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Topic: Finite subgroups of GL(2,Q) (Read 6304 times) |
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Eigenray
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Finite subgroups of GL(2,Q)
« on: Nov 18th, 2006, 8:04pm » |
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Determine all finite subgroups of GL(2,Q).
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Michael Dagg
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Re: Finite subgroups of GL(2,Q)
« Reply #1 on: Nov 24th, 2006, 8:15pm » |
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This nice problem ought to generate a fair amount of
discussion.
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Regards,
Michael Dagg
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Icarus
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Re: Finite subgroups of GL(2,Q)
« Reply #2 on: Nov 25th, 2006, 7:29am » |
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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.
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| « Last Edit: Nov 25th, 2006, 7:29am by Icarus » |
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #3 on: Nov 25th, 2006, 4:46pm » |
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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).
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Ken_Wiley
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Re: Finite subgroups of GL(2,Q)
« Reply #4 on: Nov 25th, 2006, 7:08pm » |
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I am just curious but what kind of math is this?
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #5 on: Nov 26th, 2006, 4:03am » |
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on Nov 25th, 2006, 7:08pm, Ken_Wiley wrote:| I am just curious but what kind of math is this? |
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The problem is from group theory. It deals with groups of 2x2 invertible matrices of rational (Q) numbers.
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #6 on: Nov 26th, 2006, 4:04am » |
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on Nov 25th, 2006, 4:46pm, 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). |
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How do you show this? 
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #7 on: Nov 26th, 2006, 4:10am » |
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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?
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| « Last Edit: Nov 26th, 2006, 4:26am by Barukh » |
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #8 on: Nov 26th, 2006, 9:45am » |
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Quote:
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? |
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|0 -1| (the companion matrix of the polynomial x2+x+1)
|1 -1|
has order 3 and
|0 -1|
|1 1|
has order 6.
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Icarus
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Re: Finite subgroups of GL(2,Q)
« Reply #9 on: Nov 26th, 2006, 11:25am » |
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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.
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #10 on: Nov 27th, 2006, 5:40am » |
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on Nov 26th, 2006, 9:45am, 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. |
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Interesting! Is there a matrix with order n = 5?
I still cannot find a link between this problem and linear transformations on a plane...
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #11 on: Nov 27th, 2006, 10:04am » |
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Quote:| Is there a matrix with order n = 5? |
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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.
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Icarus
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Re: Finite subgroups of GL(2,Q)
« Reply #12 on: Nov 27th, 2006, 3:59pm » |
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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.
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| « Last Edit: Nov 27th, 2006, 4:32pm by Icarus » |
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #13 on: Nov 28th, 2006, 11:06am » |
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on Nov 27th, 2006, 3:59pm, Icarus wrote:Hence the characteristic equation of A must be one of the following:
x2+2x+1
x2-2x+1 |
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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?
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #14 on: Nov 28th, 2006, 3:09pm » |
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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? |
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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.
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Icarus
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Re: Finite subgroups of GL(2,Q)
« Reply #15 on: Nov 28th, 2006, 7:39pm » |
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on Nov 28th, 2006, 11:06am, Barukh wrote:
I don't think these are relevant. |
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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. |
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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.)
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| « Last Edit: Nov 28th, 2006, 7:40pm by Icarus » |
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #16 on: Dec 3rd, 2006, 4:44am » |
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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?
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #17 on: Dec 3rd, 2006, 9:50am » |
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Quote:| It seems like all the groups were identified. |
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Not quite. There is also the dihedral group of order 6 generated by
|0 1|
|1 0|
and
|0 -1|
|1 -1|.
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Icarus
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Re: Finite subgroups of GL(2,Q)
« Reply #18 on: Dec 3rd, 2006, 1:57pm » |
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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.
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #19 on: Dec 3rd, 2006, 3:16pm » |
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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.
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #20 on: Dec 3rd, 2006, 7:42pm » |
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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).
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #21 on: Dec 3rd, 2006, 11:20pm » |
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on Dec 3rd, 2006, 1:57pm, Icarus wrote:| I'm not sure what you mean by "revertible". Do you mean "invertible"? |
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Yes, of course! 
Quote:| In that case, you can't have a group. |
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Not quite...
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Eigenray
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Re: Finite subgroups of GL(2,Q)
« Reply #22 on: Dec 6th, 2006, 5:39am » |
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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?
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #23 on: Dec 6th, 2006, 10:16am » |
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on Dec 6th, 2006, 5:39am, 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? |
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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.
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Eigenray
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Re: Finite subgroups of GL(2,Q)
« Reply #24 on: Dec 6th, 2006, 7:28pm » |
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on Dec 6th, 2006, 10:16am, 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. |
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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 |
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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. |
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The only finite noncyclic abelian subgroup is the Klein 4-group. Start of proof: Any set of commuting diagonalizable matrices can be ....
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