At 12:22 PM 12/26/03, chiachichang1123@yahoo.com.tw wrote:
generators: s1=(57)(68) s2=(56)(78) s3=(34) s4=(37)(48) s5=(26)(48) s6=(24)(68) s7=(78) s8=(68) s9=(48) S={si : i=1~9}
<S> means a permutation group generate by S
give a=(2345678)
(1) is a in <S>???
Yes; in fact, <S> is the full symmetric group on {2,3,4,5,6,7,8}.
(2) a=??
a = s9*s2*s1*s6*s3. I found the answers to these questions using GAP, available from http://www.gap-system.org/ , which describes it as "a free system for computational discrete algebra". GAP stands for "Groups, Algorithms and Programming". Here's a transcript of the GAP session, with commentary added. First, enter the generators: gap> s1 := (5,7)(6,8); # GAP has a built-in syntax for permutations (5,7)(6,8) gap> s2 := (5,6)(7,8);; # the double semicolon suppresses the response gap> s3 := (3,4);; gap> s4 := (3,7)(4,8);; gap> s5 := (2,6)(4,8);; gap> s6 := (2,4)(6,8);; gap> s7 := (7,8);; gap> s8 := (6,8);; gap> s9 := (4,8);; Create the group they generate: gap> g := Group(s1,s2,s3,s4,s5,s6,s7,s8,s9); Group([ (5,7)(6,8), (5,6)(7,8), (3,4), (3,7)(4,8), (2,6)(4,8), (2,4)(6,8), (7,8), (6,8), (4,8) ]) The order of g is 7!, so it is the full symmetric group on 2,3,...,8. gap> Order(g); 5040 Use the technique of section 37.5 of the reference manual to express a given permutation in terms of the generators. First, create a free group with the same number of generators (and give them the same names): gap> f := FreeGroup("s1","s2","s3","s4","s5","s6","s7","s8","s9"); <free group on the generators [ s1, s2, s3, s4, s5, s6, s7, s8, s9 ]> Create the homomorphism that maps f's generators onto g's ("NC" means "no check"; it promises that this will be possible so GAP needn't check): gap> hom := GroupHomomorphismByImagesNC(f,g,GeneratorsOfGroup(f),GeneratorsOfGroup(g)); [ s1, s2, s3, s4, s5, s6, s7, s8, s9 ] -> [ (5,7)(6,8), (5,6)(7,8), (3,4), (3,7)(4,8), (2,6)(4,8), (2,4)(6,8), (7,8), (6,8), (4,8) ] Find the answer we want: gap> PreImagesRepresentative(hom, (2,3,4,5,6,7,8)); s9^-1*s2^-1*s1^-1*s6^-1*s3^-1 A quirk in the algorithm has expressed the result in terms of inverses of the generators, even though they all have order two. Check the answer, skipping the inverses: gap> s9*s2*s1*s6*s3; (2,3,4,5,6,7,8) Note that GAP composes permutations from left to right; if you prefer right-to-left, your answer would be s3*s6*s1*s2*s9. The GAP reference manual includes an extensive bibliography. -- Fred W. Helenius <fredh@ix.netcom.com>