Re: The group GL_3(2)
From: Timothy Murphy (tim_at_birdsnest.maths.tcd.ie)
Date: 10/31/04
- Next message: Mos: "Re: Longest Thread Ever"
- Previous message: Kent Paul Dolan: "Re: fermat111111111111111111111111111111111111111111111111111111"
- In reply to: Van Jacques: "The group GL_3(2)"
- Messages sorted by: [ date ] [ thread ]
Date: Sun, 31 Oct 2004 02:00:18 +0000
Van Jacques wrote:
> I have long wanted to study the following group GL_3(2).
>
> I have come across it in Milne's notes on groups, but
> his methods are beyond the (very) elementary ones with which
> I am familiar. He gives the following prob and soln;
> Parts (a) and (b) are clear, but I don't even understand what
> he is talking about it (c) and (d). I don't even know what
> "Jordan canonical form" is, though it will be easy to lookup at
> Mathworld or wherever. Still, I don't see what he is even
> talking about, much less what he is doing.
> Please help--obviously help must be pretty elementary.
I think you probably do know what the Jordan canonical (or normal) form is.
It is usually applied to matrices over C,
and says (roughly) that you can find a similar matrix PTP^{-1}
with the eigenvalues along the main diagonal,
and some 1's in the diagonal line just above the main diagonal,
or in other words can find a similar matrix of the form
(A 0 0
0 B 0
0 0 C ...)
where A,B,C are submatrices of the form
(c 1 0 ...
0 c 1 ...
...
0 0 ... c).
I guess his point is that you can carry out
exactly the same reduction
_provided the eigenvalues are in the field k you are working in_,
in this case F_2.
In particular, in this case, if the eigenvalues are 1,1,1
you get 3 cases according as there are 0,1 or 2 1's above the diagonal.
On the other hand, if the eigenvalues are distinct
then the class is completely determined by the minimal equation of degree 3,
since the matrix is similar to
(1 0 b)
(0 1 c)
(0 0 1)
by taking any vector e, and the te and t^2e
(where t is the linear map in question) as basis.
Finally, if 2 eigenvalues are the same
then the third must be in F_2 by the trace,
and so must be 1.
Hence the first 2 must be 1 since the determinant is 1.
Ps I think he just forgot 42 in his list below.
>
> I know group theory thru the Sylow theorems.
>
> (c) below is completely over my head. I will study what I can find on
> this on the net, and search previous posts on sci.math.
> (I don't have access to a lib, and have to depend on sci.math and the
> net).
> -------------
> 24. Let G = GL_3(F_2).
> (a) Show that (G : 1) = 168.
> (b) Let X be the set of lines through the origin in (F_2)^3;
> show that X has 7 elements, and that there is a natural injective
> homomorphism G --> Sym(X) = S_7.
> (c) Use Jordan canonical forms to show that G has six conjugacy
> classes, with 1, 21, 42, 56, 24, and 24 elements respectively.
> [Note that if M is a free F_2[a]-module of rank one, then
> End_(F_2[a])(M) = F_2[a].]
> (d) Deduce that G is simple.
>
> ----------
>
> Here is Milne's soln.
> ------------
> 24. (a) The number of possible first rows is 2^3 ? 1; of second rows
> 2^3 ? 2; of third rows 2^3 ? 2^2; whence (G:1) = 7 × 6 × 4 = 168.
> (b) Let V = (F_2)^3. Then #V = 2^3 = 8. Each line through the origin
> contains exactly one point != origin, and so #X = 7.
> (c) We make a list of possible characteristic and minimal polynomials:
> Char poly. | Min Size Order
> 1) X^3 + X^2 + X + 1| X + 1 1 1
> 2) X^3 + X^2 + X + 1| (X + 1)^2 21 2
> 3) X^3 + X^2 + X + 1| (X + 1)^3 42 4
> 4) X^3+1=(X+1)(X^2+X+1)|Same 56 3
> 5) X^3 + X+1 (irred)| Same 24 7
> 6) X^3+X^2+1 (irred)| Same 24 7
>
> Here size denotes the number of elements in the conjugacy class.
> Case 5: Let a be an endomorphism with characteristic polynomial
> X^3 + X + 1. Check from its minimal polynomial that a^7 = 1, and so
> has order 7. Note that V is a free F_2[a]-module of rank one, and so
> the centralizer of a in G is F_2[a] /\ G = <a>. Thus the order of the
> centralizer is #C(a) = 7, and the number of elements in the conjugacy
> class of a is 168/7 = 24.
> Case 6: Exactly the same as Case 5.
> Case 4: Here V = V1 V2 as an F2[ ]-module, and
> EndF2[ ](V ) = EndF2[ ](V1) EndF2[ ](V2).
> Deduce that #CG( ) = 3, and so the number of conjugates of is 168
> 3 = 56.
> Case 3: Here CG( ) = F2[ ] \ G = h i, which has order 4.
> Case 1: Here is the identity element.
> Case 2: Here V = V_1 \+ V_2 as an F_2[a]-module, where a acts as 1 on
> V_1 and has minimal polynomial X^2 +1 on V_2. Either analyse, or simply
> note that this conjugacy class contains all the remaining elements.
> (V_1 \+ V_2 denotes the direct sum).
> (d) Since 168 = 2^3 × 3 × 7, a proper nontrivial subgroup H of G will
> have order
> 2, 4, 8, 3, 6, 12, 24, 7, 14, 28, 56, 21, 24, or 84.
> -------
> (Me; why isn't 42 in this list?)
> ----------
> If H is normal, it will be a disjoint union of {1} and some other
> conjugacy classes, and so
> |N| = 1 + Sum(c_i) with c_i equal to 21, 24, 42, or 56, but this
> doesn?t happen.
> -----------
>
> Van
-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
- Next message: Mos: "Re: Longest Thread Ever"
- Previous message: Kent Paul Dolan: "Re: fermat111111111111111111111111111111111111111111111111111111"
- In reply to: Van Jacques: "The group GL_3(2)"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
|
