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 <td
style="text-align: center; vertical-align: top; color: rgb(0, 0, 102);"><big>About HAP: Cohomology With Twisted
Coefficients 
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href="aboutGraphsOfGroups.html">next</a> 
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style="vertical-align: top; background-color: rgb(255, 255, 255); text-align: left;">The
cohomology of a group G with coefficients in a ZG-module A is defined
as: 
 
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 Hn(G,A) = 
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 <div style="text-align: center;"> 
Ker( HomZG(Rn,A) → HomZG(Rn+1,A) ) 
 <hr style="width: 100%; height: 2px;"> </div>
 <div style="text-align: center;">Image( HomZG(Rn-1,A) → HomZG(Rn,A) 
 
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When the abelian group underlying A is free of rank n we can encode A
as a group homomorphism A:G → GLn(Z). 
 
When G is a permutation group of degree n the free abelian group Zn
admits a canonical G-action defined by 
 
 <div style="text-align: center;">g·(x1, x2,
... , xn) = (xg'(1) , x_g'(2) , ... , xg'(n))
 
 </div>
 
where g'=g^-1, for g in G and x_i in Z. This
style="font-style: italic;">canonical permutation module A can
be constructed for any permutation group G using the HAP command PermToMatrixGroup().
For example: 
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style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
G:=AlternatingGroup(5);; 
 
gap> A:=PermToMatrixGroup(G,5); 
[ (1,2,3,4,5), (3,4,5) ] -> 
[ [ [ 0, 1, 0, 0, 0 ], [ 0, 0, 1, 0, 0 ], [ 0, 0, 0, 1, 0 ], 
 [ 0, 0, 0, 0, 1 ], [ 1, 0, 0, 0, 0 ] ], 
 [ [ 1, 0, 0, 0, 0 ], [ 0, 1, 0, 0, 0 ], [ 0, 0, 0, 1, 0 ], 
 [ 0, 0, 0, 0, 1 ], [ 0, 0, 1, 0, 0 ] ] ] 
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style="background-color: rgb(255, 255, 255); vertical-align: top;">The
following commands show that: 
 <ul>
 <li>the 6th cohomology of the
alternating group G=A5 with coefficients in its
5-dimensional canonical permutation module A is H6(G,A)
= Z2+Z6.</li>
 <li>The 3rd cohomology of the even subgroup B+ of
the 5-string Braid group, again with coefficients in the permutation
module A (considered as a B+-module via the quotient
homomorphism B+ → A5) is H3(B+,A)
= Z2+Z6+Z3.</li>
 </ul>
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style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
Alt5:=AlternatingGroup(5);; 
gap> A:=PermToMatrixGroup(SymmetricGroup(5),5);; 
gap> R:=ResolutionFiniteGroup(Alt5,7);; 
gap> TR:=HomToIntegralModule(R,A);; 
gap> Cohomology(TR,6); 
[ 2, 6 ] 
 
 
gap> D:=[[1,[2,3]],[2,[3,3]],[3,[4,3]]];; 
gap> R:=ResolutionArtinGroup(D,10);; 
gap> Brd5:=R!.group;; Brd5Gens:=GeneratorsOfGroup(Brd5);; 
gap>
ImGens:=[Image(A,(1,2)),Image(A,(2,3)),Image(A,(3,4)),Image(A,(4,5))];; 
gap> B:=GroupHomomorphismByImages(Brd5,Image(A),Brd5Gens,ImGens);; 
gap> EvBrd5:=EvenSubgroup(Brd5);; 
gap> S:=ResolutionSubgroup(R,EvBrd5);; 
gap> TS:=HomToIntegralModule(S,B);; 
gap> Cohomology(TS,3); 
[ 2, 6, 0, 0, 0 ] 
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style="text-align: left; background-color: rgb(255, 255, 255); vertical-align: top;">A
group G can act non-trivially on the integers Z. For example, a
permutation group G can act on Z according to the formula 
 <div style="text-align: center;">g.n= -n if g
is an odd permutation, 
 </div>
 <div style="text-align: center;"> g.n= n
if g is an even permutation. 
 <div style="text-align: left;"> 
The following commands show that, with this twisted action of S6
on Z, we have third twisted integral homology H3(S6,Z)=Z2+Z10
. 
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style="vertical-align: top; background-color: rgb(255, 255, 204);">gap>
G:=SymmetricGroup(6);; 
gap> R:=ResolutionFiniteGroup(G,4);; 
gap> C:=TensorWithTwistedIntegers(R,SignPerm);; 
gap> Homology(C,3); 
[ 2, 10 ] 
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 <td
style="vertical-align: top; background-color: rgb(255, 255, 255);">With
the analogous twisted action of S6 on Z5, the
following commands show that the twelvth homology is H12(S6,Z5)=Z5
. (The calculation relies on the fact that Hn(G,Zp)
is equal to its p-primary part Hn(G,Zp)(p)
.) 
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style="vertical-align: top; background-color: rgb(255, 255, 204);">gap>
G:=SymmetricGroup(6);; 
gap>
P:=SylowSubgroup(G,5);; 
gap> R:=ResolutionFiniteGroup(P,15);; 
gap> F:=function(R);return
TensorWithTwistedIntegersModP(R,5,SignPerm);end;; 
gap> PrimePartDerivedFunctor(G,R,F,12); 
[ 5 ] 
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