1. Not finding help here? Sign up for a free 30min 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!

Write down the cosets (right/left) of this:

  1. Apr 2, 2016 #1

    RJLiberator

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Q: In Qu, write down the elements in all the right cosets and all the left cosets of <j>.

    Screen Shot 2016-04-02 at 10.58.22 AM.png

    2. Relevant equations
    Let H ≤ G. A coset of H is a subset of G of the form Hg for some g ∈ G.
    where ≤ denotes subgroup.

    3. The attempt at a solution

    I need help understanding what all this means. Can someone give me an example of what I should be looking for / doing here?

    I don't understand what they mean by the group Qu. Is this some group u multiplied by Q? Or just the group Qu.

    etc.
     
  2. jcsd
  3. Apr 2, 2016 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    ##Qu## is just the name of the group (also known as the quaternion group). First, can you write out al the elements in ##<i>##?
     
  4. Apr 2, 2016 #3

    RJLiberator

    User Avatar
    Gold Member

    Thanks for the help. I am at work now/tonight, so I'll have to get back to this problem then.
    for my reference: https://en.wikipedia.org/wiki/Quaternion_group

    To answer your question: Write out all the elements in < i >:
    {1, -1, i, -i}
    or {e, i, -i}

    I think?

    I think it is {1, -1, i, -i} based on the quick reading of the Wiki page.
     
  5. Apr 2, 2016 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    There isn't even an element ##e## in your group??

    Anyway, can you now compute ##g<i>## for each element in the group? I'll do one for you if ##g=i##:
    [tex]i<i>=i\{1,-1,i,-i\} = \{i\cdot 1, i\cdot (-1), i\cdot i, i\cdot (-i)\} = \{i, -i, -1, 1\} = <i>[/tex]
    What for the other ##g##'s in the group?

    What about ##<i>g##?
     
  6. Apr 2, 2016 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Just wanted to point out that the original question seems to be asking for the cosets of ##<j>##, not ##<i>##.
     
  7. Apr 2, 2016 #6

    RJLiberator

    User Avatar
    Gold Member

    Thanks for the help so far. I'm on my lunch break, but here's some of my (very quick) thoughts:

    <i>g where g = i is the following:

    <1,-1,i,-i> => <1*i,-1*i,i*i, -i*i> = <i, -i, -1, 1> = <i>

    so the left coset and the right coset for this particular item came out to be the same.

    Is my analysis correct here?
    Then I would continue this process of right and left operation with the other members, 1, -1, i, -i.
     
  8. Apr 2, 2016 #7

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Be careful with your brackets. ##<i>## means the group generated by ##i##. This is right. But you clearly mean ##\{i,-i,-1,1\}## as a set, and so it shouldn't be written as ##<i, -i,1, -1>##.

    In any case, this is correct. Try to do the other ones.

    Side question, it turns out that ##i<i> = <i>## and that ##<i> i = <i>##. Should this surprise you?
     
  9. Apr 3, 2016 #8

    RJLiberator

    User Avatar
    Gold Member

    Why is it that < i > is just {1, -1, i, -i} ?
    Is < j > just {j, -j, ij, -ij} ? If so, how in the world was I supposed to know this based on the question information?

    The fact that i< i > = < i > = < i >i suprises me a bit. I don't really understand the operation of i or what it is doing.

    So to determine the right and left cosets of j all I have to do is the following

    1< j >
    -1< j >
    i< j >
    -i< j >
    j< j >
    -j< j >
    ij< j >
    -ij< j >

    For the left cosets, where < j > = { j, -j, ij, -ij }

    and then the same for the right cosets.

    And then this problem is solved.

    But how was I supposed to know that < j > = { j, -j, ij, -ij } ?
     
  10. Apr 3, 2016 #9

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    No, your ##<j>## is wrong. Can you tell us what the definition of ##<j>## is?
     
  11. Apr 3, 2016 #10

    RJLiberator

    User Avatar
    Gold Member

    Damn. I have no idea what the definition of < j > is. All I have is this table.
    If < i > = {1, -1, i, -i}

    Perhaps j is < j > = {1, -1, j, -j} ?
     
  12. Apr 3, 2016 #11

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Stop guessing. Look up the definition of ##<...>##.
     
  13. Apr 3, 2016 #12

    RJLiberator

    User Avatar
    Gold Member

    Definition of what?
    Is there a wikipedia link you can post to me?

    I will continue to look through my notes, tho.
     
  14. Apr 3, 2016 #13

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I don't get it. You're trying to solve this problem about ##<j>## without first looking up what ##<j>## even means? Who gave you this problem? He should have explained you these things.
     
  15. Apr 3, 2016 #14

    RJLiberator

    User Avatar
    Gold Member

    Well, we have 2 lectures of 50 minutes per week and then a homework assignment over the weekend. The homework assignment kind of molds/forms everything. I have never in my life seen this Qu quaternion group before, but I often learn a lot in these homeworks.

    Now, with the help you've given, if I look up the Qu wikipedia page:
    https://en.wikipedia.org/wiki/Quaternion_group

    < j > = { j, k, -j, -k } ?
     
  16. Apr 3, 2016 #15

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I don't see why your wasting your time. Don't you have a book or notes? Look up what ##<j>##means instead of guessing.
     
  17. Apr 3, 2016 #16

    RJLiberator

    User Avatar
    Gold Member

    Oi.

    < j > = {j, -j, ij, -ij}
     
  18. Apr 3, 2016 #17

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I'm not helping any more until you post the definition of ##<g>## where ##g\in G##.
     
  19. Apr 3, 2016 #18

    RJLiberator

    User Avatar
    Gold Member

    Aha! Now I have the words/hints that I needed.

    For any group G and any g ∈ G, <g> is a subgroup of G.
    < g> is called the cyclic subgroup generated by g.

    Note: A group G is called cyclic if there exists a ∈ G such that < a > = G.

    Note2: If g ∈ G, then <g> is the smallest subgroup containing the element g. Precisely, if g ∈ H ≤ G, then < g > ≤ H.

    Definition: Let G be a group and H ⊆ G. Then H is a subgroup if
    1) e ∈ H
    2) For all x,y∈ H, xy ∈ G.
    3) For all x ∈ H, x^(-1) ∈ H

    So from here, < j > = { 1, -1, j, -j}
    As you need the identity element and j and the inverses ?
     
  20. Apr 3, 2016 #19

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Yes, that is correct.
     
  21. Apr 3, 2016 #20

    RJLiberator

    User Avatar
    Gold Member

    Whew. Thanks for helping me with that. I'm not sure why, but with this being new material for me, the switching of simple letters really messes me up. I will try to keep that in mind as I go through this.

    But now, it should be pretty easy for me to figure out. Simply as stated above,

    1< j >
    -1< j >
    i< j >
    -i< j >
    j< j >
    -j< j >
    ij< j >
    -ij< j >

    for the left coset and then the opposite side for the right coset.
     
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: Write down the cosets (right/left) of this:
  1. Right Cosets (Replies: 5)

  2. Left Cosets of Z in Q (Replies: 7)

Loading...