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Factor Groups

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Let G be a group, and let H be a normal subgroup of G. Must show that every subgroup K' of the factor group G/H has the form K'=K/H, where K is a subgroup of G that contains H.


    2. Relevant equations
    I dont see how to get started.


    3. The attempt at a solution
    I wrote down the definitions of each factor group, G/H and K/H. If K'=K/H, then H must be normal in K.
     
  2. jcsd
  3. Nov 25, 2009 #2
    To prove this fact I tried to show that K' is contain in K/H for some subgroup H in G and vice versa.
     
  4. Nov 25, 2009 #3
    It might help thinking of this problem with the natural homomorphis, g: G -> G/H. From this you can show that for any subgroup of K' of G/H, g^-1(K') is a subgroup of G, containing K. Then maybe an isomorphism theorem jumps at you after that?

    Note: g^-1(K') is the pre-image of g, ie; g^-1(K') = { a in G | g(a) is in K' }
     
  5. Nov 26, 2009 #4
    I created an example where this theorem holds and observe that K must be defined as the set containing all elements in each element of K' (i.e., in each coset of G/H that is contain in K'). With this choice of K, the theorem works. I am now having alittle trouble justifying that K is indeed a subgroup.
     
  6. Nov 26, 2009 #5
    Well, this was what I was trying to say:

    Let H be a normal subgroup of G. Let g: G -> G/H. Let K' be a subgroup of G/K. We claim that g^-1(K') = { a in G | g(a) is in K' } forms a subgroup in G, which contains H (and actually we claim to show, later that this is the subgroup we wish to find).

    K = g^-1(K') is a subgroup of G:

    K is not empty, and contains H: In fact, H (which is not empty) is a subset of G, since for any h in H, g(a) = hH = eH (the identity element in G/H, which is clearly an element of K' since K' is a group).

    K is closed: let a, b be in K, then g(a) is in K', and g(b) is in K', so g(ab) = g(a)g(b) is in K' since K' is a group.

    K contains inverses: let a be in K, then g(a) is in K', so g(a)^-1 = g(a^-1) is in K', so a^-1 is in K.

    Ok, so now we show there is an isomorphism between K/H and K'. But this should be clear to you since H is the kernel of the map of g.

    So as we've shown, for each K' a subgroup of G/H is isomorphic to a subgroup K/H, K a subgroup of G.
     
  7. Nov 29, 2009 #6
    I got it, thank you. I did not consider defining this mapping. It follows by closure that g is a homomorphism and that the Kernel of g is normal to K. Then by the first isomorphism theorem it follows that the factor group K/(ker g) is isomorphic to K'.
     
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