Metamath

Left Handed And Right Handed  The Father's Throne
We have seen how under the frobenius map three of the Klein four groups form a cyclic group of order three, however it is also true that one of the Klein four groups is held static! For, if our Klein four groups in the octal are as follows,
[a,b,c],[a,d,e],[a,f,g],[b,d,f],[b,e,g,],[c,d,g],[c,e,f]
Then given a seven cycle transforming these subgroups (a,b,d,c,f,g,e) under the frobenius map when we treat a=1 in the Octal group (of the Galois field) holds the group [b,d,f] fixed under the frobenius map for that seven cycle. There is also another such cycle (a,b,f,c,d,e,g) that does the same from a separate group of seven cycles that transforms the Klein four groups similarly.
Thus, since we hold fixed the unity element "a" we may also think of this group [b,d,f] as unity for the Klein four groups.
For each element "a,b or c" etc. as our unity we can associate four groups that have the property that they do not themselves contain the unity element. (If a=1 then [b,d,f],[b,e,g],[c,e,f],[c,d,g] are candidates to be held static.) Indeed this must be so, since the groups that contain unity as subgroups of the right hand in GF(8) would form a cyclic group of order three under frobenius, yet one element in them is unity and fixed, leaving the other two elements in the "would be static group" to permute to each other  impossible in a three cycle and therefore the association to the unity they contain would make no sense, since that group could not be held static at all!
so for instance for the element "a" as unity we have the two cycles above as well as the following extra six cycles.
(a,b,e,c,g,f,d) holding [b,e,g] fixed
(a,b,g,c,e,d,f)
holding [b,e,g] fixed
(a,c,d,b,g,f,e) holding [c,d,g] fixed
(a,c,g,b,d,e,f) holding [c,d,g] fixed
(a,c,e,b,f,g,d) holding [c,e,f] fixed
(a,c,f,b,e,d,g) holding [c,e,f] fixed
We also have the three frobenius automorphisms that confirm that each of these seven cycles (which are all from the 8 different groups that transform the K4 groups properly) for example, for (a,b,f,c,d,e,g) we also have;
(a,f,d,g,b,c,e) and...
(a,d,b,e,f,g,c) from the same C7 group.
So, we have a total of 8 groups with three sevencycles (automorphisms) each for every element considered an identity, giving a total of 24 arrangements. Normal multiplication upon Klein four groups is preserved with the seven cycles as per usual and the method of addition of taking complements of the groups symmetric differences in the octal still holds for a suitable labelling of elements.
Permutation multiplication also has the property that once a unity has been chosen, it is a simple enough task to generate a static group for any unity chosen from the seven elements (effectively starting each seven cycle with any choice of element as unity) so a permutation of the field GF(8) under multiplication indeed also permutes the Klein four groups accordingly but in such a way as to not preserve the association of unity between elements and static groups.  Only the frobenius map has that property, not multiplication. This is an important fact. (Multiplication by inspection permutes the identity to the element of the product x*1 = x etc.)
(a,b,f,c,d,e,g) which holds [b,d,f] fixed may just as easily be equivalent to
(b,f,c,d,e,g,a)  where we choose the element b to be unity and [c,e,f] is held static! However, this is the very same element from the seven cycle, (simply reordered) and it is only a difference of notation being used. This seems very strange indeed! In fact it is a difference of notation and this example shows the uniqueness up to isomorphism of GF(8)  In fact all finite fields are so unique (up to isomorphism)
when taken as we have thus far, as purely "right handed".
Normally under multiplication we would "skip" elements in the cycle used for the product in higher powers, (b^{2} takes an ordered set of every second element etc), so if a=1 multiplying by b is equivalent to the above (b,f,c,d,e,g,a) yet this element is itself clearly not the identity, and if a=1 this product is equivalent to "b", as 1*b=b. Multiplying by b^{2} clearly has a different result. We note we can associate the elements in the K4 subgroups under the same cycle between the right handed GF(8)* structure and the right handed K4 subgroups easily enough. For a=1 and (a,b,f,c,d,e,g)...
a = [b,d,f] = 1
b = [c,e,f]
f = [c,d,g]
c = [a,d,e]
d = [b,e,g]
e = [a,f,g]
g = [a,b,c]
So multiplying by b if a=1 would actually make g = 1 since gb=a=1 and would hold static [a,b,c] and {f,d,e} so then we must be careful to understand this when we appreciate our notation. As {f,d,e} is not a subgroup, multiplication preserves (as held static under frobenius) static groups up to relabelling of their elements, without fixing the elements in the static group (they are cycled), with the condition of frobenius that multiplication is to the first, second and fourth power.
Since we are using a "floating unity" we can choose any of our seven associations to be unity by relabelling, but only one may truly be considered unity in the isomorphic field. As we have eight groups, and three isomorphisms we can happily read off 24 such unity associations for each element as our "1"  I will again tie this in later with the twenty four elders and the four seraphim as well as the 144,000 sealed saints of revelation.
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