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Dear Prof. Schmidt,
Thank you so much for replying!
I found an instance where I used "@" instead of "2", and I intended to
use A2 and A3 and B2 and B3 (instead of A1 and A2 and B1 and B2 [so that
no element of P1 multiplies A1 or B1]) consistently to define the
semidirect product. I apologize for my poor proofreading; I assure I did
attempt to proofread it several times.
Again, thanks for replying. Alexander Hulpke, in an independent posting,
indicated that 'SemidirectProduct' is not implemented for fp groups, so
my method of attack outlined below would not have worked; thank you for
the presentations of the semidirect products.
Sincerely,
- --
Jeffrey Rolland
<rolla...@uwm.edu>
On 01/23/2010 05:06 PM, Jack Schmidt wrote:
> I have answered your question below, but I think you should know that
> your post does not present you in a favorable light. The post has
> many typos and is much more complicated than it should be. It reads
> as if it was carelessly written.
>
>
> At any rate, a sane way to proceed:
>
> Setting A=[0,1;-1,0] and B=[0,1;-1,0], SL(2,5) is generated by A and
> B subject to the relations:
>
> P = < A, B : A^4, A^2=B^3, (A*B)^5 >
>
> There is no homomorphism taking A to B because the order of A is 4
> and the order of B is 6.
>
> The free product P * P * P is generated by A1, B1, A2, B2, A3, B3
> subject to the relations:
>
> S = < A1, B1, A2, B2, A3, B3 : Ai^4, Ai^2 = Bi^3, (Ai*Bi)^5 >
>
> The semidirect product Q |x S, with Q free on { C } acting as
> conjugation by x =(A1*A2)^q on P1 and fixing P2 and P3, is generated
> by C, A1, B1, A2, B2, A3, B3 subject to the relations:
>
> G = < C, A1, B1, A2, B2, A3, B3 : Ai^4, Ai^2 = Bi^3, (Ai*Bi)^5, A1^C
> = A1^x, B1^C = B1^x, A2^C = A2, B2^C = B2, A3^C = A3, B3^C = B3 >
>
> You can input such a group into GAP as:
>
> f := FreeGroup( "C", "A1", "B1", "A2", "B2", "A3", "B3" );
> AssignGeneratorVariables( f ); q := 3; x := (A1*A2)^q; G := f/[ A1^4,
> A1^2/B1^3, (A1*B1)^5, A2^4, A2^2/B2^3, (A2*B2)^5, A3^4, A3^2/B3^3,
> (A3*B3)^5, A1^C / A1^x, B1^C / B1^x, A2^C / A2, B2^C / B2, A3^C / A3,
> B3^C / B3 ];
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