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

1. ## 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.

2. Hmm... interesting question. Note that since the only units in the ring $\displaystyle \mathbb{Z}$ are $\displaystyle \pm 1$, the group $\displaystyle GL_2(\mathbb{Z})=\left\{\left(\begin{array}{cc}a & b\\c & d\end{array}\right)|a,b,c,d\in\mathbb{Z},ad-bc=\pm 1\right\}$. 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 $\displaystyle SL_2(\mathbb{Z})$.

But I'm not sure quite what to do next... maybe from this someone can pick something up?

3. 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...

4. Originally Posted by Neha
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 $\displaystyle \Gamma:= PSL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\}$ , then $\displaystyle \Gamma:=C_2*C_3=$ 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 $\displaystyle C_3$ . Projecting this idea back to the original group we can get the result.

Tonio

5. Originally Posted by tonio
Perhaps the following may be of some help, but it requires some knowedge of group theory: if we look at the modular

group $\displaystyle \Gamma:= PSL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\}$ , then $\displaystyle \Gamma:=C_2*C_3=$ 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 $\displaystyle C_3$ . Projecting this idea back to the original group we can get the result.

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

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).

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

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

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

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

Tonio

7. 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....

8. Originally Posted by tonio
Since $\displaystyle GL(2,\mathbb{Z})=S^*L(2,\mathbb{Z})=$ all the integer matrices with determinant $\displaystyle \pm 1$ ,

then $\displaystyle PSL(2,\mathbb{Z})$ 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 $\displaystyle PS^*L(2,\mathbb{Z}) \cong GL(2,\mathbb{Z})/\{\pm I\}$. $\displaystyle PS^*L(2,\mathbb{Z})$ and $\displaystyle PSL(2,\mathbb{Z})$ do not look the same in your link. The former refers to $\displaystyle PGL(2,\mathbb{Z})$, while the latter refers to the index 2 subgroup (a normal subgroup) of $\displaystyle PGL(2,\mathbb{Z})$.

9. Originally Posted by TheArtofSymmetry
If that is the case, it should be written $\displaystyle PS^*L(2,\mathbb{Z}) \cong GL(2,\mathbb{Z})/\{\pm I\}$. $\displaystyle PS^*L(2,\mathbb{Z})$ and $\displaystyle PSL(2,\mathbb{Z})$ 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 $\displaystyle \Gamma$" is.

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

What "THE index 2 subgroup" You mean $\displaystyle \{\pm I\}$ ? 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 $\displaystyle SL(2,\mathbb{Z})$ up to sign. This is the

definition and the notation I work with.

Tonio

.

10. What I meant by the index 2 subgroup of $\displaystyle PGL(2,\mathbb{Z})$ is $\displaystyle PSL(2,\mathbb{Z})$, i.e.,

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

Although $\displaystyle PGL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\}$ 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 $\displaystyle PGL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\}$ having the same determinant.

$\displaystyle PGL(2,\mathbb{Z})$ 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 $\displaystyle PSL(2,\mathbb{Z})$.

Back to the original question, $\displaystyle [GL(2,\mathbb{Z}):SL(2,\mathbb{Z})]=2$, that is, $\displaystyle SL(2,\mathbb{Z}) \lhd GL(2,\mathbb{Z})$. Elements in $\displaystyle GL(2,\mathbb{Z})$ whose order is 3 are some elements in $\displaystyle SL(2,\mathbb{Z})$. Since $\displaystyle SL(2,\mathbb{Z}) \lhd GL(2,\mathbb{Z})$, those elements are conjugates to each other.