Results 1 to 3 of 3

Math Help - order of a group

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    28

    order of a group

    My Question is :
    Let G=(a b , c d) such that a,b,c,d belongs to Z(3)={0,1,2} and
    det(G)=ad-bc not equal to 0
    Show that order of G=48
    In fact I know that order of a finite group is equal to order of generator of it. i.e if G is cyclic group generated by a, then the order of a is same as the order of G, namely the smallest positive integer such that a^k=e
    But it seems that this will not be helpful here. How should I start in order to proof that this group is of order 48

    Thank you in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by fuzzy topology View Post
    My Question is :
    Let G=(a b , c d) such that a,b,c,d belongs to Z(3)={0,1,2} and
    det(G)=ad-bc not equal to 0
    Show that order of G=48
    In fact I know that order of a finite group is equal to order of generator of it. i.e if G is cyclic group generated by a, then the order of a is same as the order of G, namely the smallest positive integer such that a^k=e
    But it seems that this will not be helpful here. How should I start in order to proof that this group is of order 48
    There's probably a slick group-theoretic way of doing this, but I would just use a simple counting strategy.

    There are 81=3^4 elements in \mathbb{Z}(3)^4, which I'm thinking of as the set of all 2×2 matrices over \mathbb{Z}(3). Count the number of those for which the determinant is equal to 0:
    there is one matrix with all four components zero (the zero element),
    there are 8 matrices with three components zero,
    there are 8 matrices with two zero components (in the same row),
    there are 8 matrices with two zero components (in the same column).
    There are no matrices with zero determinant and just one zero. Finally, there are 8 matrices with zero determinant and no zero components (a, b and c can all be 1 or 2, and then d is determined by the condition ad–bc=0).

    That gives a total of 33 matrices with determinant 0, hence 81 – 33 = 48 matrices with nonzero determinant.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    28
    Thank you very much> I tried to solve it by other ways but I did not succeed. Thanks alot
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. element of order II, in 2n order group
    Posted in the Advanced Algebra Forum
    Replies: 14
    Last Post: December 14th 2011, 09:32 AM
  2. Order of Group. Direct Product of Cyclic Group
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: November 19th 2011, 01:06 PM
  3. Order of a Group, Order of an Element
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 15th 2010, 06:28 PM
  4. Order of a Group, Order of an Element
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: November 15th 2010, 06:02 PM
  5. Prove that a group of order 375 has a subgroup of order 15
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 13th 2010, 11:08 PM

Search Tags


/mathhelpforum @mathhelpforum