Discussion:
Parker Vectors, Permutation Groups and Generalized Conjugacy Search
Michael Anshel
2004-08-09 14:40:10 UTC
Raw Message
In P.J.Cameron's, Permutation Groups, LMS 45 CUP (1999) p. 49 the following
problems is posed:

Problems:

(1) What does the Parker Vector of a permutation group G tell us about G.

(2) In particular,which groups or classes of groups are determined by P(G)?

Generally what progress has been made on problems ? In R.A. Parker's
paper "The Computer Calculation of Modular Characters" In Computational Group
Theory ,ed M.D. Atkinson Academic Press (1984) p.272

Parker remarks that if A1 ,B1, generate a representation and A2 ,B2 generate
another of the same dimensions then using the methods described in his
paperthat a simultaneous solution to the equations

X A1 X^(-1)= A2, X B1X^(-1)X

can be computed should one such solution exist.

This same equations arose in another context AAG 1999

I. Anshel,M. Anshel and D. Goldfeld, "An Algebraic Method for
Public-Key Cryptography Mathematical Research Letters 6,1-5,(1999)

and interested readers are invited to contact me regarding AAG 1999.

This researcher would like to thank Professor Alex Ryba for introducing
Parker's ideas and methods in a recent conversation.

Professor Michael Anshel
Department of Computer Sciences R8/206
The City College of New York
New York,New York 10031
http://www-cs.engr.ccny.cuny.edu/~csmma/
***@cs.ccny.cuny.edu
***@aol.com
Michael Amling
2004-08-16 18:03:48 UTC
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Michael Anshel wrote:

< In P.J.Cameron's, Permutation Groups, LMS 45 CUP (1999) p. 49 the following
< problems is posed:
<
< Problems:
<
< (1) What does the Parker Vector of a permutation group G tell us about G.
<
< (2) In particular,which groups or classes of groups are determined by P(G)?
<
< Generally what progress has been made on problems ? In R.A. Parker's
< paper "The Computer Calculation of Modular Characters" In Computational Group
< Theory ,ed M.D. Atkinson Academic Press (1984) p.272
<
< Parker remarks that if A1 ,B1, generate a representation and A2 ,B2 generate
< another of the same dimensions then using the methods described in his
< paperthat a simultaneous solution to the equations
<
< X A1 X^(-1)= A2, X B1X^(-1)X

Are you saying "X B1X^(-1)X" is an equation? What does it equate to?
"X B1"?

< can be computed should one such solution exist.

--Mike Amling