Results 1 to 10 of 10

Math Help - Prove that any two elements of order 3 in GL(2,Z) are conjugates.

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    5

    Prove that any two elements of order 3 in GL(2,Z) are conjugates.

    Prove that any two elements of order 3 in GL(2,Z) are conjugates.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2010
    Posts
    193
    Hmm... interesting question. Note that since the only units in the ring \mathbb{Z} are \pm 1, the group GL_2(\mathbb{Z})=\left\{\left(\begin{array}{cc}a & b\\c & d\end{array}\right)|a,b,c,d\in\mathbb{Z},ad-bc=\pm 1\right\}. Also note that by the multiplicative property of the determinant, a matrix in the group of order three must necessarily have determinant one; that is, it must be an element of SL_2(\mathbb{Z}).

    But I'm not sure quite what to do next... maybe from this someone can pick something up?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    5
    So, this i wat I could think....
    If A is a 2x2 matrix of order 3, then its characteristic and minimal polynomial are both going to b x^2+x+1... that means to say that....matrix has trace=-1 and det=1....perhaps that is wat u r also saying.
    So, the matrices have the same Rational Canonical form over Q(precisely the companion matrix)...that means they are similar(conjugate) over Q....problem is getting to GL(2,Z)....
    Lemme know if you are able to move furthur...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Neha View Post
    So, this i wat I could think....
    If A is a 2x2 matrix of order 3, then its characteristic and minimal polynomial are both going to b x^2+x+1... that means to say that....matrix has trace=-1 and det=1....perhaps that is wat u r also saying.
    So, the matrices have the same Rational Canonical form over Q(precisely the companion matrix)...that means they are similar(conjugate) over Q....problem is getting to GL(2,Z)....
    Lemme know if you are able to move furthur...

    Perhaps the following may be of some help, but it requires some knowedge of group theory: if we look at the modular

    group \Gamma:= PSL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\} , then \Gamma:=C_2*C_3= the free product of

    the cyclic groups of order 2 and 3, and by a well-known theorem in such a free product an element has finite order iff it

    conjugate to a finite order of one of the factors...

    This means that any element of order 3 has to be conjugate in the free product to one of the two non trivial elements of

    the factor C_3 . Projecting this idea back to the original group we can get the result.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2010
    Posts
    95
    Quote Originally Posted by tonio View Post
    Perhaps the following may be of some help, but it requires some knowedge of group theory: if we look at the modular

    group \Gamma:= PSL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\} , then \Gamma:=C_2*C_3= the free product of

    the cyclic groups of order 2 and 3, and by a well-known theorem in such a free product an element has finite order iff it

    conjugate to a finite order of one of the factors...

    This means that any element of order 3 has to be conjugate in the free product to one of the two non trivial elements of

    the factor C_3 . Projecting this idea back to the original group we can get the result.

    Tonio
    GL(2,\mathbb{Z})/\{\pm I\} is PGL(2,\mathbb{Z}), not PSL(2,\mathbb{Z}). The modular group you are referring to is probably PSL(2,\mathbb{Z}) \cong C_2 * C_3 whose presentation is <x, y| x^2=y^3=1>. PSL(2,\mathbb{Z}) is the index 2 subgroup of PGL(2,\mathbb{Z}).

    If the question was involving in order 3 elements in PGL(2,Z), elements like (0,1;-1,1) and (1,-1;1,0) are order 3 elements in PGL(2,Z).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by TheArtofSymmetry View Post
    GL(2,\mathbb{Z})/\{\pm I\} is PGL(2,\mathbb{Z}), not PSL(2,\mathbb{Z}). The modular group you are referring to is probably PSL(2,\mathbb{Z}) \cong C_2 * C_3 whose presentation is <x, y| x^2=y^3=1>. PSL(2,\mathbb{Z}) is the index 2 subgroup of PGL(2,\mathbb{Z}).

    If the question was involving in order 3 elements in PGL(2,Z), elements like (0,1;-1,1) and (1,-1;1,0) are order 3 elements in PGL(2,Z).

    Since GL(2,\mathbb{Z})=S^*L(2,\mathbb{Z})= all the integer matrices with determinant \pm 1 ,

    then PSL(2,\mathbb{Z}) is what I wrote...

    Definitions may vary. Check No. 1 here:Modular group - Wikipedia, the free encyclopedia

    Tonio
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2011
    Posts
    5
    Since order 3 matrix in GL(2,Z) has determinant=1....so I guess for this result, it works....hopefully I got it....thnk u guys....
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    May 2010
    Posts
    95
    Quote Originally Posted by tonio View Post
    Since GL(2,\mathbb{Z})=S^*L(2,\mathbb{Z})= all the integer matrices with determinant \pm 1 ,

    then PSL(2,\mathbb{Z}) is what I wrote...

    Definitions may vary. Check No. 1 here:Modular group - Wikipedia, the free encyclopedia

    Tonio
    If that is the case, it should be written PS^*L(2,\mathbb{Z}) \cong GL(2,\mathbb{Z})/\{\pm I\}. PS^*L(2,\mathbb{Z}) and PSL(2,\mathbb{Z}) do not look the same in your link. The former refers to PGL(2,\mathbb{Z}), while the latter refers to the index 2 subgroup (a normal subgroup) of PGL(2,\mathbb{Z}).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by TheArtofSymmetry View Post
    If that is the case, it should be written PS^*L(2,\mathbb{Z}) \cong GL(2,\mathbb{Z})/\{\pm I\}. PS^*L(2,\mathbb{Z}) and PSL(2,\mathbb{Z}) do not look the same in your link.

    Neither did I claim they do: I meant in that link are various possible definitions, according to different authors,

    of what "the modular group \Gamma" is.



    The former refers to PGL(2,\mathbb{Z}), while the latter refers to the index 2 subgroup (a normal subgroup) of PGL(2,\mathbb{Z}).


    What "THE index 2 subgroup" You mean \{\pm I\} ? And about the former

    being equal to the general linear group: I've seen somewhere (I can't remember now where) that

    some authors refer to that group as the group of all 2x2 matrices with determinant different

    from zero, though I really cannot agree with this since then most elements wouldn't

    been invertible, as somebody else already pointed.

    Anyway, for me personally, the modular group is SL(2,\mathbb{Z}) up to sign. This is the

    definition and the notation I work with.

    Tonio


    .
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    May 2010
    Posts
    95
    What I meant by the index 2 subgroup of PGL(2,\mathbb{Z}) is PSL(2,\mathbb{Z}), i.e.,

    [PGL(2,\mathbb{Z}):PSL(2,\mathbb{Z})]=2, that is, PSL(2,\mathbb{Z}) \lhd PGL(2,\mathbb{Z}).

    Although PGL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\} identifies I and -I, it does not mean that it identifies determinants 1 and -1. For example, (1,0;0,1) and (-1,0;0,-1) are the same element in PGL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\} having the same determinant.

    PGL(2,\mathbb{Z}) is partitioned into two sets with the same cardinality. One set consists of elements whose determinant is 1, and the other set consists of elements whose determinant is -1. The first set corresponds to PSL(2,\mathbb{Z}).

    Back to the original question, [GL(2,\mathbb{Z}):SL(2,\mathbb{Z})]=2, that is, SL(2,\mathbb{Z}) \lhd GL(2,\mathbb{Z}). Elements in GL(2,\mathbb{Z}) whose order is 3 are some elements in SL(2,\mathbb{Z}). Since SL(2,\mathbb{Z}) \lhd GL(2,\mathbb{Z}), those elements are conjugates to each other.
    Last edited by TheArtofSymmetry; February 13th 2011 at 04:27 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: May 19th 2011, 02:17 AM
  2. Order of elements in a group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 3rd 2010, 11:51 AM
  3. Conjugates of elements in a group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 2nd 2010, 03:17 AM
  4. Order of groups involving conjugates and abelian groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: February 5th 2009, 08:55 PM
  5. order of elements
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 27th 2007, 07:56 PM

Search Tags


/mathhelpforum @mathhelpforum