Octal Reversals, Alignments

The action of a seven cycle upon its static subgroup permutes the subgroups as if they were also elemets. We must note however that the octal structure upon those subgroups is in their place, reversed. In order to show this we must first define addition upon K4 subgroups. The operation we define is as the complement of the symmetric difference in the octal group. I.e. (A v B)^c.

So [0,a,b,c] v [0,a,d,e] = {b,c,d,e} from the octal which then has its complement [0,a,f,g]. So we simply note that the result is the remaining third subgroup that contains the intersection of the original two subgroups under addition.

In using a sven cycle to hold static one subgroup in the octal we note that if we cycle the subgroups as with a correspondence of unity(not a member of the static subgroup) then with our new definition we can se a "reversal" in the seven cycle of the additive structure.

i a = [b,d,f] VII
ii b = [c,d,g] VI
iii d = [c,e,f] V
iv c = [a,f,g] IV
v f = [b,e,g] III
vi g = [a,d,e] II
vii e = [a,b,c] I

In place here from (i) to (vii) we see that adding (on the left) say (ii) and (iii) from the octal singletons gives (v) wheras adding the corresponding subgroups (II) and (III) (from the right) gives (V) from the right, and not as were it (V) and (VI) which gives (I) and not (III). The result is a reversal of the labels, or what we simply term in place as an "octal reversal".

Then if we alter the octal structure to the left hand with its similarly static subgroup corresponding to "a" we notice;

i a = {c,e,g} I
ii b = {a,e,f} II
iii d = {a,b,g} III
iv c = {b,d,e} IV
v f = {a,c,d} V
vi g = {b,c,f} VI
vii e = {d,f,g} VII

And the left hand octal is in alignment as under a similar definition of addition of K4 subgroups, (A v B)^c.

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