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Math Help - [SOLVED] intersection of cosets of different subgroups

  1. #1
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    [SOLVED] intersection of cosets of different subgroups

    Let H and K be subgroups of the group G, and let a,b\in G. Show that either aH\cap bK=\emptyset or else aH\cap bK is a left coset of H\cap K.

    I've been working on this one for over a day, now, and I'm still stumped. Assistance would therefore be much appreciated!

    EDIT: I finally solved it, and it is indeed extremely simple. I can't believe I missed it for so long.

    Proof: Suppose aH\cap bK\neq\emptyset, and let x\in aH\cap bK. Then there are h\in H, k\in K with x=ah=bk. So a=bkh^{-1}, and therefore aH\cap bK=bkh^{-1}H\cap bK=bkH\cap bK=bkH\cap bkK=bk(H\cap K). The conclusion follows. \blacksquare
    Last edited by hatsoff; February 1st 2010 at 11:32 AM.
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  2. #2
    Senior Member Dinkydoe's Avatar
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    Suppose \emptyset\neq aH\cap bK

    First: show that c(H\cap K) = cH\cap cK.

    Secondly: Observe that ah=bk \Longleftrightarrow h = (a^{-1}b)k

    Let c = (a^{-1}b) and show that c(H\cap K) = aH\cap bK
    Last edited by Dinkydoe; January 25th 2010 at 08:26 AM.
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  3. #3
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    Quote Originally Posted by Dinkydoe View Post
    Suppose \emptyset\neq aH\cap bK

    First: show that c(H\cap K) = cH\cap cK.

    Secondly: Observe that ah=bk \Longleftrightarrow h = (a^{-1}b)k

    Let c = (a^{-1}b) and show that c(H\cap K) = aH\cap bK
    Thanks, I suspected a^{-1}b(H\cap K) might be the coset in question, but unfortunately I cannot prove this.

    To show that cH\cap cK=c(H\cap K) is quite straightforward: Let x\in cH \cap cK. Then there are h\in H,k\in K with x=ch=ck. So h=k=c^{-1}x, and therefore x\in c(H\cap K). Now let x'\in c(H\cap K). Then there is h'\in H\cap K with x'=ch' and therefore x'\in cH\cap cK. So cH\cap cK=c(H\cap K). The conclusion follows.

    However, I am unable to show that a^{-1}bH\cap a^{-1}bK= aH\cap bK.

    If anyone has a solution, or the outline of a solution, it would be much appreciated!
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Perhaps my last post didn't really clarify that much:

    I hope this will help:

    Observe that aH\cap bK = H\cap a^{-1}bK. This is easy to prove.

    Let c= a^{-1}b\neq e
    One can show that H\cap cK = [H\setminus (H\cap K)]\cap c[K\setminus (H\cap K)] .

    This follows from the fact that different cosets are disjunct. (for any H, we have H \cap cH = \emptyset)

    I'll try to give a better explanation later.
    Last edited by Dinkydoe; January 26th 2010 at 12:05 AM.
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  5. #5
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    Quote Originally Posted by Dinkydoe View Post
    Perhaps my last post didn't really clarify that much:

    I hope this will help:

    Observe that aH\cap bK = H\cap a^{-1}bK. This is easy to prove.

    Let c= a^{-1}b\neq e
    One can show that H\cap cK = [H\setminus (H\cap K)]\cap c[K\setminus (H\cap K)] .

    This follows from the fact that different cosets are disjunct. (for any H, we have H \cap cH = \emptyset)

    I'll try to give a better explanation later.
    Thank you very much for your assistance, but unfortunately I'm still stuck.

    It helps to recall that cosets partition a group, which I had forgotten. Since aH=bH or aK=bK implies aH\cap bK is b(H\cap K) or a(H\cap K), respectively, then we can assume aH\cap bH=aK\cap bK=\emptyset, which means

    H\cap a^{-1}bK=(H\setminus K)\cap a^{-1}bK=(H\setminus[H\cap K])\cap a^{-1}b(K\setminus [H\cap K]),

    as you have already noted. However, I don't see how this implies aH\cap bK = H\cap a^{-1}bK. I worked on it for several hours this morning without further progress.
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  6. #6
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    So, just to recap, we may assume that a^{-1}b\notin H\cup K, because if a^{-1}b is in either subgroup, the result is easy to prove.
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