Metamath

666 Outside of Christ  (Antichrist)
There are 7 * 6 * 5 / (1 * 2 * 3) = 35 ways to choose a triple of three elements from seven. Then there are seven subgroups of order four in an octal group. So If we restrict ourselves to examining those triples of elements sent to themselves under the frobenius map in GF(8) that are not subgroups, (ie our antichrist "bows") we have to start 35  7 = 28 triples. (removing the seven in our particular reference octal or GF(8)+) We also associate one of four such triples with the unity element.
So given our 28 remaining triples aside from the original octal (which are all different) We may for each choose one "choice of unity " or equivalently "one choice of static subgroup" from a total of four, giving us a total number of 112 combinations of subgrouptriple pairs. However one of these combinations aligns perfectly with our original octal, thereby "flling up the cosets" of the original octals, K4 subgroups.
That leaves us with 111 possible deviations from the original octal. In choosing one subgrouptriple pair from the 112 we fix unity to correspond to one of seven elements in our original octal. Whilst this is self referential, we will justify this at the bottom of the page. (We may only remove one and not seven, because each choice of subgrouptriple pair fixes unity in the octal. The other 111 are essentially refer to different octals, or properly different fields GF(8).) Likewise there is no need to drop from 28*4 to 28*3, since of the 28 bows aligning to the original octal, only one is fixed with one particular choice of unity in the original octal if indeed we choose that one bow, we remove only one of the 28*4 = 112, leaving 111.
We could merely observe that of the 112 possible bows, the reference octal must align to one and only one of them; since whatever the state of the GF(8) underneath these 112 possibilities, one of the 112 must be "correct", so we may subtract one, and consider only the remaining 111.
We have several cases. Depending on the seven cycle used and the choice of unity we can change or shift the octal underneath our induced triplesubgroup pairs (on the "left" and "right" hands respectively) shifting several possible octal groups within which these pairs appear. We find then that in seven symbols holding fixed say [b,d,f] as our subgroup and then with the remaining static elements {c,e,g} (with a=1) We have twelve seven cycles which preserve these pairs.
(a,b,d,g,f,c,e) shares the same octal with (a,b,f,g,d,e,c)
(a,b,d,g,f,e,c) shares the same octal with (a,b,f,g,d,c,e)
(a,b,d,c,f,g,e) shares the same octal with (a,b,f,c,d,e,g)
(a,b,d,c,f,e,g) shares the same octal with (a,b,f,c,d,g,e)
(a,b,d,e,f,g,c) shares the same octal with (a,b,f,e,d,c,g)
(a,b,d,e,f,c,g) shares the same octal with (a,b,f,e,d,g,c)
These are also in rows the 6 octals that contain [0,b,d,f] as a subgroup.
Then these twelve seven cycles are the only ones that hold fixed that subgroup and that bow with that choice of unity. However in pairs, these seven cycles act only on six octals: so we have to divide twelve by two to get a factor of six
We can then associate with our choice of subgrouptriple pairs, twelve seven cycles over six octals that hold static those particular pairs over that octal. Then as we also associate four possible left handed triples (as {c,e,g}) with each possible static subgroup and vice versa, (i.e. we have two seven cycles that hold fixed each static subgrouptriple pair over a given octal, we were not counting "multiplication" to begin with.) we have six octals per static pair, giving;
((28 triples * 4 choices of "fixed bow" with each unity)  1 (corresponding to the reference octal))) * 6 underlying octals = 666 possible triples from all octals.
ie (112  1) * 6 = 666
Thus removing one from 112 to make 111 is only reasonable! We then immediately have the full 666 possible arrangements for the "pale horse" or "Death" spirit of the fourth seal of revelation. This is also the number of the beast (the USA) that was constructed by the same devices of the horses, as well as the number of the "image" to the beast that gives the image its "life"  the intent of the false prophet spirit..
In essence we are not excluding six octals as we would use (672  6), but those octals that may hold a particular subgroup static. In removing the six octals that hold fixed one chosen static subgroup with a particular choice of unity, we either subtract 112 1 = 111 as before multiplication, or 672  6 = 666 as after.
Ignore the original or "starting" octal for one moment, and restrict yourself to considering the structure of the first four seals. We are counting the number of configurations of the cosets to whatever subgroups Christ may be considered to "sit at the right hand" within. (We subtracted these 7 subgroups from the initial 35 since they are all such possiblities in the reference octal.)
Each K4 subgroup is in six octals, and we consider only those subgroupbow pairs in those octals that hold that subgroup static: and we do so considering only one element as unity.
From the perspective of the static (as Christ) subgroup, the cosets of "other or 'anti' christs" are configured in 666 alternate configurations, aside from the one configuration in the original "sun octal" or field that is genuine as the "right hand". Thus "anti" takes its proper meaning "in place of". The meaning of "against" is implicit with the nature of the cosets themselves.
Thus 666 is the number of "antichrist" rather than that of God in GF(8). (We take Christ to correspond to one possible subgroup with one particular unity.)
To see that the six sets of 111 subgrouptriple pairs on those octals are "disjoint" we may note that there are 28 bows per octal, each under a disjoint set of seven cycles. There are four possible choices of unity for each static subgroup whcih generate four disjoint left and octals, giving us our 4*7 = 28 remaining "bows". Back referencing each of these 28 disjoint sets creates 28*4 possible octals in which these separate bows may occur: These 112 cases as octals are not all "disjoint" (there are repeats) but under frobenius only one of those 112 bowsubgroup pairs in them must be static: That gives us 112 cases (Remember there are only 30 octals), rather than a smaller number that represent those underlying octals.
Then 6 possible octals per subgroupbow pair over four choices of unity for each "bow" gives us a total number of 672. We then subtract the original "Christ" subgroup's six octals leaving us 666. That there are different (or even repeats of the 30) octals underneath clouds the issue, but the fact that the possible seven cycles over them induce with frobenius 112 distinct states over six octals shows they are essentially different cases
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