Page 91 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.

mož Moravec: Some Topics in the Theory of Finite Groups 79

1 a
b ba

ab aba

bab baba

Cayley graphs can be constructed within GAP using a package called GRAPE. This package
has to be loaded into GAP using LoadPackage. After that all the commands of the pack-

age are available. One can then construct Cayley graphs Cay(G ,S); the result is a record

that contains several attributes of the graph; we refer to GAP’s manual for further details
on records, and GRAPE’s manual for further commands. Here we show how to construct

a Cayley Graph of A4 with respect to the generating set {(1 2 3), (1 2 4)}, and compute its
adjacency matrix.

gap> LoadPackage("grape");;
------------------------------------------------------------------------
Loading GRAPE 4.6.1 (GRaph Algorithms using PErmutation groups)
by Leonard H. Soicher (http://www.maths.qmul.ac.uk/~leonard/).
Homepage: http://www.maths.qmul.ac.uk/~leonard/grape/
------------------------------------------------------------------------
gap> cay := CayleyGraph(AlternatingGroup(4), [(1,2,3),(1,2,4)]);
rec( adjacencies := [ [ 5, 6, 7, 10 ] ], group := Group([ (1,5,7)(2,4,8)

(3,6,9)(10,11,12), (1,2,3)(4,7,10)(5,9,11)(6,8,12) ]), isGraph := true,
isSimple := true,
names := [ (), (2,3,4), (2,4,3), (1,2)(3,4), (1,2,3), (1,2,4), (1,3,2),

(1,3,4), (1,3)(2,4), (1,4,2), (1,4,3), (1,4)(2,3) ], order := 12,
representatives := [ 1 ],
schreierVector := [ -1, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1 ] )
gap> CollapsedAdjacencyMat(cay);
[ [ 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0 ],
[ 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0 ],
[ 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1 ],
[ 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0 ],
[ 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0 ],
[ 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1 ],
[ 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1 ],
[ 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1 ],
[ 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0 ],
[ 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0 ],
[ 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0 ],
[ 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0 ] ]

3.2.8 Problems

1. Supply the missing proofs in this section.