Metamath

Octal Reversal and Left/Right Equivalence
Starting with a seven cycle and a corresponding set of subgroups with an appropriate unity we have;
(a,b,d,c,f,g,e) acting upon [b,d,f]=>[c,d,g]=>[c,e,f]=>[a,f,g]=>[b,e,g]=>[a,d,e]=>[a,b,c]
On the right hand we have 24 possible multiplications for a = 1. These corrrespond to all the possible such cycles that correctly transform the octal group. However, the actual additive octal structure is curiously reversed in such a case. For every choice of four groups, corresponding to the four Seraphim (The four beasts in revelation) We have a complete reversal of the octal structure.
What do I mean? well, if 'a' to 'g' correspond to the right hand elements in the singletons and 'A' to 'G' correspond to the right hand subgroups under the same multiplication, then the additive octal structure of subgroups is reversed to that of the singletons.
(a,b,d,c,f,g,e)
[A,B,D,C,F,G,E]
The additive subgroup [a,b,c] corresponds to [E,G,C] and the subgroup [b,d,f] to [G,F,D] etc. The structure of the additive subgroup is reversed by the corresponding cycle. In listing them we see in (#) and (*) the reversal of strucure clearly.
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 can relabel elements and find 7 x 24 = 168 possible labelings that multiply upon the same underlying octal structure  but as before we require our "floating unity" to perform this. If we take away a unity of these seven, then the six remaining elements add up to 144 possible combinations. To repeat, 144. (A significant number in revelation.)
Alternatively on the right hand subgroup approach: If we fix one seven cycle of Klein four groups; for instance:
[b,d,f]=>[c,d,g]=>[c,e,f]=>[a,f,g]=>[b,e,g]=>[a,d,e]=>[a,b,c]
Without any particular preference to any seven cycles that preserve the multiplcation of K4 subgroups we can simply permute the set of elements to find combinations that align in such a way that the subgroups above do not contain their corresponding unity element, we can decisively show that the total number of relabelings of elements in subgroups also show 144 such permutations
However, instead of relabelling elements to preserve all the structure of the underlying octal group  relabeling but keeping isomorphism intact  in this latter case we may relabel the elements to also include different octal groups. There are 5040 possible permutations of seven symbols, 720 possible elements in C7 groups under relabelling, 120 different C7 groups, there are 8 C7 groups over each octal, but 30 possible octal group labelings. Each seven cycle is therefore a transformation on the klein four groups of 2 octals.
When we make K4 groups of subgroups and choose naturally the intersection of three K4 subgroups, we recover the right hand singleton elements, but we don't need to reverse the octal structure all over again! Rather than give God four hands, we take the natural intersection and find that we have not relabeled our elements, they naturally correspond to their singletons. Because the octal structures are reversed again after taking the intersection, and we have used the same seven cycle throughout  we must have obtained not one of our isomorphic relabelings but the very same underlying octal group due to the 2nd reversal. Instead of only going from right hand to right hand through the subgroups, we can go from subgroups to subgroups through singletons equivalently.  See the page "The 144 Octal Reversals".
if a = 1 then we can associate [b,d,f] = a as previously. We can also associate the elements {c,e,g} = 1  the complement triple "bow" is static also.
If we make K4 groups of these "bows" in what is become a "left handed octal" (a,b,d,c,f,g,e) would send;
[c,e,g]=>[a,e,f]=>[a,b,g]=>[b,d,e]=>[a,c,d]=>[b,c,f]=>[d,f,g]
which is a valid but different octal group. It would get very lengthy and complicated to expound further across all 30 octals at this point.
So what of these 144 permutations of symbols for permuting these triples (bows) without introducing new K4 multiplication? Well, each group in the fixed octal structure can correspond to one of four elements only (as unity) from the remaining coset. That there are 144 such symbols does not surprise. We could order the subgroups any way we wished and therefore obtain the same number of combinations. This makes a bijection, alike to traversing from right to right through the left or vice versa which is concrete.
There are four possible symbols for each choice of unity and consequent static triple  but there are 144 arrangements on the octal under multiplication. This would appear to be a simple combinatoric problem rather than one concerning field structure: It would seem that the 24 + 144 on the right hand would allow more on the right than the 144 on the left! Or, rather the 24 are the elements that express the isomorphism  the uniqueness and sovereignty of God  whereas the 144 are relabelings that are isomorphic on the right, and on the left hand there are 144 relabelings in total with regards to sets of elements, that have no real structure unless the right hand seven cycle is taken into account as a multiplication over hat is now, left handed addition on left handed sugroups..
There are again, 144 on the right when relabeling the elements within the octal, and 144 on the left in a relabeling that is not also now coherent with both addition and multiplication. This is not equivalence of left and right, but it shows that there is a similarity between the fields on the right hand and the sets on the other that is not only numerical, but if on the left hand the symbols can be reconfigured to subgroups that align with those on the right hand: then the corresponding bijection between aligned octals through the left and right should appear to be well made.
The combinatoric problem for sets under relabelling would hold were the groups reordered. We can simply see that every relabeling on the right hand is also a consequent relabeling on the left hand for a choice of untiy. We can reorder the groups to match a particular cycle  every right handed relabeling would have to be in this complete set of permutations somewhere. It is easily shown with a simple maple program that there are 144, so the left hand must simply indicate the same 144 wider relabelings of the 24 initial elements for each unity in the right hand.
But what of the 24 elements in the right hand unity not in the left? It would appear that the only identity element corresponds to the right handed elements and is of the seven cycle. The sets do not appear to have an identity element unless unity is "shared": They are not without the ultrafilter of the Holy Trinity however. The bijections of right and left hands for a given unity correspond the seven cycle elements to the left hand: but the Holy Trinity finds identity in a union of both. Or rather, just right hand to right hand in GF(8) on the right, and left to left in GF(8) on the left. There is only an equivalence of left to right if the permuted symbols are reordered appropriately to match a unique cycle with a particular unity. There are far more seven cycles than relabelings: but a bijective union of left and right holds only if the right hand corresponds to the finite field, and the left hand corresponds to cosets in the right rather than groups under specific seven cycles.
Thus we find to sit with God in heavenly places as an elder is to be one of 24, there are also 144 separate isomorphic labelings, and there are 144 such relabelings in the left hand that match these from a total of 5040. It would appear then that there is a seperation of groups and cosets, and to be in the right hand singeltons of the Holy Spirit includes you somehow in the left also. Since only 144 such cycles can preserve the octal addition, we truly see that there is some importance to the right hand in singletons indeed! There must be something very special concerning the 6 x 24 . Ideally we would see the twenty four elders as an identity for both left and right: (Since singletons then naturally correspond to left handed subgroups also!)
We find the answer easily enough  we can choose one of three elements belonging to a K4 subgroup  an alignment rather than a reversal, and for our 8 C7 groups per octal, we have our beloved 24 elements for our identity. Simple enough, but gladly treated seperately to show the difference! Truly Christ is sat at the right hand of God which was our initial metamath construction!
Of the corresponding K4 group to unity, there are three elements in the group: none of which may be the unity element as described before.Do we indeed obtain a reversal?
(a,b,d,c,f,g,e) would correspond to [b,d,f]=>[c,d,g]=>[c,e,f]=>[a,f,g]=>[b,e,g]=>[a,d,e]=>[a,b,c]
We could try to align the subgroups to (b,d,c,f,g,e,a) or (d,c,f,g,e,a,b) or (f,g,e,a,b,d,c) but these are reversals! However;
[c,e,g]=>[a,e,f]=>[a,b,g]=>[b,d,e]=>[a,c,d]=>[b,c,f]=>[d,f,g] aligns perfectly!
Indeed, This cycle of left handed groups is properly aligned with the initial right handed cycle of singeltons and is also one of 144 relabelings! In each case respectively for [b,d,f] = 1 in the right hand we require say, a=1. Therefore these left hand alignments agree completely with the right hand singletons of the seven cycle.
They also share the correspondences with the twenty four automorphisms upon each unity in the right hand, as this has the exact same unity element as the original cycle of right handed subgroups, merely started from a different set.We now may associate singletons directly to "bows" (now made subgroups) which are a consequence of the choice of the right hand static groups,.. we can travel from right to left and vice versa. The two octals that are valid for each seven cycle are clearly after this manner.
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