-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Am 15.02.2013 10:02, schrieb rahul kitture: > Given two subgroups $H$ and $K$ of a finite group (say Symmetric/ > Alternating Group), how do we compute the product $HK$ in the > group? I couldn't find anything from Help or topics in online > library.
This may not be the most elegant way, but gap> Group(Concatenation(List([H,K], GeneratorsOfGroup)), Identity(H)); should do the trick. > Also, given finite number of square (invertible) matrices over a > finite field, how do we get the (multiplicative) group, generated > by them? (The command "GroupWithGenerators(); " does not work with > the elements(generators) to be matrices. > Works for me. gap> GL(3, GF(2)); SL(3,2) gap> List([1..3], x -> Random(last)); [ <an immutable 3x3 matrix over GF2>, <an immutable 3x3 matrix over GF2>, <an immutable 3x3 matrix over GF2> ] gap> Group(last); <matrix group with 3 generators> gap> GroupWithGenerators(last2); <matrix group with 3 generators> gap> last=last2; true Are the entries of your matrices elements of the finite field? If not, multiply the matrices by One(field). HTH Sven Reichard -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.5 (GNU/Linux) Comment: Using GnuPG with Thunderbird - http://www.enigmail.net/ iQEVAwUBUR4BdmFjB3Gki4XVAQLFEggAlJmqyz7renijLE4RJnSgnctPxrtZtOq/ F9AradxaWZFP0BGKOqSUSE0TOLc0lextSs48h53gffQOGZfOhnQyjn1krx9aZBdz vkJ7XbcVBV7IrKvgWQK37TaesxYlHZKn5ac1Ub+o0uEyfJ76OmDDIGvTyjHFDSZm QsBS2S3bJxYTR0eyC/7YA5V6p6uz8TQxolDtmZEqVD6tEOSqZYOGQnEZnnUGrEK5 c7VbUn9UcaHPOM58oWOFZQFbCc5V+NHqxJ8/riepy+oTzhTe+JiSPXuZRkQn4RAo /Mo1Eo1+33eVSl1ul6zpCPCezNQFNqMzrdKyzow/akcq4uWb9xErfg== =VVZt -----END PGP SIGNATURE----- _______________________________________________ Forum mailing list Forum@mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum
