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Dear Prof. Hulpke,
Thank you very much! This was very helpful!
Sincerely,
- --
Jeffrey Rolland
<rolla...@uwm.edu>
On 01/25/2010 11:24 PM, Alexander Hulpke wrote:
> Dear Forum,
>
> Jeffrey Rolland asked:
>
>> The semidirect product group G is Q |x S, where Q = Z, the integers
>> (i.e., the free group on 1 generator), P = SL(2,5), and S = P_1*P_2*P_3
>> (three copies of P). (I have successfully coded the groups into GAP.)
>>
>> Now, the matrices A = [[0 1][4 0]] and B = [[0 1][4 1]] (row vectors) in
>> SL(2,5) have the property that \Xi(A) is not equal to B or B^{-1} for
>> any \Xi \in Aut(P). (This result can be proven using the Jordan normal
>> forms of A and B, or verified directly using Maple, e.g., and the fact
>> that Aut(P) is isomorphic to PGL(2,5).)
>>
>> Let A_2 be the copy of A in P_2 and A_3 be the copy of A in P_3, and
>> similarly for B_@ and B_3.
>>
>> An automorphism \phi of S can be defined by setting \phi(X_1) =
>> (A_1A_2)^{-1}X_1(A_1A_2) for X_! in P_1 and \phi(X_2) = X_2 and
>> \phi(X_3) = X_3 for X_2 \in P_2 and X_3 \in P_3.
>>
>> Similarly, an automorphism \psi of S can be defined by setting \psi(X_1)
>> = (B_1B_2)^{-1}X_1(B_1B_2) for X_! in P_1 and \psi(X_2) = X_2 and
>> \psi(X_3) = X_3 for X_2 \in P_2 and X_3 \in P_3.
>>
>> Now, I wish to create a homomorphism \Phi: Q -> Aut(S) by sending
>> \Phi(q) = \phi^q (\phi composed with itself q times) and \Psi: Q -> S by
>> sending \Psi(q) = \psi^q (\psi composed with itself q times).
>
> This is (see below) not yet a full solution, but it might help you further:
>
> Firstly, as you get the FP group anyhow from a routine, the cost of using
> *your* generators A and B is neglegible. (This holds for SL(2,5), clearly for
> other groups there can be a substantial benefit for some generators).
>
> f:=GF(5);
> A:=[[0,1],[4,0]]*One(f);
> B:=[[0,1],[4,1]]*One(f);
> SL25:=Group(A,B);
> # ensure p fgenerators correspond to A,B
> fphom:=IsomorphismFpGroupByGenerators(SL25,[A,B]);
> SLfp:=Image(fphom);
> S:=FreeProduct(SLfp,SLfp,SLfp);
>
> Now the generators of S are simply A1,B1 etc. This makes creating phi and psi
> easy.
>
> gens:=GeneratorsOfGroup(S);
> # now gens[1,2] are the first copy etc.
> A1:=gens[1];
> B1:=gens[2];
> A2:=gens[3];
> B2:=gens[4];
> phi:=GroupHomomorphismByImagesNC(S,S,gens,
> [(A1*A2)^-1*gens[1]*A1*A2,
> (A1*A2)^-1*gens[2]*A1*A2,
> gens[3],gens[4],gens[5],gens[6]]);
> SetIsBijective(phi,true);
> psi:=GroupHomomorphismByImagesNC(S,S,gens,
> [(B1*B2)^-1*gens[1]*B1*B2,
> (B1*B2)^-1*gens[2]*B1*B2,
> gens[3],gens[4],gens[5],gens[6]]);
> SetIsBijective(psi,true);
>
> Here I'm using `GroupHomomorphismByImagesNC' simply to have the system trust
> me that this is really a homomorphism. (Otherwise it would have to check
> whether the generator images fulfill the relators, which would force it into
> solving the word problem for S, something that is not yet implemented.
> Similarly I'm telling the system that both maps really *are* automorphisms.
>
> Constructing the homomorphisms from Q now is easy:
> Q:=FreeGroup(1);
> phihom:=GroupHomomorphismByImages(Q,Group(phi),[Q.1],[phi]);
> psihom:=GroupHomomorphismByImages(Q,Group(psi),[Q.1],[psi]);
>
> Well, now for the bad news: There currently is no method for
> `SemidirectProduct' for fp groups. This is not a fundamental obstacle, so far
> just nobody wanted to construct them. Indeed, it is rather easy to write down
> a presentation by taking presentations for Q and S and simply adding the
> appropriate conjugation relations. Let me know if you need help with this.
>
> Best wishes,
>
> Alexander Hulpke
>
> -- Colorado State University, Department of Mathematics,
> Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA
> email: hul...@math.colostate.edu, Phone: ++1-970-4914288
> http://www.math.colostate.edu/~hulpke
>
>
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