Multiplying permutations with GAP?

stf92 · 07-03-2017, 04:50 PM

Is anyone familiar with the GAP system for computational discrete algebra? In the affirmative case, I would like to know how to multiply permutations. I could learn how to do it by reading the manual but that would involve several weeks, thanks.

norobro · 07-04-2017, 09:20 PM

Not familiar with the GAP system, but being curious I took a look at the docs. Multiplying simple permutations is pretty straight forward.

Docs: https://www.gap-system.org/Manuals/d...66C8707B5DE10A

Example from here: https://math.stackexchange.com/quest...yclic-notation

Code:

gap> a := (1,3,5,2);
(1,3,5,2)
gap> b := (2,5,6);
(2,5,6)
gap> c := (1,6,3,4);
(1,6,3,4)
gap> Display(b*a);;
(1,3,5,6)
gap> Display(c*a);;
(1,6,5,2)(3,4)

And reading commands from a file:

Code:

$ cat perms.g
a := (1,3,5,2);
b := (2,5,6);
c := (1,6,3,4);
Display(b*a);;
Display(c*a);;

Code:

gap> Read("perms.g");
(1,3,5,6)
(1,6,5,2)(3,4)

Please post an example of what you are trying to do.

stf92 · 07-05-2017, 07:52 PM

Very simple. There's a problem: Prove every element of A_5, the alternate group of order 5, is a commutator. A_5 has only three conjugacy classes, A, B and C: (ij)(kl), (ijk) and (ijklm), plus (1) of course. In internet I saw them all three written as commutators. Well, that's a straightforward proof, but how do you guess the commutators? Instead I did the following: take one class, say that to which (ijl) belongs, that is, the class of all 3-cycles, which is B above. Certainly, x^{-1}x^g, where x^g is x conjugated by g, is a commutator. Now as x sweeps A, x will sweep A too, hummm... I think I made a mistake. x --> x^g is not a bijection for g fixed.

Never mind. I want to send x^{-1}x^g, for every x in A, to a set, say set Ma. Do the same with B and C, creating Mb and Mc. And then ask the program what the cardinal of Ma U Mb U Mc is (U stands for union). I don't think that is very difficult to do, is it?

norobro · 07-06-2017, 01:04 PM

Well, that's beyond my meager math skills.

Maybe someone will be along to help.

stf92 · 07-06-2017, 04:50 PM

Thanks anyway.