Page 9.

(AU3)(U-3B) = U[(U-1AU)(U-1BU)]U-1

The result above is quite useful. If we define

Ea = U ea U-1,

then it is not difficult to prove that

(EaU3)(U-3Ea+1) = Ea+5.

That is,

A --> UAU-1

is an isomorphism from our starting copy of the octonions (D(+)) to the copy altered
by the X-product (AU3)(U-3B).

Ok, now I want to introduce some notation from my book, which, while not common, I find very useful.
For any octonions A and B, define

AL[B] = AB;
AR[B] = BA.

That is, AL is the left multiplication operator, and AR is the right multiplication operator.
Then the isomorphism A --> UAU-1 implies that the group of operators


is the automorphism group of the octonions endowed with the product (AU3)(U-3B). So it too is a copy of G2.
This is a special case of a more general result that is on one of the following pages.

Note that if U is a 6th root of unity, then

G2 = UR-1ULG2UL-1UR,

since in that case the initial product is not altered by the X-product. The calculator below
is the same one that I put on the previous page.

Inputs: e0e1 e2e3 e4e5 e6e7

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