[GAP Forum] action on Schur multipliers

0002319s Graham.Ellis at NUIGALWAY.IE
Mon Mar 13 12:34:54 GMT 2006


Dear GAP Forum,

Deepak Naidu wrote:

> Let H be a normal subgroup of a finite group G. The action of G on H, by 
conjugation,
> induces an action of G on the Schur multiplier H^2(H, C*) of H.
> Is there any way I can use GAP to find the number of orbits of H^2(H, C*) 
under the action of G?


The following commands (which use the HAP package) show that the there are 4 
orbits in the Schur multiplier of N:=AltenratingGroup(6) under the conjugation 
action of G:=SymmetricGroup(6).

Note that the Schur multiplier of N is isomorphic to the Second integral 
homology of N, and that N acts trivial on its Schur multiplier/second homology 
(for any group N). So we are really interested in orbits under an action of 
the quotient group G/N.


############## CUT ######################
gap> G:=SymmetricGroup(6);;
gap> N:=AlternatingGroup(6);;
gap> R:=ResolutionFiniteGroup(N,3);;

gap> SchurMultiplier:=Homology(TensorWithIntegers(R),2);
[6]

gap> #So the Schur multiplier of N is cyclic of order 6.

gap> ConjugationHomomorphism:=function(g);
> return GroupHomomorphismByFunction(N,N,x->g*x*g^-1);
end;;

gap> HomologyMapInducedByConjugation:=function(g)
> local f;
> f:=EquivariantChainMap(R,R,ConjugationHomomorphism(g));
> f:=TensorWithIntegers(f);
> f:=Homology(f,2);
> return f;
> end;;

gap> #Note that (1,2) represents a nontrivial element in G/N.

gap> HM:=HomologyMapInducedByConjugation((1,2));
[ f1, f2, f3, f4, f5, f6 ] -> [ f1^4*f2^-3, f1^2*f2^-1, f3, f3^-2*f4^2*f5,
f3^-2*f4^2*f5, f6 ]

gap> IsomorphismPermGroup(Source(HM));
[ f1, f2, f3, f4, f5, f6 ] -> [ (), (1,2,3)(4,5,6), (1,4)(2,5)(3,6), (), (),
((1,4)((2,5)(3,6) ]
##################### CUT ##################

Here Source(HM) is a finitely presented group isomorphic to the cyclic group 
H_2(N) of order 6.We see that conjugation by (1,2) in G induces an 
endomorphism on H_2(N)=C_2 x C_3 which fixes the generator f3 of order two, 
and inverts the generator f2 of order three.


Graham



More information about the Forum mailing list