Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - abstract algebra - dihedral groups

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    166

    abstract algebra - dihedral groups

    Can someone help me with this proof? It was an exam problem, but I can't get it.

    D_2n = < r, s | r^n = s^2 = 1, rs = sr^(-1) >
    If n = 2k is even and n>= 4, show that z = r^k is an element of order 2 which commutes with all elements of D_2n. Show also that z is the only nonidentity element of D_2n which commutes with all elemtns of D_2n.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    Can someone help me with this proof? It was an exam problem, but I can't get it.

    D_2n = < r, s | r^n = s^2 = 1, rs = sr^(-1) >
    If n = 2k is even and n>= 4, show that z = r^k is an element of order 2 which commutes with all elements of D_2n. Show also that z is the only nonidentity element of D_2n which commutes with all elemtns of D_2n.
    Hint: Use the fact that r^n s = r^{-n} s
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2008
    Posts
    166
    For the first proof, should I show that r^k is an element of order 2 or should I show that r^k commutes with all elements of D_2n?

    Can you help me to get started? Thank you.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    For the first proof, should I show that r^k is an element of order 2 or should I show that r^k commutes with all elements of D_2n?
    Note that r^k \not = e -- why?
    Now (r^k)^2 = r^{2k} = e.
    Thus the order of r^k is two.
    This means that (r^k)(r^k)^{-1} = e.
    Thus, r^{-k} = r^k.

    Any element in the dihedral group has form r^as^b.
    Now (r^as^b)r^k = r^a(s^br^k) = r^a(r^{-k}s^b) = r^a(r^ks^b) = r^k(r^as^b).
    Thus r^k commutes with everything.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2008
    Posts
    166
    Quote Originally Posted by ThePerfectHacker View Post
    Note that r^k \not = e -- why?
    Now (r^k)^2 = r^{2k} = e.
    Thus the order of r^k is two.
    So I know that r^k \not = e and (r^k)^2 = r^{2k} = e by the definition of D_{2n}, because n = 2k, so r^n = r^{2k} = (r^k)^2 = e?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    So I know that r^k \not = e and (r^k)^2 = r^{2k} = e by the definition of D_{2n}, because n = 2k, so r^n = r^{2k} = (r^k)^2 = e?
    Yes
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2008
    Posts
    166
    How do I show that z is the only nonidentity element of D_{2n} which commutes with all elements of D_{2n}?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    How do I show that z is the only nonidentity element of D_{2n} which commutes with all elements of D_{2n}?
    Hint: The element which commutes with all element if and only if it commutes with r,s. Thus find which element commutes with r,s.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2008
    Posts
    166
    Obviously, the identity element commutes with all elements in D_{2n}. But I am looking for a nonidentity element, so from the first proof, r^k is the nonidentity element that commutes with all elements in D_{2n}. But how do I know that it is the ONLY nonidentity that commutes with those elements?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    But how do I know that it is the ONLY nonidentity that commutes with those elements?
    Look at what I wrote above! Let x be an element in this group then we can write it as x=r^as^b where 0\leq a\leq n-1, 0\leq b\leq 1. Now r^as^b commute with all the element if and only if it it commutes with the generating set i.e. it commute with r and s. Thus, find the conditions on a,b so that r^as^b commute with r and s.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Sep 2008
    Posts
    166
    So I let x = r^as^b where 0<=a<=n-1, 0<=b<=1, then (r^as^b)r = r^a(s^br) = r^a(r^{-1}s^b) = r^ars^b = r(r^as^b) and (r^as^b)s = (r^as)s^b = sr^{-a}s^b = s(r^as^b). So r^as^b commutes with all elements of D_{2n} no matter what a and b are, isn't it? Then the nonidentity element in the form of r^as^b also commutes with all elements of D_{2n}, so r^a is not the only nonidentity element that commutes with the elements in D_{2n}...?
    Am I thinking this right?...
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    So I let x = r^as^b where 0<=a<=n-1, 0<=b<=1, then (r^as^b)r = r^a(s^br) = r^a(r^{-1}s^b) = r^ars^b = r(r^as^b) and (r^as^b)s = (r^as)s^b = sr^{-a}s^b = s(r^as^b). So r^as^b commutes with all elements of D_{2n} no matter what a and b are, isn't it? Then the nonidentity element in the form of r^as^b also commutes with all elements of D_{2n}, so r^a is not the only nonidentity element that commutes with the elements in D_{2n}...?
    Am I thinking this right?...
    Let us consider two cases: (i) x=r^a, 0\leq a\leq n-1 (ii) x=r^as, 0\leq s\leq 1.

    In case (i) we have that x=r^a commutes with s,r. It definitely commutes with r no matter what a is but to commute with s we need to have r^a s = sr^a = r^{-a}s and so a\equiv -a ~ (n) this forces a = k (where k is such an integer so 2k=n). Thus, we find that x=r^k commutes with all the elements.

    You try case (ii).
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Sep 2008
    Posts
    166
    So, (r^as)s = (sr^{-a})s = (sr^a)s = s(r^as). So r^as commutes with s. And (r^as)r = (sr^{-a})r = (sr^a)r . Here is what I wonder, if I am given r^as = sr^{-a}, does that mean sr^a = r^{-a}s?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dori1123 View Post
    So, (r^as)s = (sr^{-a})s = (sr^a)s = s(r^as). So r^as commutes with s. And (r^as)r = (sr^{-a})r = (sr^a)r . Here is what I wonder, if I am given r^as = sr^{-a}, does that mean sr^a = r^{-a}s?
    (r^a s)s = r^a s^2 = r^a.
    s(r^a s) = (sr^a)s = (r^{-a} s)s = r^{-a}.
    This means if we want commutavity then r^a = r^{-a} i.e. a=0,k.

    Now there are two cases: (i) x=r^0 s = s (ii) x=r^k s.
    Which of these commute with the remaining element r?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Sep 2008
    Posts
    166
    Quote Originally Posted by ThePerfectHacker View Post
    Now there are two cases: (i) x=r^0 s = s (ii) x=r^k s.
    Which of these commute with the remaining element r?
    So for x = r^0s = s, xr = sr does not equal to rx = rs = sr^{-1} (unless r = r^{-1}?). For x = r^ks, xr = r^ksr = sr^{-k}r = sr^{-k+1} and rx = rr^ks = r^{k+1}s, so r^ksr does not equal to rr^ks. Right?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Abstract Algebra-Groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 30th 2009, 12:39 PM
  2. Abstract Algebra: Groups
    Posted in the Advanced Algebra Forum
    Replies: 14
    Last Post: April 6th 2008, 07:52 PM
  3. Abstract Algebra: Groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 5th 2008, 03:13 AM
  4. Abstract Algebra: Groups
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: February 11th 2008, 09:39 AM
  5. Abstract Algebra: Groups and Symmetry
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 11th 2008, 05:43 AM

Search Tags


/mathhelpforum @mathhelpforum