Results 1 to 6 of 6

Thread: [SOLVED] intersection of cosets of different subgroups

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    410

    [SOLVED] intersection of cosets of different subgroups

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

    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 $\displaystyle aH\cap bK\neq\emptyset$, and let $\displaystyle x\in aH\cap bK$. Then there are $\displaystyle h\in H$, $\displaystyle k\in K$ with $\displaystyle x=ah=bk$. So $\displaystyle a=bkh^{-1}$, and therefore $\displaystyle aH\cap bK=bkh^{-1}H\cap bK=bkH\cap bK=bkH\cap bkK=bk(H\cap K)$. The conclusion follows. $\displaystyle \blacksquare$
    Last edited by hatsoff; Feb 1st 2010 at 11:32 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    Suppose $\displaystyle \emptyset\neq aH\cap bK$

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

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

    Let $\displaystyle c = (a^{-1}b)$ and show that $\displaystyle c(H\cap K) = aH\cap bK$
    Last edited by Dinkydoe; Jan 25th 2010 at 08:26 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by Dinkydoe View Post
    Suppose $\displaystyle \emptyset\neq aH\cap bK$

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

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

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

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

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

    If anyone has a solution, or the outline of a solution, it would be much appreciated!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    Perhaps my last post didn't really clarify that much:

    I hope this will help:

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

    Let $\displaystyle c= a^{-1}b\neq e$
    One can show that $\displaystyle 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 $\displaystyle H$, we have $\displaystyle H \cap cH = \emptyset$)

    I'll try to give a better explanation later.
    Last edited by Dinkydoe; Jan 26th 2010 at 12:05 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by Dinkydoe View Post
    Perhaps my last post didn't really clarify that much:

    I hope this will help:

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

    Let $\displaystyle c= a^{-1}b\neq e$
    One can show that $\displaystyle 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 $\displaystyle H$, we have $\displaystyle H \cap cH = \emptyset$)

    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 $\displaystyle aH=bH$ or $\displaystyle aK=bK$ implies $\displaystyle aH\cap bK$ is $\displaystyle b(H\cap K)$ or $\displaystyle a(H\cap K)$, respectively, then we can assume $\displaystyle aH\cap bH=aK\cap bK=\emptyset$, which means

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

    as you have already noted. However, I don't see how this implies $\displaystyle aH\cap bK = H\cap a^{-1}bK$. I worked on it for several hours this morning without further progress.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    So, just to recap, we may assume that $\displaystyle a^{-1}b\notin H\cup K$, because if $\displaystyle a^{-1}b$ is in either subgroup, the result is easy to prove.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cosets and Normal Subgroups
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Mar 2nd 2011, 11:15 AM
  2. Cosets and subgroups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 4th 2010, 10:05 PM
  3. Cosets of Subgroups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 18th 2009, 01:29 PM
  4. subgroups and cosets
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Apr 22nd 2009, 02:26 PM
  5. Normal subgroups and cosets
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Feb 23rd 2009, 02:29 PM

Search Tags


/mathhelpforum @mathhelpforum