Page 9.

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

The result above is quite useful. If we define

E_a = U e_a U^-1,

then it is not difficult to prove that

(E_aU³)(U^-3E_a+1) = E_a+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

A_L[B] = AB;
A_R[B] = BA.

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

U_R^-1U_LG₂U_L^-1U_R

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 = U_R^-1U_LG₂U_L^-1U_R,

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: e0 e1 e2 e3 e4 e5 e6 e7 U V W A UAU*: V(UAU*)V*: W(V(UAU*)V*)W*:

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