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I've been reading up on the sextonions and was wondering if it would be possible to construct a Cayley table for the split sextonions the same way as one would do so for the split quaternions and split octonions. That is, the first element times itself is 1, the second and third should be -1, and the last three would again be 1.

If the above holds (which I'd imagine so), would it be also possible to produce a 'symmetric sextonion', where the first three elements times themselves result in 1, and the last three result in -1? Would such a group retain the lovely properties of the sextonions, much the same as the quaternions, or would this eliminate the rotation properties of such groups? How would the group's geometric properties be changed by demanding this type of symmetry?

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Let $S$ be a real $6$-dimensional vector space with basis $1,i,j,k,r,s$. We can define the multiplication on this basis as we want and then extend bilinearly to $S$. This yields "sextonians", as you wish, with $i^2=j^2=k^2=-1, r^2 =s^2=1$. The question is which identities we would require form this bilinear product. We know that there is no real normed division algebra of dimension $6$, so necessarily "some" properties will get lost. However, there are many $6$-dimensional commutative algebras (but not necessarily associative), see for example here.

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