)(ak'^-1a^-1) = a(kk'^-1)a^-1,

    and since K is a subgroup kk'^-1 is in K whenever k,k' are, so xy^-1 is in aKa^-1.

    thus aKa^-1 is certainly a subgroup of G.

    what we need to do is show that aKa^-1 must be a subset of H.

    note that K is a subgroup of H, hence aKa^-1 is a subset of aHa^-1. but H is normal in G, so aHa^-1 = H.

    thus aKa^-1 is contained in H.


    How does this look?
     
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