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Re: too many special linear matrices


The order of the special linear group with entries in F_q --the field  
with q elements (where q is a power of a prime) is well known to be:

f[n_, q_] := (1/(q - 1))*Product[q^n - q^i, {i, 0, n - 1}]

This gives:


Table[f[n,2],{n,1,5}]

{1,6,168,20160,9999360}


Table[f[n,3],{n,1,5}]

{1,24,5616,12130560,237783237120}

The list of results you quote in your post seems to be a union of  
these two (corresponding to taking coefficients in the integers  
module 2 and modulo 3).   Bu how on earth did you get 60 using the  
formula you quote from Blyth and Robertson?

In any case, this formula seems to be simply  f[2,p] (p any prime)  
which you have somehow managed to misstate.

Simplify[f[2, p]]


p*(p^2 - 1)



Andrzej Kozlowski




On 16 Aug 2006, at 09:36, Roger Bagula wrote:

> In an old group theory book they talk about special linear groups over
> the modulo  of prime
> Integers: SL[2,P]
> The formula given is
> number of matrices in the group  =If [n=2,6,Prime[n]*(Prime[n]^2-1)]
> (Essentual Student Algebra, Volume 5 ,Groups, T.S. /Blyth and E.F.
> Robertson,1986, Chapman and Hall,New York, page 14)
> So I tried to  generate the elements of the group in Mathematica by a
> search program for Determinant one
> matrices.
> I get:
> 6,24,124,348
> instead of what I should get:
> 6,12,60,168
> Since the famous Klein group SL[2,7] is one of these ,
> it would help to have a set of elements for that group!
>
> Mathematica code:
> Clear[M, k, s]
> M = {{l, m}, {n, o}};
> k = 3
> s =
> Union[Delete[Union[Flatten[Table[Flatten[Table[Table[If[Mod[Abs[Det 
> [M]],
> k] - 1 == 0, M , {}], {l, 0,k - 1}], {m, 0, k - 1}], 1], {n, 0, k -  
> 1},
> {o, 0, k - 1}], 2]], 1]]
> Dimensions[s]
>


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