That is, suppose that a subset S of the octonions $\mathbb{O}$ is a group under octonionic multiplication. Does it follow that S is contained in the Quaternions $\mathbb{H}$?
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Are you assuming that the subset is multiplicatively closed? If not, exactly what do you mean by "associative"? Do you mean associations of length 3, or of all lengths?
If $S$ is associative, then the $\mathbb R$-linearity of multiplication shows that the vector subspace generated by $S$ is associative (simply break it up into terms and use associativity on each one). If you mean all associations, then that generates an associative subalgebra and you're done. (Being contained in a division algebra is enough to prove Frobenius.)
I'm not sure what to do if only associations of length 3 are allowed. There is probably still a proof of some kind.
answered Feb 10, 2012 at 3:10
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4$\begingroup$ So that this answer really answers the question, one should probably add to observation that the maximal associative subalgebras of $O$ are all isomorphic to $H$ (but they are many such subalgebras isomorphic to $H$ —transitively permuted by the action of $G_2$, iirc— so the statement «$S$ is contained in $H$» needs to be tweaked into «contained up to conjugation») $\endgroup$ Feb 10, 2012 at 4:46