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Help Understanding Quotient Groups? (Dummit and Foote)

  1. Jun 4, 2013 #1
    The definition given is...

    "Let ##\phi: G \rightarrow H## be a homomorphism with kernel ##K##. The quotient group ##G/K## is the group whose elements are the fibers (sets of elements projecting to single elements of H) with group operation defined above: namely if ##X## is the fiber above ##a## and ##Y## is the fiber above b then the product of ##X## and ##Y## is defined to be the fiber above the product ##ab##."


    But what if we have a homomorphism ##\alpha: G \rightarrow A## and a homomorphism ##\beta: G \rightarrow B## that both have kernel ##K##?

    Wouldn't this mean ##G/K## is not unique because ##\alpha: G \rightarrow A## requires ##G/K## to be fibers consisting of elements in ##A## whereas ##\beta: G \rightarrow B## requires ##G/K## to be fibers consiting of elements in ##B##?


    I'm confused. :(
     
  2. jcsd
  3. Jun 4, 2013 #2

    micromass

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    That's a pretty horrible definition.

    But anyway, you can show that ##G/K## as defined by the map ##\alpha## will be isomorphic to ##G/K## as defined by ##\beta##.
     
  4. Jun 7, 2013 #3
    I left out one part of the definition. It should say

    "Let ϕ:G→H be a homomorphism with kernel K. The quotient group G/K is the group whose elements are the fibers of ϕ (sets of elements projecting to single elements of H) with group operation defined above: namely if X is the fiber above a and Y is the fiber above b then the product of X and Y is defined to be the fiber above the product ab."


    WANT TO SHOW: "If ##\alpha: G \rightarrow A## and ##\beta \rightarrow B## have the same kernel, K, then the quotient group whose elements are the fibers of ##\alpha## is isomorphic to the quotient group whose elements are the fibers of ##\beta##."

    ... I don't know how to show this. :(
    This would mean ##|image(\alpha)|=|image(\beta)|##, but I don't know how to show that either.

    Could I have a push in the right direction?

    EDIT: I'm working on Office_Shredder's response right now...
     
    Last edited: Jun 7, 2013
  5. Jun 7, 2013 #4

    Office_Shredder

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    The more normal definition (haha pun intended as will become clear) is typically this:

    If K is a subgroup of G with the following property: for all [itex] g\in K[/itex] and [itex] h\in G[/itex], [itex] h^{-1} g h \in K[/itex] we call K a normal subgroup.

    G/K is then defined as the set
    [tex] \{ hK\ |\ h\in G \}[/tex]
    is the set of all cosets of G, with the multiplication operation (hK is the set of all elements hg with g in K)
    [tex] \left(hK \right) \cdot \left( gK \right) = \left( (hg) K \right) [/tex]
    K being normal is exactly the property required for this to be a group (exercise: prove this)

    Cool facts: The kernel of a homomorphism is always a normal subgroup (easy exercise to prove). Also, with the above construction every normal subgroup is also the kernel of a group homomorphism, in particular the projection [itex] p: G \to G/K [/itex] with [itex] p(g) = gK[/itex] (the kernel is precisely K)

    The (first?) isomorphism theorem of group theory says that if [itex] \phi:G\to H [/itex] is a group homomorphism. then [itex] G/(ker(\phi)) = \phi(G) [/itex] i.e. the groups on the left and right are isomorphic, and phi is an isomorphism between them - phi is defined on G/(ker(phi)) by [itex] \phi( h ker(\phi)) = \phi(h) [/itex] (you can check this is well defined and still defines a group homomorphism)

    Now given K we could have picked an arbitrary [itex] \phi[/itex] with kernel K, [itex] \phi: G\ to H [/itex] and we see that the fiber definition in the original post gives a group isomorphic to the definition of G/K at the top of this post. Dummit and Foote decided on a different path for defining quotient groups but I think it makes it a lot harder to do calculations than this one (and is a lot less intuitive)
     
  6. Jun 7, 2013 #5
    I don't see why K needs to be normal?

    In order for ##(G/K,\cdot)## to be a group we need:

    i) ##g_1K\cdot(g_2K\cdot{g_3K})=(g_1K\cdot{g_2K})\cdot{g_3K}##.
    ii) ##\exists{e\in{\frac{G}{K}}}## s.t. for all g in ##\frac{G}{K}##, ##e\cdot{gK}=gK##.
    iii) for every ##gK\in{\frac{G}{K}}## we need there to exist a ##(gK)^{-1}## s.t. ##(gK)(gK)^{-1}=e##.
    iv) we need the set to be closed under the operation ##\cdot##.


    i) is satisfied because ##g_1K\cdot(g_2K\cdot{g_3K})=g_1K\cdot((g_2g_3)K)=(g_1(g_2g_3))K=((g_1g_2)g_3)K=(g_1g_2)K\cdot{g_3K}=(g_1K\cdot{g_2K})\cdot{g_3K}##.

    ii) is satisfied because ##e_g{K}\cdot{gK}=(e_gg)K=gK##.

    iii) is satisfied because ##g^{-1}K\cdot{gK}=(g^{-1}g)K=e_gK##, which is the identity of G/K.

    iv) the set is closed under the operation ##\cdot## because if ##g_1K## and ##g_2K## are any two arbitrary elements in the set, then since ##g_1K\cdot{g_2K}=(g_1g_2)K##, and since ##(g_1g_2)\in{G}##, we know ##(g_1g_2)K## is in the set "G/K".





    So why do we need K to be normal?
     
    Last edited: Jun 7, 2013
  7. Jun 7, 2013 #6

    micromass

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    Because you want that if ##aK = a^\prime K## and ##bK = b^\prime K##, then ##(aK)(bK) = (a^\prime K) (b^\prime K)##. This is precisely guaranteed if ##K## is normal.
     
  8. Jun 7, 2013 #7
    I don't understand, why can't we just say:

    If ##aK=a'K## and ##bK=b'K##, then ##(aK)(bK)=(a'K)(bK)=(a'K)(b'K)##.


    ?
     
  9. Jun 7, 2013 #8

    Office_Shredder

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    Pick some g, and h and h' such that hK = h'K. Is (hg)K = (h'g)K?
    This is the same as h(gK) = h'(gK) (this part conforms to your intuition for how these things work) which is NOT immediately implied by hK = h'K (because gK has different elements than K does)
     
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