lhsoicher
commented
1 year ago Are you looking for a list of defining relators for your group with respect to a given list of generators? If so, then you should proceed as follows in GAP (not using GRAPE):
gap> G:=SymmetricGroup(4); Sym( [ 1 .. 4 ] ) gap> gens:=[(1,2),(2,3),(3,4)]; [ (1,2), (2,3), (3,4) ] gap> iso:=IsomorphismFpGroupByGenerators(G,gens); [ (1,2), (2,3), (3,4) ] -> [ F1, F2, F3 ] gap> RelatorsOfFpGroup(Image(iso)); [ F2^2, F1^2, F3^2, (F3F1)^2, (F3F2)^3, (F1*F2)^3 ] gap>
See the GAP documentation for more info.
Best, Leonard
From: hongyi-zhao @.***> Sent: 11 November 2022 08:18 To: gap-packages/grape Cc: Subscribed Subject: [gap-packages/grape] Utilizing the minimal cycles in Cayley graph to identify the power product expression of group elements equal to identity element of the group. (Issue #35)
Hi here,
I want to find the power product expression of group elements equal to identity element of the group. For this purpose, Dima Pasechnik @.**@.>) gives me the following ideahttps://lists.uni-kl.de/gap/arc/forum/2022-11/msg00011.html:
consider the Cayley graph of the group, w.r.t. the generators you want. Now, the minimal cycles in the graph are exactly the minimal, indecomposable, words in generators (i.e. power product experssions) you want.
Based on the above comment, I tried the following method but still can't figure out how to do the trick with the help of grapehttps://github.com/gap-packages/grape package:
gap> LoadPackage("grape"); true gap> C:=CayleyGraph(SymmetricGroup(4),[(1,2),(2,3),(3,4)]); rec( adjacencies := [ [ 2, 3, 7 ] ], group := Group([ (1,10,17,19)(2,9,18,20)(3,12,14,21)(4,11,13,22)(5,7,16,23)(6,8,15,24), (1,7)(2,8)(3,9)(4,10)(5,11)(6,12)(13,15) (14,16)(17,18)(19,21)(20,22)(23,24) ]), isGraph := true, isSimple := true, names := [ (), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4), (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2), (1,3), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3) ], order := 24, representatives := [ 1 ], schreierVector := [ -1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2 ] ) gap> Girth(C); 4 gap> Diameter(C); 6
Any tips will be appreciated.
Regards, Zhao
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hongyi-zhao
Hi here,
I want to find the power product expression of group elements equal to identity element of the group. For this purpose, Dima Pasechnik (dima@pasechnik.info) gives me the following idea:
Based on the above comment, I tried the following method but still can't figure out how to do the trick with the help of grape package:
Any tips will be appreciated.
Regards, Zhao