Dear GAP Forum,
Bruce Colletti asked
What is the command (if any) that returns (in a list) the disjoint = cyclic factors of a permutation? For instance, (1,2,3)(4,5,6) returns = [(1,2,3), (4,5,6)]. =20
As far as I know, there is no special GAP command for that.
I would use the following function.
gap> cyclicfactors:= p -> List( Cycles( p, MovedPoints( p ) ), > x -> RestrictedPerm( p, x ) );; gap> cyclicfactors( (1,2)(3,4,5)(6,7) ); [ (1,2), (3,4,5), (6,7) ]
All the best,