[GAP Forum] Re:Orbit Lengths

[Dmitrii Pasechnik]

I suppose you mean
V := FullRowSpace(GF(2),3);
Then OrbitLengths(m,V) gives the correct answer:
[ 1, 3, 3, 1 ]

(it's puzzling that you got [2,4,4,4] somehow...)


On 11/14/07 4:45 PM, "Thekiso Seretlo" <207525917 at ukzn.ac.za> wrote:

> Let V := FullRowSpace(GF(2),2)
> m1:=One(GF(2))*[[0,1,0],[1,0,0],[0,0,1]];
> m2:=One(GF(2))*[[0,1,0],[0,0,1],[1,0,0]];
> m:=Group(m1,m2);
> this is a matrix representation of $S3$ in GF(2)
> If we say
> OrbitLengths(m,V)
> we get [2,4,4,4] that is we show that  this has four orbits and of the lenghts
> given that is the conjugacy classes here. Moving to irreducible characters by
> Brauer we know that the number of orbits is four, my problem is how do we get
> the orbitlengths of the orbits of irreducible characters ?
> Yours faithfully
> TT Seretlo 
> 
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-- 
Dima Pasechnik
http://www.ntu.edu.sg/home/dima/