# Proof that the subalgebra generated by any two elements of $\mathbb{O}$ is isomorphic to $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$

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.

• Can you prove that a subalgebra of $\mathbb H$ generated by any non-real element is isomorphic to $\mathbb C$? – Grigory M May 18 '14 at 14:22
• 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 $\mathbb{C}$ . – stefan-niculae May 18 '14 at 14:31
• 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... – Grigory M May 18 '14 at 16:41
• ...that was the plan I had in mind — but actually, proving 'diassociativity' seems to be not that easy, sorry. – Grigory M May 18 '14 at 16:44
• No worries, I understood the idea. I have also found the formal proof using Artin and Frobenius' theorems, thanks for the help. – stefan-niculae May 18 '14 at 17:01

## 1 Answer

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 $\Bbb R$, $\Bbb C$, $\Bbb H$.

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