[r-t] (no subject)

[RTF]

"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 well-studied 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."?