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Topic: Finite subgroups of GL(2,Q) (Read 6303 times) |
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #25 on: Dec 6th, 2006, 9:00pm » |
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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.
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| « Last Edit: Dec 9th, 2006, 7:52pm by ecoist » |
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #26 on: Dec 6th, 2006, 11:22pm » |
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on Dec 6th, 2006, 5:39am, Eigenray wrote:| There is still one more group to be found. |
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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!) |
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #27 on: Dec 8th, 2006, 8:46am » |
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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).
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| « Last Edit: Dec 9th, 2006, 7:35pm by ecoist » |
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #28 on: Dec 8th, 2006, 5:12pm » |
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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 2m3n, 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.
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| « Last Edit: Dec 10th, 2006, 9:36am by ecoist » |
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Eigenray
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Re: Finite subgroups of GL(2,Q)
« Reply #29 on: Dec 13th, 2006, 4:07am » |
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All the groups have been found. They are:
C2, C3, C4, C6,
D2, D3, D4, D6
(and of course the trivial group).
on Dec 8th, 2006, 5:12pm, ecoist wrote:| Since the centralizer of a matrix of order 3 has order at most 6 |
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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 = +/- Ak for some k.
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Here's a crude argument to eliminate the quaternion group as a subgroup of GL(2,Q) |
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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.)
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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. |
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Looks nice. But why can't N(H)=H?
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #30 on: Dec 13th, 2006, 10:29am » |
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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? |
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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? |
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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).
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| « Last Edit: Dec 14th, 2006, 9:39am by ecoist » |
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Michael Dagg
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Re: Finite subgroups of GL(2,Q)
« Reply #31 on: Dec 13th, 2006, 8:20pm » |
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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.
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| « Last Edit: Dec 13th, 2006, 8:22pm by Michael Dagg » |
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Michael Dagg
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ecoist
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Re: Finite subgroups of GL(2,Q)
« Reply #32 on: Dec 13th, 2006, 9:58pm » |
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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.) |
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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.
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Eigenray
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Re: Finite subgroups of GL(2,Q)
« Reply #33 on: Dec 14th, 2006, 6:50am » |
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on Dec 13th, 2006, 10:29am, ecoist wrote:| Every proper p-subgroup of a finite p-group is not its own normalizer. |
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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.
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Eigenray
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Re: Finite subgroups of GL(2,Q)
« Reply #34 on: Dec 14th, 2006, 7:29am » |
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on Dec 13th, 2006, 8:20pm, Michael_Dagg wrote:| 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. |
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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 Cn, Dn (n=1,2,3,4,6), A4, or S4. (Conversely, all of these are realized in GL(3,Q).)
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Michael Dagg
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Re: Finite subgroups of GL(2,Q)
« Reply #35 on: Dec 15th, 2006, 5:22pm » |
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I am seeing the missing post syndrome again. I think 3
here for me and 1 in another place.
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Michael Dagg
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ThudnBlunder
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Re: Finite subgroups of GL(2,Q)
« Reply #36 on: Dec 15th, 2006, 5:34pm » |
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on Dec 15th, 2006, 5:22pm, Michael_Dagg wrote:I am seeing the missing post syndrome again. I think 3
here for me and 1 in another place. |
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Yes, I noticed that your acknowledgement of the solution to y'' = yy' (by Icarus) also disappeared.
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Michael Dagg
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Re: Finite subgroups of GL(2,Q)
« Reply #37 on: Dec 15th, 2006, 5:51pm » |
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That is the other 1.
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Michael Dagg
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Michael Dagg
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Re: Finite subgroups of GL(2,Q)
« Reply #38 on: Dec 15th, 2006, 6:46pm » |
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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) ).
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| « Last Edit: Dec 16th, 2006, 8:29am by Michael Dagg » |
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Michael Dagg
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #39 on: Dec 16th, 2006, 5:31am » |
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on Dec 3rd, 2006, 4:44am, Barukh wrote:| BTW, what happens if we relax the restriction for the matrix to be invertible? |
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I know it’s kind of off-topic…
Let Eij be a 2x2 matrix whose only non-zero element is aij = 1. (Clearly, it’s not in GL(2,Q)). Then, the sets (E11, -E11) and (E22, -E22) are groups under matrix multiplication.
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Michael Dagg
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Re: Finite subgroups of GL(2,Q)
« Reply #40 on: Dec 16th, 2006, 8:38am » |
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See also the paper by Newman in the Monthly, Feb. 2002.
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Michael Dagg
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Barukh
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Re: Finite subgroups of GL(2,Q)
« Reply #41 on: Dec 17th, 2006, 2:11am » |
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on Dec 16th, 2006, 8:38am, Michael_Dagg wrote:| See also the paper by Newman in the Monthly, Feb. 2002. |
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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?
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Michael Dagg
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Re: Finite subgroups of GL(2,Q)
« Reply #42 on: Dec 17th, 2006, 8:30am » |
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Yes Barukh that's the paper. I don't know why I typed Newman (but
he is one of the references in that paper).
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Michael Dagg
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