Metamath

Right And Left Hand Invariance
In our K4 subgroups of the Octal we have noted that the structure of the group is preserved under multiplcation by elements of the multiplicative group of GF(8). We then limit ourselves to a simple structure of "floating unity" in the ultrafilter.
In this (left) hand of subgroups of the right hand elements "a through g", we simply note that if G is our K4 group  a subgroup of H our Octal, Then the structure that states G < H is invariant under product with any element of C7. We could label each H and G as H[i] or G[i] (Even G[i] is as permuted to itself under the element of C7 used.) We just need to note that unity (and the static subgroup under frobenius) is also changed under multiplcation by an element of C7.
However we are simply interested in that G < H in product with any element x of C7 leads us to xG < xH in other words. (although x is in H (more accurately GF(8)*) so xH=H).
The element unity in xH was shifted from the element (as before multiplication) that was the inverse of x in C7. We shall look a little deeper into this.
But for right hand invariance we simply put it no better than as in this diagram.
Now multiplicatively our elements "a through g" are not considered as fixed elements of the Octal, and under frobenius we have the elements {x, x^{2}, x^{4}} forming a group C3. Fixing one element as unity immediately permutes the elements of the left and right hands such that a single element (as Christ slain) in the right hand is fixed (unity), and this corresponds to a fixed subgroup in the left hand. The elements of this left hand subgroup correspond through the seven cycle back to those subgroups containing unity in the right hand. (subgroups in the right correspond to elements in the left, and vice versa.)
Now, for the left handed subgroup corresponding to the right handed "b" is {a,e,f} here. We note that the fixed left handed subgroup here, {c,e,g} which is "a" in the right hand, corresponds to the subgroups in the right hand containing 'a'. The sum {c,e,g} + {a,e,f} = {b,d,e} has its analogue when using the right hand operation, a+b = c = {b,d,e}. This is the consequence of the alignment to the singleton elements rather than a reversal as between the left and right hand's subgroups.
Similarly for an element of {c,e,g} say "c" would in the right hand give us a + c = b = {a,e,f} And we see that {c,e,g} + {b,d,e} = {a,e,f} as desired.
Likewise from Frobenius we know that if "c" is now become 'unity' as opposed to "a" then in the right hand the subgroup {a,f,g} is now static as is the set {b,d,e} in the left handed set. And we have the same invariant correspondence of left and right subgroups to elements.
Basically We have that 1 (unity) in C7 is invariant (free to float) under addition in the right handed octal when in alignment using the subgroups of the left hand which results in GF(4)* aligning with the fixed left handed subgroup coresponding to the right hand unity.
So the only remaing case is to see whether the correspondence is correct for "a" added to "a".
Adding {c,e,g} to {c,e,g} in the left hand results in the octal or 0. (as does a +
a = 0 in the right hand.)
Since 0 is actually a member of the octal we also have {c,e,g} + 0 = {c,e,g} and a + 0 = a.
We have nothing "special" here other than to state that addition by any member of the octal (including 0) holds fixed the relationship that (as with frobenius) the elements of C7 {x, x^2, and x^4} are equivalent to those of GF(4)* in the alternate hand's octal.
So if G = GF(4)* and H is C7
Then the relationship G < H is invariant under addition by an element of the octal in a similar respect. ie x+G < x+H. However this is only the case when G is a group in the left hand if H is the right hand octal and vice versa.
It is true that x+G would only be a subgroup if x is in G. However if x is an element from the left handed octal, it corresponds to a singleton in the right as in alignment: and we should realise that aside from the restriction of the notation or symbols used, the result is that x+G is the sum of two subgroups, which will result in a third subgroup or the octal itself.
Now we see that xG
a = [b,d,f] = {c,e,g} = 1 (static)
b = [c,d,g] = {a,e,f}
d = [c,e,f] = {a,b,g}
c = [a,f,g] = {b,d,e}
f = [b,e,g] = {a,c,d}
g = [a,d,e] = {b,c,f}
e = [a,b,c] = {d,f,g}
under multiplication by the seven cycle (a,b,d,c,f,g,e) would lead to an invariance of the relationships,.. there is a simple bijection to the result. (We simply move the row corresponding to unity.)
b = [c,d,g] = {a,e,f}
d = [c,e,f] = {a,b,g}
c = [a,f,g] = {b,d,e}
f = [b,e,g] = {a,c,d}
g = [a,d,e] = {b,c,f}
e = [a,b,c] = {d,f,g}
a = [b,d,f] = {c,e,g} = 1 (static)
Every correspondence is still valid.
Under addition things are a little different.
0 = [a,b,c,d,e,f,g]
a = [b,d,f] = {c,e,g} = 1 (static)
b = [c,d,g] = {a,e,f}
d = [c,e,f] = {a,b,g}
c = [a,f,g] = {b,d,e}
f = [b,e,g] = {a,c,d}
g = [a,d,e] = {b,c,f}
e = [a,b,c] = {d,f,g}
We could add say 'f' to each row.. However we must apply it to the left hand in groups and right hand by singletons, not the right hand subgroups. This is only possible by virtue of the twofold reversals.
0 + f = f = {a,c,d}
a + f = g = {c,e,g} +{a,c,d} = {b,c,f}
b + f = d = {a,e,f} + {a,c,d} = {a,b,g}
d + f = b = {a,b,g} + {a,c,d} = {a,e,f}
c + f = e = {b,d,e} + {a,c,d} = {d,f,g}
f + f = 0 = [a,b,c,d,e,f,g]
g + f = a = {b,c,f} + {a,c,d} = {c,e,g}
e + f = c = {b,d,e} + {a,c,d} = {b,d,e}
The relationships of element to element(s) are the important part: The elements as before align to;
a = [b,d,f] = {c,e,g} = 1 (static)
b = [c,d,g] = {a,e,f}
d = [c,e,f] = {a,b,g}
c = [a,f,g] = {b,d,e}
f = [b,e,g] = {a,c,d}
g = [a,d,e] = {b,c,f}
e = [a,b,c] = {d,f,g}
What is vital to observe are the three elements that form the static group under frobenius are exactly the same before and after since multiplication by the same seven cycle is preserved (under action of itself or addition, a consequence of the distributive law.) if a = 1 before multiplication and 'xy = 1' after multiplication by x, then we only need to observe that xy = a, or if 'x+ y = 1' under addition by x , then 'x + y = a' is the result. The static subgroup is then determined by the seven cycle used. (The structure is invariant, the same static subgroup occurs before and after multiplication or addition by virtue that a = 1 is our free choice of the seven, and the multiplication holds static the same subgroup.)
Where does this correspondence preserving a = 1 come from? The finite fields used are almost always written in polynomials of one variable  permuting them with addition or multiplication shift the elements, but there will always be a unity element "1" since the additive and multiplicative parts of the field are closed groups. One element must be unity, and all we have really done is relabelled the elements in another combination though in a manner that the previous relationships still hold. The structure is related by a bijection: it does not form an isomorphism but there are no new elements, or old elements lost in the process.
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