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

**Neha** · February 11th 2011, 08:28 AM

**topspin1617** · February 11th 2011, 04:55 PM

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?

**Neha** · February 11th 2011, 09:52 PM

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...

**tonio** · February 12th 2011, 01:13 AM

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 $\Gamma:= PSL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\}$ , 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

**TheArtofSymmetry** · February 12th 2011, 03:15 AM

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 $\Gamma:= PSL(2,\mathbb{Z})=GL(2,\mathbb{Z})/\{\pm I\}$ , 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).

**tonio** · February 12th 2011, 05:05 AM

Originally Posted by TheArtofSymmetry

$GL(2,\mathbb{Z})/\{\pm I\}$ 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

**Neha** · February 12th 2011, 05:15 AM

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....

**TheArtofSymmetry** · February 12th 2011, 06:22 AM

Originally Posted by tonio

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

**tonio** · February 12th 2011, 08:14 AM

Originally Posted by TheArtofSymmetry

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

.

**TheArtofSymmetry** · February 13th 2011, 03:42 AM

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.

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