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In the wikipedia article http://en.wikipedia.org/wiki/Octonion it is stated that "one can show that the subalgebra generated by any two elements of $\mathbb{O}$ is isomorphic to $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$" as a result of $\mathbb{O}$ being alternative.

I understand that any octonion is a linear combination of the elements in the basis {1, $e_1$, $e_2$, $e_3$, $e_4$, $e_5$, $e_6$, $e_7$} and that:

  • the subalgebra generated by {1} is isomorphic to $\mathbb{R}$

  • the subalgebra generated by {1, $e_i$} for any $i = 1,...,7$ is isomorphic to $\mathbb{C}$
    (the one generated by $e_i$ is also isomorphic to $\mathbb{C}$ because $e_i^2 = -1$)

  • the subalgebra generated by {$1, e_i, e_j, e_k$} is isomorphic to $\mathbb{H}$
    (the one generated by {$e_i, e_j, e_k$}, again because $e_i=e_j=e_k=-1$)

  • the subalgebra generated by {$1, e_i, e_j, e_k, e_l$} is actually $\mathbb{O}$ (because you can obtain the other three elements in the basis by multiplying $e_i, e_j, e_k, e_l$)
    (the one generated by {$1, e_1, ...,e_p$} for $p > 3$ is $\mathbb{O}$,
    as is any subalgebra generated by {$e_1, ...,e_p$}, again because $e_i^2=-1$)

But the article states that the subalgebra generated by ANY two elements is isomorphic to $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$, not $\mathbb{O}$.
Can someone help me understand this, or where I am wrong (preferrably in simple terms)?

Thank you.

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    $\begingroup$ Can you prove that a subalgebra of $\mathbb H$ generated by any non-real element is isomorphic to $\mathbb C$? $\endgroup$ – Grigory M May 18 '14 at 14:22
  • $\begingroup$ Well yes, because $\mathbb{H}$ = <1, i, j, k>. Any non-real element $x$ would satisfy $x^2 = -1$ so we obtained 1. And {1,x} can be seen as {1,i} which generates $\mathbb{C}$ . $\endgroup$ – stefan-niculae May 18 '14 at 14:31
  • $\begingroup$ Indeed. Using a similar argument one can show that any associative division algebra is $\mathbb R$, $\mathbb C$ or $\mathbb H$. So if we can show the well-known fact that a subalgebra of $\mathbb O$ is associative, we're done... $\endgroup$ – Grigory M May 18 '14 at 16:41
  • $\begingroup$ ...that was the plan I had in mind — but actually, proving 'diassociativity' seems to be not that easy, sorry. $\endgroup$ – Grigory M May 18 '14 at 16:44
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    $\begingroup$ No worries, I understood the idea. I have also found the formal proof using Artin and Frobenius' theorems, thanks for the help. $\endgroup$ – stefan-niculae May 18 '14 at 17:01
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That two elements of an alternative algebra generate an associative algebra is Artin's theorem; a proof can be found on page 10 of these notes by Pete Clark. Subsequently the real subalgebra of the octonions generated by any two elements is an associative subalgebra $A$. Let $a\in A$ be a nonzero element. Since the map $a\mapsto ax$ is injective on $\Bbb O$ it is injective on $A$, which is a finite-dimensional real vector space, and as such must also be surjective so that $ab=1$ for some $b\in A$. We then have that $(ba)^2=(ba)(ba)=b(ab)a=ba$ hence $ba$ is idempotent, but nonzero as $\Bbb O$ has no zero divisors, hence $ba=1$ as well and $b$ is a two-sided inverse of $a$. Thus $A$ is a real division algebra, and the Frobenius theorem (proof on Wikipedia) says that the only real division algebras are $\Bbb R$, $\Bbb C$, $\Bbb H$.

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    $\begingroup$ Wow, I'd never heard of this result (Artin's theorem) Thanks for linking this! $\endgroup$ – rschwieb May 19 '14 at 16:40

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