Metamath

The 144 Octal Reversals
Taking an octal reversal as follows;
a = [b,d,f]
b = [c,d,g]
d = [c,e,f]
c = [a,f,g]
f = [b,e,g]
g = [a,d,e]
e = [a,b,c]
We see immediately that the corresponding missing elements (in each row) we could add form the alternate octal for which the seven cycle also preserves subgroups.
now, {a,b,c} = [[b,d,f],[c,d,g],[a,f,g]] = ? (not a group) we must remember we have a reversal!
So {c,e,g} = [a,b,c],[a,d,e],[a,f,g] = a since a=1
{d,f,g} = [[a,d,e],[b,e,g],[c,e,f]] = e
{b,c,f} = [[a,f,g],[b,e,g],[c,d,g]] = g
{a,c,d} = [[a,f,g],[b,d,f],[c,e,f]] = f
{b,d,e} = [[a,b,c],[c,e,f],[c,d,g]] = c
{a,b,g} = [[a,d,e],[b,d,f],[c,d,g]] = d
{a,e,f} = [[a,b,c],[b,e,g],[b,d,f]] = b
{c,e,g} = [[a,b,c],[a,d,e],[a,f,g]] = a
So traversing or translating between left and right hands does indeed preserve the octal group! It also preserves the seven cycle. Not only are we looking at an apparently different and reversed octal group, it is actually clear that the seven cycle is the same. So, we actually do not have a different octal per se, because taking intersections also preserves the inversion of the octal structure: we simply havent used that fact in out notation yet!
Were we to consider the intersection (the rightmost column above) as actually mapping to the multiplicative inverse of the cycle at that point, we would actually have a a product equating 'b' on the left hand to 'e' on the right and vice versa, with "a = 1" in that case. There are elements that do not provide a reversal  we will examine these later..
if a = 1 = [b,d,f] and our seven cycle (a,b,d,c,f,g,e) is X
then a*X = X but X*X^{1} = 1 = a : which is why the elements appear to be reversed on their inverses.
For applying X twice sends 'a' to 'd', but' 'g' *X*X = 1 which appears to be the element reversed, ( X^{1} corresponds to 'e' in reality) because we have multipled through the set. In reality, taking the intersection of three groups in the ultrafilter actually gives us back a structural reversal and aligns the elements the correct way round.
As previously, an octal has eight such groups of seven cycles; two each preserving Klein four subgroup correspondences to single elements (static group to unity). One singleton corresponds in each cycle with a choice of unity and each C7 group provides three automorphisms of GF(8) with that choice. (Providing 8 x 7 x 3 = 168.) However, six of those seven choices of unity elements are relabelings of that one GF(8)  they are obtained without holding the unity element "fixed" which produces 8 x 3 x 6= 24 x 6 = 144 elements.
The floating unity provides a cyclic group of order seven under these relabelings with the 24 states acting in the identity position in the right hand with no change on the unity element. The other (reversal of) structures in the right hand that map the right hand unity to another element, correspond to the 144 relabellings of the field GF(8) with those 24 states found through the multiplication on the field, and the identity is more or less arbitrary up to isomorphism, but as unity is not fixed, it is not a "true" isomorphism.
There are three isomorphisms of course of this new set, but I have yet to find any extra significance in the operation of addition or multiplication  the identities of the left and right hands, the octal and zero also serve their purpose across each C7 group, though I have no significant operation across C7 groups other than the mystery of the seven lampstands in revelation:  each serves to keep their operations intact individually.
The threefiold application of the frobenius map cycles over each static subgroup (corresponding to a fixed unity say a = 1) and cycle just three elements as [b,d,f] , or also subgroups that correspond to one static triple in the left hand: as [a,b,c],[a,d,e],[a,f,g] to {c,e,g}.
Originally in the right hand we had a = 1 say, and frobenius sent a K4 group through [a,b,c]=>[a,d,e],=>[a,f,g] So we have actually to some extent recovered a cyclic group of order three spanning the whole of the Holy trinity! If we can find a simple correspondence between a single triple and the elements of the Octal, we should be finished, indeed we are already: Yet there is as yet no direct function that does the job.
We start with the whole octal and can map down to three subgroup elements that span it, then again to one element by intersection: and then we have through similarity a congruence between 3 (as automorphisms of GF(8)) of our 24 "elders" and the whole octal group.We may start the cycle again.
We should also note that there must be automorphisms: but that we must also hold our field fixed. For the identity of our additive group (when we deal with three K4 subgroups spanning the octal) we use the octal group, for three elements in K4 we may use a K4 group itself. For unity only, we are back to a left/right correspondence between the octal and zero for additive identity. It turns out that the identity needed is always the group being used: or zero under definition as such: but the operation for the complement of the symmetric difference within the subgroups works fine. In this manner we may place in the K4 form from the last page
0 = [0,a,b,c]
1 = a = [a,b,c]
b = [a,d,e]
c = [a,f,g]
where [0,a,b,c] is an identity for (A v B )^{c} upon the subgroups in {[0,a],[0,b],[0,c]}
and [0,a,b,c,d,e,f,g] is an appropriate identity for products on elements from {[a,b,c],[a,d,e],[a,f,g]} or even the whole octal of subgroups.
In this manner, The K4 form is able to govern itself as an identity  to "sit on my throne even as I sat down with my Father on his throne" as it were.
So we have the Holy trinity pretty much mapped out. We shall see precisely how seals, trumpets and vials of wrath fit in to all of this! There are plenty of verses that mention the 24 elders and the 144,000... though on an individual basis I think covering the sabbath rest in the body of Christ as marriage to Christ is full treatment enough... perhaps more threads will draw themselves together later.
Continue To Next Page
Return To Section Start
Return To Previous Page 