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        riddles >> putnam exam (pure math) >> Finite subgroups of GL(2,Q)
        
(Message started by: Eigenray on Nov 18^th, 2006, 8:04pm)

Title: Finite subgroups of GL(2,Q)
Post by Eigenray on Nov 18^th, 2006, 8:04pm

Determine all finite subgroups of GL(2,Q).

Title: Re: Finite subgroups of GL(2,Q)
Post by Michael_Dagg on Nov 24^th, 2006, 8:15pm

This nice problem ought to generate a fair amount of
discussion.

Title: Re: Finite subgroups of GL(2,Q)
Post by Icarus on Nov 25^th, 2006, 7:29am

A great many interesting problems come by when I have no time to examine them, I'm afraid. When I do have time, I've usually forgotten about them.

If A is in gl(2,F), for any field F, then there is a Q in GL(2,F) such that D = QAQ_-1 is diagonal. If A is in a finite subgroup of GL(2,F), then Aⁿ = I for some N. Equivalently, Dⁿ = I, and therefore d₁₁ⁿ = 1 & d₂₂ⁿ = 1. For F a subfield of R, both are either 1 or -1.

Hence any finite subgroup of GL(2,Q) must consist of elements of the form QDQ^-1, where D is one of the 4 matrices (1, 0; 0, 1), (-1, 0; 0, 1), (1, 0; 0, -1), (-1, 0; 0, -1).

Which is all I have time for now.

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Nov 25^th, 2006, 4:46pm

The matrix

| 0 1|
|-1 0|

has order 4 in GL(2,Q) and so is not diagonalizable over Q (unless I am miss-reading you, Icarus).

Title: Re: Finite subgroups of GL(2,Q)
Post by Ken_Wiley on Nov 25^th, 2006, 7:08pm

I am just curious but what kind of math is this?

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Nov 26^th, 2006, 4:03am

on 11/25/06 at 19:08:00, Ken_Wiley wrote:

I am just curious but what kind of math is this?

The problem is from group theory. It deals with groups of 2x2 invertible matrices of rational (Q) numbers.

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Nov 26^th, 2006, 4:04am

on 11/25/06 at 16:46:31, ecoist wrote:

The matrix

| 0 1|
|-1 0|

has order 4 in GL(2,Q) and so is not diagonalizable over Q (unless I am miss-reading you, Icarus).

How do you show this? ;)

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Nov 26^th, 2006, 4:10am

Here're my first 2 cents (S will denote finite subgroup):

|S| = 2. The following matrices generate a subgroup of order 2:

a) d₁₁² = d₂₂² = 1, d₁₂ = d₂₁ = 0.

b) d₁₁ = d₂₂ = 0, d₁₂d₂₁ = 1.

|S| = 4. The matrix given by ecoist and its negation generate a subgroup of order 4. I don't know if these are the only possible, though...

I also wonder is there a group with |S| = 3?

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Nov 26^th, 2006, 9:45am

Quote:

How do you show this?

If the matrix were rationally diagonalizable, one of the, thus rational, diagonal entries of the diagonal matrix would have multiplicative order 4, impossible for a rational number.

Quote:

I also wonder is there a group with |S| = 3?

|0 -1| (the companion matrix of the polynomial x²+x+1)
|1 -1|

has order 3 and

|0 -1|
|1 1|

has order 6.

Title: Re: Finite subgroups of GL(2,Q)
Post by Icarus on Nov 26^th, 2006, 11:25am

I thought I replied to this yesterday, but my response is missing. I forgot that F has to be algebraicly complete for diagonalization to always be possible. The eigenvalues for ecoist's matrix are i & -i. You can always replace F with it's closure in my remarks, but in this case, the closure has infinitely many matrices whose diagonal elements are roots of unity. So the utility is much reduced.

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Nov 27^th, 2006, 5:40am

on 11/26/06 at 09:45:00, ecoist wrote:

|0 -1| (the companion matrix of the polynomial x²+x+1)
|1 -1|

has order 3 and

|0 -1|
|1 1|

has order 6.

Interesting! Is there a matrix with order n = 5?

I still cannot find a link between this problem and linear transformations on a plane...

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Nov 27^th, 2006, 10:04am

Quote:

Is there a matrix with order n = 5?

No. Eigenray pointed this out in the trinity, pentagram puzzle (for 3x3 matrices over Q). If a 2x2 matrix A over Q has order n, then it satisfies the polynomial equation xⁿ-1=0. But, since it is also 2x2 over Q, A must have rational minimum polynomial m(x) of degree at most 2, since the characteristic polynomial of A has degree 2. But then m(x) must divide xⁿ-1. Since x⁵-1 equals x-1 times a quartic irreducible over Q, and m(x) cannot equal x-1, since this means A has order 1, no 2x2 matrix over Q can have order 5.

Title: Re: Finite subgroups of GL(2,Q)
Post by Icarus on Nov 27^th, 2006, 3:59pm

Despite my gaff, my approach is not without merit: If A is in GL(2,Q), then in GL(2,C), A is diagonalizable: A = QDQ^-1, and if A has order n, then so does the diagonal matrix D: Dⁿ = I. Therefore the eigenvalues of A must be n-th roots of unity. Since the characteristic equation of A is quadratic, the eigenvalues must be of 1st, 2nd, 3rd, 4th, or 6th roots of unity, since higher order roots do not satisfy quadratic equations with rational coefficients (essentially the same argument as Eigenray's that ecoist reproduces above, though dressed a bit differently). Hence the characteristic equation of A must be one of the following:
x²+2x+1
x²-2x+1
x²-1
x²+1
x²+x+1
x²-x+1
since these are the only unital rational quadratics constructable from those 8 roots. The choice of any of these as characteristic equation imposes 2 constraints on the coordinates of A, which means that the set of all matrices in GL(2,Q) of finite order is two-dimensional.

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Nov 28^th, 2006, 11:06am

on 11/27/06 at 15:59:52, Icarus wrote:

Hence the characteristic equation of A must be one of the following:
x²+2x+1
x²-2x+1

I don't think these are relevant.

So, the following was established so far: Every element in the finite subgroup of GL(2,Q) has order 1, 2, 3, 4 or 6.

What next? Surely, that means cyclic groups of above orders are possible. These are the only groups with orders 2, 3. What about |S| = 4?

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Nov 28^th, 2006, 3:09pm

Quote:

What next? Surely, that means cyclic groups of above orders are possible. These are the only groups with orders 2, 3. What about |S| = 4?

There's the Klein-four group generated by

|1 0|
|0 -1|

and

|-1 0|
| 0 1|.

If you add a third generator

|0 1|
|1 0|,

you get the dihedral group of order 8. I think it is the only group of order 8 in GL(2,Q) up to isomorphism, but I don't know how to prove it.

Title: Re: Finite subgroups of GL(2,Q)
Post by Icarus on Nov 28^th, 2006, 7:39pm

on 11/28/06 at 11:06:57, Barukh wrote:

I don't think these are relevant.

They are if your group is going to contain I (kind of hard to do without that) or -I. Admittedly, that's all they are needed for.

Quote:

So, the following was established so far: Every element in the finite subgroup of GL(2,Q) has order 1, 2, 3, 4 or 6.

Also, the groups are at most 2 dimensional. There are few other technical facts to be gleaned as well from knowing that det(A) = 1 or -1, and trace A = -1, 0, or 1, if A =/= I, -I.

(Of course |det(A)| = 1 follows more easily from the requirement that Image(det) be a finite subgroup of Q*: {1} and {1, -1} are the only candidates.)

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Dec 3^rd, 2006, 4:44am

It seems like all the groups were identified. Does Eigenray have something to add? :-/

BTW, what happens if we relax the restriction for the matrix to be revertible?

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 3^rd, 2006, 9:50am

Quote:

It seems like all the groups were identified.

Not quite. There is also the dihedral group of order 6 generated by

|0 1|
|1 0|

and

|0 -1|
|1 -1|.

Title: Re: Finite subgroups of GL(2,Q)
Post by Icarus on Dec 3^rd, 2006, 1:57pm

Even if these are all the groups, no one has demonstrated as much, yet.

I'm not sure what you mean by "revertible". Do you mean "invertible"? In that case, you can't have a group.

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 3^rd, 2006, 3:16pm

I tried the following to test whether or not a given finite group G can be a subgroup of GL(2,Q). Find a faithful representation of G in GL(n,Q), for some n and see if there is a G-invariant subspace of Qⁿ of dimension 2 on which G acts faithfully.

Example 1: The symmetric group S on 3 letters has a faithful representation on Q³ as a group of 3x3 permutation matrices. Since all row sums of these matrices equal 1, S acts trivially on the subspace of Q³ spanned by the vector (1,1,1). The orthogonal complement of this 1-dimensional subspace is S-invariant and has dimension 2. Hence S can be realized as a subgroup of GL(2,Q).

Example 2: The dihedral group G generated by the permutations (1,2,3,4) and (2,4) has a faithful representation as 4x4 permutation matrices over Q. G acts faithfully on the 3-dimensional subspace of Q⁴ consisting of all vectors whose coordinates sum to zero. With a little experimentation, one can see that G acts faithfully on the 2-dimensional subspace spanned by the vectors (1,0,-1,0) and (0,1,0,-1). Hence G can be realized as a subgroup of GL(2,Q).

Example 3: On the other hand, the symmetric group P on 4 letters cannot be a subgroup of GL(2,Q). For, P is generated by (1,2,3,4), (2,4), and (1,2,3). Since (1,2,3) maps (1,0,-1,0) to (-1,1,0,0), a vector independent of (1,0,-1,0) and (0,1,0,-1), it follows that P, in its action on Q⁴, has no 2-dimensional P-invariant subspace on which P acts faithfully.

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 3^rd, 2006, 7:42pm

I goofed in Example 3. I ignored the possibility that a P-invariant subspace might be reducible. The action of P must split Q⁴ into either two or three irreducible P-invariant subspaces, 1-dimensional and 3-dimensional, or two 1-dimensionals and a 2-dimensional. Since 1-dimensional G-invariant subspaces and those of the subgroup generated by (1,2,3) intersect only in the subspace spanned by the vector (1,1,1,1), it follows that P cannot have two independent 1-dimensional P-invariant subspaces; so the action of P must split Q⁴ into an irreducible P-invariant subspace of dimension 3 and one of dimension 1. Hence the symmetric group P on 4 letters is not a subgroup of GL(2,Q).

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Dec 3^rd, 2006, 11:20pm

on 12/03/06 at 13:57:58, Icarus wrote:

I'm not sure what you mean by "revertible". Do you mean "invertible"?

Yes, of course! ;D

Quote:

In that case, you can't have a group.

Not quite... ;)

Title: Re: Finite subgroups of GL(2,Q)
Post by Eigenray on Dec 6^th, 2006, 5:39am

There is still one more group to be found. (And, of course, the proof of this fact!)

ecoist - Why can't a subgroup of GL₂(Q) have an irreducible faithful representation of dimension > 2?

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 6^th, 2006, 10:16am

on 12/06/06 at 05:39:40, Eigenray wrote:

There is still one more group to be found. (And, of course, the proof of this fact!)

ecoist - Why can't a subgroup of GL₂(Q) have an irreducible faithful representation of dimension > 2?

I don't see why not. I don't think I used this in my (clumsy) arguments. I used only specific representations on specific spaces over Q.

Is the last group isomorphic to the alternating group on 4 letters, A₄, generated by

|0 1|
|1 0|

and

|0 -1|
|1 -1|?

Have no idea how to prove that there aren't any more. Can't even prove that there is no abelian subgroup of order 8.

Title: Re: Finite subgroups of GL(2,Q)
Post by Eigenray on Dec 6^th, 2006, 7:28pm

on 12/06/06 at 10:16:53, ecoist wrote:

I don't see why not. I don't think I used this in my (clumsy) arguments. I used only specific representations on specific spaces over Q.

Given that P=S₄ acts faithfully on Q⁴, splitting it into irreducible subspaces of dimensions 1,3, you concluded that P is not a subgroup of GL₂(Q). How?

Quote:

Is the last group isomorphic to the alternating group on 4 letters, A₄, generated by

Those matrices generate D₃ (= S₃), not A₄. In fact A₄ is not a subgroup of GL₂(Q).

Quote:

Can't even prove that there is no abelian subgroup of order 8.

The only finite noncyclic abelian subgroup is the Klein 4-group. Start of proof: Any set of commuting diagonalizable matrices can be ....

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 6^th, 2006, 9:00pm

Ah, yes! I could have started with a faithful representation of S₄ on Q⁸ and somehow split it into irreducible representations that included a faithful one of dimension 2. Thanks, Eigenray. My scheme was not only clumsy, it was faulty.

Sloppy too! Didn't bother to check my phony construction of A₄.

The centralizer of a matrix with distinct eigenvalues must have the same eigenspaces as the matrix. Hence, for a non-scalar matrix of order 2, its centralizer can have order at most 4. Also, there can be no 2x2 matrix of order 8. Hence GL(2,Q) contains no abelian groups of order 8.

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Dec 6^th, 2006, 11:22pm

on 12/06/06 at 05:39:40, Eigenray wrote:

There is still one more group to be found.

Is it the D₆ group generated by a matrix of order 6 and the matrix of order 2?

| 0 1|, |0 1|
|-1 1| |1 0|

Quote:

(And, of course, the proof of this fact!)

:-/

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 8^th, 2006, 8:46am

I show that A₄ cannot be a subgroup of GL(2,Q). Since S₄ contains A₄, S₄ cannot be a subgroup of GL(2,Q) either.

Since A₄ has no subgroup of index two, all matrices of A₄ have determinant 1. But only the scalar matrix of order 2 has determinant 1, contradicting that A₄ has three matrices of order 2. Hence A₄ is not a subgroup of GL(2,Q).

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 8^th, 2006, 5:12pm

Thanks to eigenray's hint, I get some of the picture now. Since the only 2x2 rational matrices of prime order have orders 2 or 3, a finite subgroup of GL(2,Q) must have order 2^m3ⁿ, for some nonnegative integers m and n. Since the centralizer of a matrix of order 3 has order at most 6 and the center of a 3-group must contain an element of order 3, it follows that n=1. We show that m=3 after tackling the quaternion group.

Here's a crude argument to eliminate the quaternion group as a subgroup of GL(2,Q) (although the quaternion group is a subgroup of GL(2,C), where C is the complex field). Let A and B be 2x2 rational matrices of order 4 such that A^-1BA=B^-1. Let u be any nonzero vector in Q². Then, since A has no eigenspaces over Q, Au=v is independent of u, whence {u,v} is a basis for Q². Also, since A has order 4, A²u=-u. Similarly, Bu=au+bv, for some rationals a and b, with b=/=0.

-u=B²u=B(Bu)=B(au+bv)=aBu+bBv=a(au+bv)+bBv=a²u+abv+bBv.

Hence Bv=(-1-a²)b^-1u-abv. Since A^-1BA=B^-1=B³, we have

A^-1BAu=A^-1Bv=A^-1((-1-a²)b^-1u-av)=(-1-a²)b^-1(-v)+aA^-1u=
=(1+a²)b^-1v-au,

and

B³u=B(B²u)=B(-u)=-Bu=-au-bv. Hence (1+a²)b^-1=-b, or a²+b²=-1, impossible, for a and b rational. Therefore, GL(2,Q) cannot contain the quaternion group.

Now suppose that GL(2,Q) contains a subgroup G of order 16. G cannot be elementary abelian because the centralizer of a non-scalar matrix of order 2 has order at most 4. Hence G contains elements of order 4. Let H be a cyclic subgroup of G of order 4. Since the centralizer of an element of order 4 cannot exceed 4, it follows that N(H) has order 8. By the previous argument above, N(H) cannot contain any elements of order 4 outside H. Hence H is unique in N(H). But then, since G normalizes N(H), G must also normalize H, contradiction. Hence GL(2,Q) has no subgroup of order 16 and so m=3.

Title: Re: Finite subgroups of GL(2,Q)
Post by Eigenray on Dec 13^th, 2006, 4:07am

All the groups have been found. They are:
C₂, C₃, C₄, C₆,
D₂, D₃, D₄, D₆
(and of course the trivial group).

on 12/08/06 at 17:12:24, ecoist wrote:

Since the centralizer of a matrix of order 3 has order at most 6

Do you have a nice way to show that? The "brute force" way is to assume A = [0,1], [1,1]. The equations AB=BA, det(B)=d, and tr(B)=t give
3x² - 3xt + t² - d = 0
where x is the upper-left entry of B. If B is nonscalar, then (t,d) = (0, +/- 1) or (+/- 1, 1), and the only rational solutions end up being when B = +/- A^k for some k.

Quote:

Here's a crude argument to eliminate the quaternion group as a subgroup of GL(2,Q)

The argument can be cleaned up some: Suppose BA = AB^-1. If B has order 4, it is conjugate (over Q) to [0, -1], [1, 0], and then BA = AB^-1 gives A = [a, b], [b, -a], which can't have determinant 1. So in fact A has order 2. (This works over R just as well.)

Quote:

Now suppose that GL(2,Q) contains a subgroup G of order 16. G cannot be elementary abelian because the centralizer of a non-scalar matrix of order 2 has order at most 4. Hence G contains elements of order 4. Let H be a cyclic subgroup of G of order 4. Since the centralizer of an element of order 4 cannot exceed 4, it follows that N(H) has order 8. By the previous argument above, N(H) cannot contain any elements of order 4 outside H. Hence H is unique in N(H). But then, since G normalizes N(H), G must also normalize H, contradiction. Hence GL(2,Q) has no subgroup of order 16 and so m=3.

Looks nice. But why can't N(H)=H?

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 13^th, 2006, 10:29am

Quote:

ecoist wrote:Since the centralizer of a matrix of order 3 has order at most 6

Do you have a nice way to show that?

Well, let's try this. Let A and B be commuting matrices of order 3. Then Q(A) is a field extension of Q in which the polynomial x³-1 splits completely. Hence, adjoining B to Q(A), its minimum polynomial over Q(A) splits completely and so B lies in Q(A).

Very nice argument to eliminate the quaternion group!

Quote:

Looks nice. But why can't N(H)=H?

Sorry, I didn't point out a property of p-groups, where p is a prime. Every proper p-subgroup of a finite p-group is not its own normalizer. That's because, when you represent a p-group G on the cosets of a proper subgroup H, the number of fixed points of H must be a nonzero multiple of p since the orbits of H are powers of p and the index of H in G is a power of p. Further, the number of fixed points of H equals the index of H in N(H).

Title: Re: Finite subgroups of GL(2,Q)
Post by Michael_Dagg on Dec 13^th, 2006, 8:20pm

Nice discussion (and it did go a fair amount)!

I thought someone might mention that each finite group is compact and
so fits inside a maximal compact subgroup of GL(2,R), but all those
maximal compact subgroups are conjugate to O(2,R), which has the circle
group SO(2,R) as a subgroup of index 2, and therefore all these finite
groups have a cyclic group of index at most 2. (Actually if not cyclic
they have to consist of a cyclic group of rotations and the other elements
will be reflections, i.e. all of order 2. So they will then be dihedral.)

You guys have noted that if the group has a 2-dimensional _rational_
representation, its elements' orders have to be small, so you get C_n
or D_n for a few values of n.

Title: Re: Finite subgroups of GL(2,Q)
Post by ecoist on Dec 13^th, 2006, 9:58pm

Quote:

I thought someone might mention that each finite group is compact and
so fits inside a maximal compact subgroup of GL(2,R), but all those
maximal compact subgroups are conjugate to O(2,R), which has the circle
group SO(2,R) as a subgroup of index 2, and therefore all these finite
groups have a cyclic group of index at most 2. (Actually if not cyclic
they have to consist of a cyclic group of rotations and the other elements
will be reflections, i.e. all of order 2. So they will then be dihedral.)

Nice, but does this approach eliminate cyclic of orders 8 and 12? Well, I suppose it does if one knows what SO(2,R) is.

Title: Re: Finite subgroups of GL(2,Q)
Post by Eigenray on Dec 14^th, 2006, 6:50am

on 12/13/06 at 10:29:48, ecoist wrote:

Every proper p-subgroup of a finite p-group is not its own normalizer.

Ah, right right. I'd forgotten about that characterization of nilpotency.

So it looks like you're done then: for |G|=12, you just need to rule out C₃ x| C₄, and then if |G|=24, you have D₆ < SL(2,Q), impossible.

Another proof goes:
(1) G is conjugate to a subgroup of GL(2,Z)
(2) GL(2,Z) embeds into GL(2,Z₃)
(3) Any non-cyclic subgroup of SL(2,Z₃) contains Q₈.

Title: Re: Finite subgroups of GL(2,Q)
Post by Eigenray on Dec 14^th, 2006, 7:29am

on 12/13/06 at 20:20:17, Michael_Dagg wrote:

I'm not familiar with maximal compact subgroups, but if the same reasoning works for n=3, then any finite subgroup of GL(3,Q) is isomorphic to G or GxC₂, where G is a finite subgroup of SO(3). Since G contains elements of orders 1,2,3,4,6 only, G is either C_n, D_n (n=1,2,3,4,6), A₄, or S₄. (Conversely, all of these are realized in GL(3,Q).)

Title: Re: Finite subgroups of GL(2,Q)
Post by Michael_Dagg on Dec 15^th, 2006, 5:22pm

I am seeing the missing post syndrome again. I think 3
here for me and 1 in another place.

Title: Re: Finite subgroups of GL(2,Q)
Post by THUDandBLUNDER on Dec 15^th, 2006, 5:34pm

on 12/15/06 at 17:22:46, Michael_Dagg wrote:

I am seeing the missing post syndrome again. I think 3
here for me and 1 in another place.

Yes, I noticed that your acknowledgement of the solution to y'' = yy' (by Icarus) also disappeared.

Title: Re: Finite subgroups of GL(2,Q)
Post by Michael_Dagg on Dec 15^th, 2006, 5:51pm

That is the other 1.

Title: Re: Finite subgroups of GL(2,Q)
Post by Michael_Dagg on Dec 15^th, 2006, 6:46pm

What I said in the other posts was nothing really important (actually they were trivial).
I already know that a couple members read them and so, that's that unless someone has copies.

I will restate what I said in the last post about the book by Weyl entitlted Symmetry.
Therein some aspect of the conclusion that the finite subgroups of GL(2,R) are cyclic or
dihedral are discussed. In fact, it might surprise you to note something called Leonardo da Vinci's
theorem and is about this same problem (therein as well). There's another called The Enormous Theorem
which paves a nice path to this same topic and is certainly worth looking at.

(e.g. pretty easy to see that Q_8 is definitely not an element of GL(2,Q) ).

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Dec 16^th, 2006, 5:31am

on 12/03/06 at 04:44:59, Barukh wrote:

BTW, what happens if we relax the restriction for the matrix to be invertible?

I know it’s kind of off-topic…

Let E_ij be a 2x2 matrix whose only non-zero element is a_ij = 1. (Clearly, it’s not in GL(2,Q)). Then, the sets (E₁₁, -E₁₁) and (E₂₂, -E₂₂) are groups under matrix multiplication.

Title: Re: Finite subgroups of GL(2,Q)
Post by Michael_Dagg on Dec 16^th, 2006, 8:38am

See also the paper by Newman in the Monthly, Feb. 2002.

Title: Re: Finite subgroups of GL(2,Q)
Post by Barukh on Dec 17^th, 2006, 2:11am

on 12/16/06 at 08:38:44, Michael_Dagg wrote:

See also the paper by Newman in the Monthly, Feb. 2002.

Michael, the AMM 02/2002 issue has the following relevant paper:

"Finite Groups of Matrices Whose Entries Are Integers" by J. Kuzmanovich and A. Pavlichenkov.

Is this what you meant?

Title: Re: Finite subgroups of GL(2,Q)
Post by Michael_Dagg on Dec 17^th, 2006, 8:30am

Yes Barukh that's the paper. I don't know why I typed Newman (but
he is one of the references in that paper).