# Unit group of octonions over finite fields

One can define the algebra $$A(K)$$ of octonions over an arbitrary field $$K$$, see for example the command OctaveAlgebra in GAP: https://www.gap-system.org/Manuals/doc/ref/chap62.html . When $$K$$ is a finite field, this is a finite dimensional $$K$$-algebra and thus has finitely many elements. Let $$A_q$$ denote the octonions over a field with $$q$$ elemetns.

Question 1: What is the number of units in $$A_q$$? Can one even describe the (possibly non-associative) group of units up to isomorphism?

For $$q=2$$ the order is 120 and for $$q=3$$ the order is 4320. In both cases it is indeed a group according to GAP.

Question 2 is motivated by Will Sawin's comment (I forgot the unit "group" might not be associative):

Question 2: For which $$q$$ is the unit "group" of $$A_q$$ associative?

It would be interesting to see what the smallest $$q$$ is such that the unit "group" is not associative.

• I guess it should be a (Moufang) loop of units instead of a group, because it's not associative (except maybe in characteristic 2 it is). I don't know how easy those are to describe... – Will Sawin Aug 5 at 19:37
• @WillSawin Thanks, I forgot about that. For $q=2$ and $q=3$ it is indeed a group according to GAP. – Mare Aug 5 at 19:51
• Units are exactly octonions with norm different from $0$, so you need to count the number of solutions of $\sum_{k=1}^8x_i^2\neq 0$, where $x_i\in \mathbb{F}_q$. – GreginGre Aug 5 at 20:36
• Just to add that GAP has a bug in dealing with the non-associativity here. This will be fixed. – ahulpke Aug 19 at 16:06

This is all worked out in the article "A class of simple Moufang loops" by L.J. Paige. The short answer is that the loop of units has size $$q^3(q^4-1)(q-1)$$, and is not associative for any $$q$$. The example given by Paige (lemma 3.5) is given in terms of Zorn vectors as $$\left[\begin{pmatrix} 1 & (0,0,1)\\ (0,0,0) & 1 \\ \end{pmatrix}\begin{pmatrix} 1 & (1,0,0)\\ (0,0,0) & 1 \\ \end{pmatrix}\right]\begin{pmatrix} 0 & (0,1,0)\\ (0,-1,0) & 1 \\ \end{pmatrix}=\begin{pmatrix} 0 & (1,1,1)\\ (-1,-1,1) & 2 \\ \end{pmatrix}$$ and $$\begin{pmatrix} 1 & (0,0,1)\\ (0,0,0) & 1 \\ \end{pmatrix}\left[\begin{pmatrix} 1 & (1,0,0)\\ (0,0,0) & 1 \\ \end{pmatrix}\begin{pmatrix} 0 & (0,1,0)\\ (0,-1,0) & 1 \\ \end{pmatrix}\right]=\begin{pmatrix} 1 & (1,1,1)\\ (-1,0,1) & 1 \\ \end{pmatrix}$$ so these two products cannot be equal over any characteristic. For $$q=2$$ we obtain the smallest simple nonassociative Moufang loop, which has order 120.
The article actually shows that a certain subloop modulo its center is a simple Moufang loop. At the time before Paige's result, the only simple Moufang loops known where the simple groups. Liebeck later proved the converse: Every finite simple Moufang loop that is not a group corresponds to such a subloop of octonions over some $$\mathbb F_q$$. In particular we shouldn't expect a simple classification.
• In fact Paige loops are not associative for $q\in \{2,3\}$ either (despite the claim in the OP), as stated in Theorem 4.1 of the linked article. – pregunton Aug 6 at 7:14
• @Mare, it's a little hard to tell from the documentation whether Units is meant to handle a non-associative ring. The definition of a ring allows it, but then the documentation says that Units always returns a group, so it's not so clear. – LSpice Aug 6 at 15:21
• I edited the answer to correctly reflect that Paige loops are always nonassociative. Some of the lemmas in the paper require being careful in the cases $q\in \{2,3\}$, but the part that shows nonassociativity works over all $q$. – Gjergji Zaimi Aug 6 at 15:59