I am a computer scientist using the geometric (Clifford) algebras $G(n,0)$ over $\mathbb{Z}_3 = \{0,1,-1\}$ to describe distributed computations. My question concerns $G(5,0)$ with generators $\{a,b,c,d,e\}$. It is a (novel) fact that the forms $vw+xyz$ in this algebra (dubbed "TauQuinions") form a 40-member five-fold group whose elements either form quaternion-isomorph triplets or are various versions of $-1$.
The 40 can be reduced to 8 by insisting that the bivector components $\{vw\}$ be chosen from $G(4,0) = \{a,b,c,d\}$ such that $(vw+vx+vy+wx+wy+xy)^2=0$. [This nilpotence may not be necessary, but this is the example I've used in my various calculations.] Then an instance of the tauquinion group is $T = \{ab+cde, ac-bde, ad+bce, -bc-ade, bd-ace, -cd-abe, 1-abcde, -e+abcd\}$, where $1-abcde$ is a $-1$, as is $-e+abcd$ (tho with a twist). Both square to $-1+abcde$, which is idempotent and plays the role of $+1$ in this group. [My calculations were carried out by a custom symbolic algebra system, so if your hand-checking disagrees, recall that $vw=-wv$ and $1+1+1=0$ here.]
I think that the tauquinion group $T$ is isomorphic to $SU(3)$ - the number of elements ($8 = 3^2-1$) is right, as is the quaternion connection - but all the (understandable) descriptions use 3x3 matrices and definitions over their i,j components, which is just not relevant (or relateable) to the tauquinion version of things.
So, I am looking for some other, hopefully straightforward, way to either establish that the tauquinion group is isomorphic to $SU(3)$, or not. A proof is welcome, as are suggestions.
Also, I've just learned that there are two versions of $SU(3)$ around, subC and subF, so if true, which one is $T$?