Results 1 to 8 of 8

Math Help - 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 c \in G such that cH=bH, then  Hc = Hb
    3. If  f \in G such that Hf=Hb , then fH=bH

    My Proof.

    1 to 2:

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

    Now, suppose that y is in Hc, then there exist an element d in H such that 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. bH=Hb). Let c be such that cH=bH. Hence cH=bH=Hb.

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

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

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

    Claim: uH \subseteq Hu

    Let  x \in uH , then there exist an element a \in H such that 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 u\in bH, so you want to prove that u\in Hb.

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

    Two remarks:
    1) There is something we should have mentioned first: if xH\subset yH, then xH=yH. We already proved this once or twice: we have x=yh for some h\in H, hence y=xh^{-1} and 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 bH\subset Hb. There remains to prove that Hb\subset bH. However, if you look at the proof of (i) implies (ii), we only made use of the inclusion bH\subset Hb. Hence we have proved that:
    bH\subset Hb is equivalent to: for every c\in G, 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: bH\subset Hb is equivalent to Hb\subset bH, or bHb^{-1}\subset H is equivalent to bHb^{-1}=H.
    If the subgroup H is finite, this is OK ( bH and 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 bH\subset Hb and not Hb\subset bH.

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

    A possible correct statement would be: For any b\in G, the following properties are equivalent:
    (i') bH\subset Hb;
    (ii) For any c\in G, if cH=bH, then 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 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; September 7th 2008 at 03:02 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: October 30th 2010, 06:28 PM
  2. Coset help
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 12th 2010, 06:25 AM
  3. Coset Representatives
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 31st 2010, 04:15 AM
  4. Coset Decoding
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: May 25th 2010, 05:01 PM
  5. Subgroup coset problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 12th 2008, 10:38 PM

Search Tags


/mathhelpforum @mathhelpforum