Results 1 to 8 of 8

Thread: Coset problem

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    Coset problem

    Let H be a subgroup of G, and b a fixed element of G.

    Prove the following statements are equviaent.

    1. bH = Hb
    2. If $\displaystyle c \in G $ such that $\displaystyle cH=bH$, then $\displaystyle Hc = Hb$
    3. If $\displaystyle f \in G $ such that $\displaystyle Hf=Hb $, then $\displaystyle fH=bH$

    My Proof.

    1 to 2:

    Let x be in cH, then there exist an element u in H such that $\displaystyle x=cu$. Since $\displaystyle cH=bH$, there exist an element v in H such that $\displaystyle x=cu=bv$. Since $\displaystyle bH=Hb$, there exist an element w in H such that $\displaystyle x=cu=bv=wb$.

    Now, suppose that y is in Hc, then there exist an element d in H such that $\displaystyle y=dc$.

    I want to get y to look like Hb, but can't really get it...

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    About 1 to 2:

    Assume 1 (i.e. $\displaystyle bH=Hb$). Let $\displaystyle c$ be such that $\displaystyle cH=bH$. Hence $\displaystyle cH=bH=Hb$.

    Notice that $\displaystyle c\in cH=bH=Hb$, so that there are $\displaystyle h,g\in H$ such that $\displaystyle c=bh=gb$.

    Now, if $\displaystyle y\in Hc$, $\displaystyle y=dc$ for some $\displaystyle d\in H$, hence $\displaystyle y=d(gb)=(dg)b\in Hb$ (because $\displaystyle H$ is a subgroup), and $\displaystyle Hb=cH$, so that $\displaystyle y\in cH$. We've proved that $\displaystyle Hc\subset cH$. Let's prove the reverse inclusion. If $\displaystyle y\in cH$, then $\displaystyle y=db$ for some $\displaystyle d\in H$ (because $\displaystyle cH=Hb$), hence $\displaystyle y=db=d(g^{-1}c)=(dg^{-1})c\in Hc$ (because $\displaystyle H$ is a subgroup, again). This proves that $\displaystyle cH\subset Hc$. Finally, $\displaystyle Hc=cH$.

    Laurent.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2
    Thanks for the tips, and I have proved from i to iii similiarly.

    Now, I'm trying to prove from ii to i:

    Let c be in G such that cH=bH, and we have Hc=Hb.

    c is in cH and Hc, so there exist elements x,y in H such that c=bx=yb. Implies that $\displaystyle b=cx^{-1}=y^{-1}c$

    Suppose that u is in bH, then there exist h in H such that u=bh, I'm trying to get this to look like (xxx)b, am I still on the right track?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Don't forget that (ii) is "for every c such that ..." and not "there exists c such that...", so you'll probably need to apply (ii) to some well chosen c.
    The most obvious choice would be to try applying (ii) to the element $\displaystyle u$ you introduced. You can check that it indeed works, and conclude the proof.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2
    Here is what I have so far:

    Let $\displaystyle u=bh$ then $\displaystyle uH=(bh)H=bH$, then I have $\displaystyle Hu=Hb$

    I want to show that $\displaystyle uH = Hu $, which would mean that $\displaystyle bH=Hb$

    Claim: $\displaystyle uH \subseteq Hu$

    Let $\displaystyle x \in uH $, then there exist an element $\displaystyle a \in H $ such that $\displaystyle x = ua = (bh)a$

    Am I on the right track here?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Remember you let $\displaystyle u\in bH$, so you want to prove that $\displaystyle u\in Hb$.

    Once you have proved $\displaystyle Hu = Hb$, the trivial identity $\displaystyle u\in Hu$ shows you that $\displaystyle u\in Hb$. Hence $\displaystyle bH\subset Hb$.

    Two remarks:
    1) There is something we should have mentioned first: if $\displaystyle xH\subset yH$, then $\displaystyle xH=yH$. We already proved this once or twice: we have $\displaystyle x=yh$ for some $\displaystyle h\in H$, hence $\displaystyle y=xh^{-1}$ and $\displaystyle yH=xh^{-1}H\subset xH$. Using this, the proof could have been shortened a bit.

    2) More important: I wonder if the statement of your problem is correct. The reason is: we proved that (ii) implies $\displaystyle bH\subset Hb$. There remains to prove that $\displaystyle Hb\subset bH$. However, if you look at the proof of (i) implies (ii), we only made use of the inclusion $\displaystyle bH\subset Hb$. Hence we have proved that:
    $\displaystyle bH\subset Hb$ is equivalent to: for every $\displaystyle c\in G, $ $\displaystyle cH=bH\Rightarrow Hc=Hb.$
    The same way, there is an equivalence between the inverse inclusion and (iii).
    So if your statement were exact, (ii) and (iii) would be equivalent, which means that: $\displaystyle bH\subset Hb$ is equivalent to $\displaystyle Hb\subset bH$, or $\displaystyle bHb^{-1}\subset H$ is equivalent to $\displaystyle bHb^{-1}=H$.
    If the subgroup $\displaystyle H$ is finite, this is OK ($\displaystyle bH$ and $\displaystyle Hb$ have same cardinality). Otherwise I feel like this is not always true but I can't find a counter-example. I'm going to ask the question in a new thread.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    As a conclusion:

    The answer (by NonCommAlg) to this thread proves that it is possible to have $\displaystyle bH\subset Hb$ and not $\displaystyle Hb\subset bH$.

    For that reason, (ii) and (iii) can't be equivalent.

    A possible correct statement would be: For any $\displaystyle b\in G$, the following properties are equivalent:
    (i') $\displaystyle bH\subset Hb$;
    (ii) For any $\displaystyle c\in G$, if $\displaystyle cH=bH$, then $\displaystyle Hc=Hb$.

    And the same with reverse inclusion and (iii).

    I spent a bunch of time yesterday trying to prove (i) from (ii) before I realized this was possibly false... Where did you get this exercise from? (or did you forget to mention that $\displaystyle H$ is supposed to be finite?)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2
    This is from a modern algebra class, a written problem presented by our professor.

    I think I missed something from the problem, here is the whole thing:

    Let H be a finite subgroup of a group G, and let b be a fixed element of G. Prove that the following statements are equivalent. (G need not be a finite group)

    Sorry, my notes are pretty bad so I kept making mistake here.

    Thanks for your help!
    Last edited by ThePerfectHacker; Sep 7th 2008 at 02:02 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Coset Representatives
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Oct 30th 2010, 05:28 PM
  2. Coset help
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 12th 2010, 05:25 AM
  3. Coset Representatives
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Aug 31st 2010, 03:15 AM
  4. Coset Decoding
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: May 25th 2010, 04:01 PM
  5. Subgroup coset problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Sep 12th 2008, 09:38 PM

Search Tags


/mathhelpforum @mathhelpforum