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" <[EMAIL PROTECTED]> 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 > > Please find our Email Disclaimer here: http://www.ukzn.ac.za/disclaimer/ > > _______________________________________________ > Forum mailing list > [email protected] > http://mail.gap-system.org/mailman/listinfo/forum -- Dima Pasechnik http://www.ntu.edu.sg/home/dima/ _______________________________________________ Forum mailing list [email protected] http://mail.gap-system.org/mailman/listinfo/forum
