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Math Help - [G : H],[G : K] finite -> [G:H n K] finite

  1. #1
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    Smile [G : H],[G : K] finite -> [G:H n K] finite



    I'm not looking for the solution, but a helpful first step.

    So far, I realize that if H \cap K = \{e\}, then [G : H \cap K ] is infinite. More generally, if o(H \cap K) < \infty, then [G : H \cap K] is infinite.

    Problem is, I don't see how to show anything more than e \in H \cap K

    Thanks for your help.
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  2. #2
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    Quote Originally Posted by davismj View Post


    I'm not looking for the solution, but a helpful first step.

    So far, I realize that if H \cap K = \{e\}, then [G : H \cap K ] is infinite. More generally, if o(H \cap K) < \infty, then [G : H \cap K] is infinite.

    Problem is, I don't see how to show anything more than e \in H \cap K

    Thanks for your help.

    Let us denote by G_N the set of left cosets of G wrt some subgroup N .

    Define now \phi:\,G_{H\cap K}\rightarrow G_H\times G_K by \phi(x(H\cap K)):=(xH,xK) . Now show this map is a well-defined 1-1 function so...

    Tonio
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    Quote Originally Posted by tonio View Post
    Let us denote by G_N the set of left cosets of G wrt some subgroup N .

    Define now \phi:\,G_{H\cap K}\rightarrow G_H\times G_K by \phi(x(H\cap K)):=(xH,xK) . Now show this map is a well-defined 1-1 function so...

    Tonio
    If phi is well-defined and one-to-one, then o(G_{H\cap K}) = o(G_H)o(G_K) \implies o(G_{H\cap K}) < \infty?
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  4. #4
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    Quote Originally Posted by davismj View Post
    If phi is well-defined and one-to-one, then o(G_{H\cap K}) = o(G_H)o(G_K) \implies o(G_{H\cap K}) < \infty?
    Yes, but be careful. These sets of cosets are...well....sets. In general the notation for order implies there is a group structure on this. This doesn't change the result, just a remark.
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  5. #5
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    Quote Originally Posted by tonio View Post
    Let us denote by G_N the set of left cosets of G wrt some subgroup N .

    Define now \phi:\,G_{H\cap K}\rightarrow G_H\times G_K by \phi(x(H\cap K)):=(xH,xK) . Now show this map is a well-defined 1-1 function so...

    Tonio
    Okay. I'm still not getting it. Maybe this example is too abstract.

    Lets consider the integers mod 12. Let H = {0,3,6,9} and K = {0,4,8}. Thus, H \cap K = \{0\}.

    Thus \phi (x) would be defined as

    0 = ({0,3,6,9},{0,4,8})
    1 = ({1,4,7,10},{1,5,9})
    2 = ({2,5,8,11},{2,6,10})
    3 = ({0,3,6,9},{3,7,11})
    4 = ({1,4,7,10},{0,4,8})
    5 = ({2,5,8,11},{1,5,9})
    6 = ({0,3,6,9},{2,6,10})
    7 = ({1,4,7,10},{3,7,11})
    8 = ({2,5,8,11},{0,4,8})
    9 = ({0,3,6,9},{1,5,9})
    10 = ({1,4,7,10},{2,6,10})
    11 = ({2,5,8,11},{3,7,11})

    Now, the claim is that for any H \cap K, this function is one-to-one. I don't see how you could make this claim if you don't know anything about H \cap K?

    I notice, though, that in the above example, \phi(x) returns the cosets xH, xK \implies x \in xH \cap xK. If this function is indeed one-to-one, there is one and only one g \in G with x \in gH \cap gK = g(H \cap K). Therefore, since there are finitely many such cosets, there are finitely many such g with \cup_{i \in I}g_i(H \cap K) = G.

    I guess I understand where this is going, but I don't know how to prove the injectivity if you don't know H \cap K.
    Last edited by davismj; May 31st 2010 at 11:28 AM.
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  6. #6
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    After looking at this for a little while longer, I realized that maybe I'm overcomplicating this; maybe I see the logic about why you don't really need to know the intersection of H and K to prove anything.

    Does this look about right?
    Last edited by davismj; May 31st 2010 at 12:22 PM.
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