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,
Thomas

Miles-Receive-Header: reply