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

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

A paper on "the group of units of the integral group ring ZS3" by Hughes and Pearson refers to this result and says that it has been picked up from "Representation Theory of finite groups and associated algebras" by Curtis and Reiner but I am not able to trace the result...can someone help???
 

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  • #2
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That means that all order-3 elements of that group reside in the same conjugacy class.

a' = g*a*inv(g)

This problem stumped me at first, so I decided to try a simpler problem -- all order-2 elements. They have forms

{{a,b},{b,-a}} with a2 + b2 = 1
{{a,b},{-b,a}} with a2 - b2 = 1
{{0,b},{1/b,0}} with b nonzero

The first two are identical for a field with characteristic 2, though not otherwise.

By trying to find g's that relate different solutions by conjugacy, I find these conjugacy classes:

Identity matrix I

- I
for characteristic != 2

{{a,b},{-b,a}} and {{a,-b},{b,a}}
for characteristic != 2
a separate class for each set of nonzero a, b satisfying a2 - b2 = 1

A single class of:
{{a,b},{b,-a}}
for every nonzero a, b satisfying a2 + b2 = 1
and also
{{0,b},{1/b,0}}
for every nonzero b


For the integers, GL(2,Z), one gets these conjugacy classes:
I
-I
{{0,1},{-1,0}} and {{0,-1},{1,0}}
 
  • #3
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To follow up on the previous post, I note that conjugacy classes share the values of the traces of powers of their members. However, more than one class can share a set of trace-of-power value.

For the order-2 case, I get:
I: 2 (really order 1)
-I: -2
all the sets {{a,b},{-b,a} and {{a,-b},{b,a}}: 2a
The remaining one: 0


For the order-3 case, working from {{a11,a12},{a21,a22}}, the possible solutions are

a*I where a3 = 1

{{a11,a12},{a21,a22}}
where a112 + a11*a22 + a222 + a12*a21 = 0
and at least one of a22 != a11, a21 != 0, and a21 != 0 is true

For a12 = a21 = 0, we get
a113 = a223 = 1
a11 and a22 are thus any of the field's cube roots of unity, either {1} or {1,w,w2}

For at least one of a12 and a21 nonzero, we get
(a11 + a22)3 = -1

The conjugacy classes can be distinguished by their traces of powers. For matrix A = {{a's}}:
Tr(A) = a11 + a22
Tr(A2) = a112 + a222 + 2*a12*a21

For a12 = a21 = 0 (diagonal),
Tr(A) = a11 + a22
Tr(A2) = a112 + a222
Their conjugacy classes are thus
a*I for each a in {1,w,w2}
diag(a,b) and diag(b,a) for each set of distinct a,b in {1,w,w2}

For at least one of a12 and a21 nonzero (non-diagonal),
Tr(A) = a11 + a22
Tr(A2) = - (a11 + a22)2 = - (Tr(A))2
Thus, Tr(A) is any of {-1} or {-1, -w, -w2}
Every one of these matrices that shares a value of Tr(A) falls into the same conjugacy class.

Is Tr(A2) = - (Tr(A))2 for any of the non-identity-multiple diagonal matrices? If so, then that matrix falls into the appropriate non-diagonal-matrix class.

{1,w} -- -w2, -w
{1,w2} -- -w, -w2
{w,w2} -- -1, -1

That's the case for all of them.

Thus, the order-3 matrices fall into these conjugacy classes:
w*I
w2*I
the rest with Tr(A) and Tr(A2) values:
-1, -1
-w, -w2
-w2, -w

If there is only one cube root of unity in the field, then there is only one class of order-3 elements, which proves what neham was asking about.
 

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