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!

Group theory

  1. May 9, 2009 #1
    Show that the Cardinality of a set which doesn't include inverse elements from a group G is always even.

    So the set with all g in G, which includes no inverse elements from G. (g=!g^-1)

    I can get this in every example I've done, checking mainly with dihedral groups, it's always been true but I can't find a pattern.

    I know that the neutral element can't be in the set, so thats one down, then I thought maybe halfing it, which is wrong I know, as I kept finding cases where the element itself was it's own inverse.

    I've really no idea where to go from here.
     
  2. jcsd
  3. May 9, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you mean the set H of all g in G such that g!=g^(-1), just think about grouping the elements in H in pairs, like {h,h^(-1)}.
     
  4. May 9, 2009 #3
    Yeah that's what I mean.

    But how do I group an element in a pair if the inverse of that element is itself?
     
  5. May 9, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    H doesn't contain any elements that are inverses of themselves, does it?
     
  6. May 9, 2009 #5
    Well H doesn't, but G does, and those are the elements (like the neutral element) that I'm trying to exclude.

    Like the dihedral group D6 of a triangle, the symmetry element is it's own inverse.. I think.

    In effect I'm trying to find a way to count all the elements in H after I've excluded all the inverse elements from G, and see if it's divisible by 2.
     
  7. May 9, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are confusing me. You are trying to count H, right? Not G. The definition of H excludes all elements of G that are their own inverse. Why are you worried about them?
     
  8. May 9, 2009 #7
    Well the question is this (part b hint)

    t0q642.jpg

    If I don't exclude them somehow from G and then count I don't see how I can prove it's even.
     
  9. May 9, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You know t(G) has even order, right? If G has even order then G-t(G) also has even order, right? You know e is in G-t(G), right? That means G-t(G) must contain an element g that is not equal to e. What's the order of g?
     
  10. May 9, 2009 #9
    Hmm. How did you get the order of t(G) was even? I'm not familiar with the order of a set, I know the order of a group is its cardianlity, and I know the definition of the order of an element, but not that lol.

    By G-t(G) do you mean literally G minus t(G)?
     
  11. May 9, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I know t(G) is even because I grouped the elements in pairs, like we were discussing earlier. G-t(G) just means the set of elements in G that aren't in t(G). By order of a set, I mean it's cardinality.
     
  12. May 9, 2009 #11
    That's what I'm not getting, you grouped the elements of t(G) as (h,h^-1), but I don't see any h^-1 in the set, as I thought that because we excluded all the inverses then the elements dont have inverses in the new set, so I don't see how they can be grouped.
     
  13. May 9, 2009 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    t(G) is the set of all elements that AREN'T their own inverses. If h isn't it's own inverse then h^(-1) isn't it's own inverse either. If h is in t(G), so is h^(-1).
     
  14. May 9, 2009 #13
    Oooh now I get it, I thought it meant there were no inverses, not specifically only the inverses that were the same element!

    I get this line now:

    'You know t(G) has even order, right? If G has even order then G-t(G) also has even order, right? You know e is in G-t(G), right?'

    As for this:

    'That means G-t(G) must contain an element g that is not equal to e. What's the order of g?'

    I realise that there has to be at least one g in G-t(G) that isn't equal to e (as it's even), but it could be of any order surely, depending on the group.
     
  15. May 9, 2009 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If g is in G-t(G) and isn't e, then it is it's own inverse. How could it be of any order? Repeat, it is it's own inverse.
     
  16. May 9, 2009 #15
    I know you're probably hinting the order of g is 2 lol, but still..

    I can see how it applys to this dihedral group D6, as I have s.r (rotation then symmetry) as my element of order 2, but I can't see why that order is special.
     
  17. May 9, 2009 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If g is it's own inverse, what is g^2? Sigh.
     
  18. May 9, 2009 #17
    Oh then it's g again!

    Sorry.. no excuses

    Thanks for all the help!
     
  19. May 9, 2009 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No, it's not g, g^2=e, isn't it??? That's what you meant, right?
     
  20. May 9, 2009 #19
    Yes that's what i meant!

    Sorry it's half past midnight here :D
     
  21. May 9, 2009 #20

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I guess that explains a certain slowness. Good job on checking the dihedral groups though. It's always a good idea to check concrete examples before trying to prove something.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Group theory
  1. Group Theory (Replies: 2)

  2. Group Theory (Replies: 1)

  3. Group Theory (Replies: 16)

  4. Group Theory (Replies: 1)

  5. Group theory (Replies: 1)

Loading...