[rt] (no subject)
RTF
mapc01 at bangor.ac.uk
Sun Sep 26 21:14:06 UTC 2004
"Robin Woolley" <robin at robinw.org.uk> writes:
> Consider the first four changes of Cambridge suitably labelled:
>
> A 12345678
> B 21436587
> C 12463857
> D 21648375
>
> The basic ratio is one of the pair multiplied by the inverse of the other.
> (The inverse is that permutation resulting in rounds). So:
>
> X(AC) = inv(C).A X(BD) = inv(D).B
Since your convention is that "ca" means "a first, then c", it might
be better to define X by
X(A,C) = C .inv(A)
the reason is that X measures the "distance" between A and C (A is
false against C iff the distance is zero, i.e. the identity
permutation), so we have the picture
A
*>*

C

v
*
and the "vector" going from the end of the line marked A to the end of
the line marked C is "go backwards along A, and then forwards along
C", which is
C.inv(A)
in your notation.
[I'm sure this X has been wellstudied by group theorists, and must
have a "proper name", does anyone know what it is usually called?
Also, is there an action of G on itself that makes X a homomorhpism
from the semidirect product to G?]
Either that or X(AC) means "from C to A" which seems backwards to me!
The group generated by the values of X is the same, [I think I am
reading things right, and that you use the values of X to generate the
'falseness group' for the section?].so this is notational only
Once advantage of looking it at it from this more geoetric point of
view is that it is then completely obvious (just by drawing a picture)
that
X(A.Z,B.Z) = X(A,B)
X(1,A) = A
And then your symmetric section is
A
B=pA
C=qpA
D=pqpA
then using the two rules above we get
X(A,C) = X(A,qpA) = X(1,qp)=qp
X(B,D) = X(pA,pqpA) = X(1,pq)=pq
then, as you note, inv(pq)=inv(q).inv(p)=qp so we only need one of
X(A,C) and X(B,D) to generate the falseness group.
> Inverting the place notation, e.g., 38x38 for x38x gives C = pqA, so
> X'(AC) = inv(C).A = inv(A).qpA = inv(inv(A)pqA)
> = inv(X(AC))
> which is the inverse of the original ratio, thus giving the same falseness.
So in the "inverted section" we've just swapped p and q? in that case
the new X values will be X'(A,C)=pq, and we've already shown that this
is the inverse of the original X(A,C)
> Finally, an asymmetric section has place notation of the form pqr, so D =
> rpqA. X(AC) is only one of eight ratios giving the complete falseness.
> X(AC).J (where J = 12436587) is another.
I didnt really follow this... most likely I'm missing the obvious as
usual!
> We look at X(AC)= X(BD).J:
>
[so this is the definition of J?]
> X(AC) = inv(A).pqA = X(BD).J = inv(A)pqr.pA.J
> => pq = pqrpJ (on letting A = e (rounds))
> => e = rpJ => rp = J
Here you meant to type "e = rpJ => pr = J" I suppose, but I don't see
how this tells us that "An asymmetric section may also have just one
falseness group."?
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