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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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up vote 2 down vote accepted

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.

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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») –  Mariano Suárez-Alvarez Feb 10 '12 at 4:46
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