If I have the following presentation for $A_5$ $$\langle x,y,z\mid x^3 = y^3= z^3 =(xy)^2=(xz)^2= (yz)^2= 1\rangle$$ with subgroup $$ H = \left\langle {x,y} \right\rangle$$ and let GAP apply coset enumeration to my generators and relations, as with the code below, is there a command I can use to check whether this presentation is indeed for $A_5$?
<free group on the generators [ x, y, z ]>
gap> x:=F.x;
x
gap> y:=F.y;
y
gap> z:=F.z;
z
gap> rels:=[x^3,y^3,z^3,(x*y)^2,(x*z)^2,(y*z)^2];
[ x^3, y^3, z^3, (x*y)^2, (x*z)^2, (y*z)^2 ]
gap> G:=F/rels;
<fp group on the generators [ x, y, z ]>
gap> gens:=GeneratorsOfGroup(G);
[ x, y, z ]
gap> xG:=gens[1];
x
gap> yG:=gens[2];
y
gap> zG:=gens[3];
z
gap> H:=Subgroup(G,[xG,yG]);
Group([ x, y ])
gap> ct:=CosetTable(G,H);
[ [ 1, 3, 4, 2, 5 ], [ 1, 4, 2, 3, 5 ], [ 1, 3, 5, 4, 2 ],
[ 1, 5, 2, 4, 3 ], [ 2, 3, 1, 4, 5 ], [ 3, 1, 2, 4, 5 ] ]
gap> Display(TransposedMat(ct));
[ [ 1, 1, 1, 1, 2, 3 ],
[ 3, 4, 3, 5, 3, 1 ],
[ 4, 2, 5, 2, 1, 2 ],
[ 2, 3, 4, 4, 4, 4 ],
[ 5, 5, 2, 3, 5, 5 ] ]
I'm asking because I'm doing research where I will enter candidate presentations into GAP and check if the presentation is equal to a certain alternating group.
Best Answer
Yes.
Use
IdGroup(G);. IfGis indeed (a presentation for a group isomorphic to) $A_5$, the output of this command is[60, 5].