Prove that any two elements of order 3 in GL(2,Z) are conjugates.
Hmm... interesting question. Note that since the only units in the ring are , the group . 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 .
But I'm not sure quite what to do next... maybe from this someone can pick something up?
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 , then 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 . Projecting this idea back to the original group we can get the result.
What I meant by the index 2 subgroup of is , i.e.,
, that is, .
Although 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 having the same determinant.
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 .
Back to the original question, , that is, . Elements in whose order is 3 are some elements in . Since , those elements are conjugates to each other.