1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Showing an operator is well-defined

  1. Nov 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##H## be a normal subgroup of a group ##(G, \star)##, and define ##G/H## as that set which contains all of the left cosets of ##H## in ##G##. Define the binary operator ##\hat{\star}## acting on the elements of ##G/H## as ##g H \hat{\star} g' K = (g \star g') H##.

    2. Relevant equations


    3. The attempt at a solution
    I am having difficulty demonstrating that ##\hat{\star}## is well-defined. I understand that it is two cosets to be equal. So, for instance, we could have

    ##g_1 H = g_2 H##

    and

    ##g_3 H = g_4 H##

    So, to show that the operator is well-defined, I would have to show that

    ##g_1 H \hat{\star} g_3 H = g_2 H \hat{\star} g_4 H##

    is true. However, I am having difficulty with this. If I understand correctly, ##g_1 H = g_2 H## does not necessarily imply that ##g_1 = g_2##.
     
  2. jcsd
  3. Nov 21, 2014 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    ##g H \hat{\star} g' H = (g \star g') H## presumably.
    Yes, but write out what that means according to the definition of ##\hat{\star}##. Then consider what ##(g_1 \star g_3) H## is as a set.
     
  4. Nov 21, 2014 #3
    So, let me see if I understand correctly. ##g_1 H~ \hat{\star} ~g_3 H = (g_1 \star g_3) H## is simply some set, as well as ##g_2 H~ \hat{\star}~ g_4 H = (g_2 \star g_4) H##; and we are trying to establish the equality of the sets ##(g_1 \star g_3) H## and ##(g_2 \star g_4) H##. Would we do this by considering an arbitrary element in each set, and then showing that they are equivalent?
     
  5. Nov 21, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes.
    No. It will not in general be true that ##g_1 \star g_3 \star g = g_2 \star g_4 \star g##.
    Write out what ##g_3 H## means in set-theoretic notation (like, {some expression involving h : h element of H}), then extend that to what ##(g_1 \star g_3) H## means.
     
  6. Nov 21, 2014 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Use that H is NORMAL. That's very important.
     
  7. Nov 22, 2014 #6
    Well, the set ##g_3H## is ##\{g_3 h | h \in H\}##, and ##(g_1 \star g_3) H## would be ##\{g_1 g_3 h | h \in H \}##. Because ##H## is normal, that means every ##g \in G## and every ##h \in H##, there exists and ##h_1 \in H## such that ##h## can be written as ##h = g h_1 g^{-1}##.

    Does this mean that there are many representations of ##h##? Does this mean that I should be able to write ##h## in a similar manner as above for any ##g##? For instance, I could write ##h = g_1 h_1 g_1^{-1}##, or ##h = g_2 h_1 g_2^{-1}##?
     
  8. Nov 22, 2014 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Not with the same ##h_1##. I usually think of the definition of ##H## being normal as ##gH=Hg## for any element ##g##. Do you see how H being normal implies that if ##g_1 H=g_2 H## then there are elements of ##H##, ##h## and ##h'## such that ##g_1=g_2 h## and ##g_1=h' g_2##?
     
  9. Nov 22, 2014 #8
    I realize that if ##H## is normal, then, as you said, ##gH = Hg## ##\forall g \in G##. In particular, ##g_2 H = H g_2##. Returning to the assumption that ##g_1 H = g_2 H##, this can be written as ##g_1 H = Hg_2##.

    So, every element in ##g_1 H## will also be in ##g_2 H## and ##Hg_2##. So, because ##g_1 \in g_1 H##, then it must be true that there exist ##h_1 , h_2 \in H## such that ##g_1 = g_2 h_1## and ##g_1 = h_2 g_2##, as you have written. But I am having difficulty connecting that with what we have already covered.
     
    Last edited: Nov 22, 2014
  10. Nov 22, 2014 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You want to show that if ##g_1 H=g_2 H## and ##g_3 H=g_4 H## then ##g_1 g_3 H=g_2 g_4 H##. Try to convert one side into the other.
     
    Last edited: Nov 22, 2014
  11. Nov 23, 2014 #10
    I am sorry, but this problem is giving me great difficulty. We know that ##g_1 = g_2 h_1## and ##g_1 = g_2 h_2##, allowing me to make the substitutions:

    ##(g_1 \star g_3 ) H = (g_2 h_1 \star g_3) H##

    or

    ##(g_1 \star g_3 ) H = (h_2 g_2 \star g_3)H##

    But I do not see how either would be helpful, unless I could somehow eliminate ##h_1## and ##h_2##.
     
  12. Nov 23, 2014 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What is ##h H## where ##h## is an element of ##H##?
     
  13. Nov 23, 2014 #12
    Wouldn't it be that ##hH = H##? No, because if ##h \ne e_H##, then ##hH## would not contain the identity element.
     
  14. Nov 23, 2014 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Think again. ##h^{-1}## is in ##H##.
     
  15. Nov 23, 2014 #14
    Oh, of course! So, then how can I use this fact to simplify ##(g_2 h_1 \star g_3)H##? Would I write ##(g_2 h_1 \star g_3)H = g_2 H ~ \hat{\star} ~h_1 H~ \hat{ \star}~ g_3 H##. And then I can use the fact that ##h_1 H = H = e_H H##. Finally, I would do a similar substitution for ##g_3##. Does sound correct?
     
  16. Nov 23, 2014 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's a little awkward. How about ##g_2 h_1 g_3 H=g_2 h_1 H g_3=g_2 H g_3=g_2 g_3 H##? Can you justify those steps?
     
  17. Nov 23, 2014 #16
    Yes, I can. You seem to use the fact that ##g_3 H = H g_3## twice, and the fact that ##h_1 H = H## in between using the aforementioned fact. Is that right?
     
    Last edited: Nov 23, 2014
  18. Nov 23, 2014 #17

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Exactly.
     
  19. Dec 1, 2014 #18
    Hold on. I don't understand why we have to use the fact that ##H## is normal. We know that ##g_1 \in g_1 H##, and because ##g_1 H = g_2##, we know that ##g_1 = g_2 h_1##; similarly, ##g_3 = g_4h_2##

    ##g_1 H ~ \hat{\star} ~ g_3 H = g_2 h_1 H ~ \hat{\star} ~ g_4h_2 H##

    Because ##hH = H## ##\forall h \in H##, we get

    ##g_1 H~ \hat{\star}~ g_3 H = g_2 H~ \hat{\star}~ g_4 H##

    Thus, the different representations get mapped to the same thing.
     
  20. Dec 1, 2014 #19

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Indeed. But you also need that ##g_1 H g_3 H=g_1 g_3 HH=g_1 g_3 H##. That shows that the product of two cosets is a coset. If ##H## isn't normal that's not necessarily true.
     
  21. Dec 1, 2014 #20
    ##g_1 H g_3 H=g_1 g_3 HH=g_1 g_3 H##

    I don't understand what your wrote in the middle. In the problem statement they define ##\hat{\star}## as ##gH \hat{\star} g' H = (g \star g')H = gg'H##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Showing an operator is well-defined
  1. Well defined (Replies: 12)

Loading...