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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.

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    $\begingroup$ 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... $\endgroup$ – Will Sawin Aug 5 at 19:37
  • $\begingroup$ @WillSawin Thanks, I forgot about that. For $q=2$ and $q=3$ it is indeed a group according to GAP. $\endgroup$ – Mare Aug 5 at 19:51
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    $\begingroup$ 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 $x_i\in \mathbb{F}_q$. $\endgroup$ – GreginGre Aug 5 at 20:36
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    $\begingroup$ Just to add that GAP has a bug in dealing with the non-associativity here. This will be fixed. $\endgroup$ – ahulpke Aug 19 at 16:06
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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.

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  • $\begingroup$ 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. $\endgroup$ – pregunton Aug 6 at 7:14
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    $\begingroup$ The article of Liebeck: The classification of finite simple Moufang loops (MSN). $\endgroup$ – LSpice Aug 6 at 15:18
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    $\begingroup$ @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. $\endgroup$ – LSpice Aug 6 at 15:21
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    $\begingroup$ 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 $q$. $\endgroup$ – Gjergji Zaimi Aug 6 at 15:59
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    $\begingroup$ @LSpice using the command StructureDescription(U) seems to calculate forever (even though U has only order 120 for q=2). So the developers also might forgot about the fact the units do not need to be associative. $\endgroup$ – Mare Aug 6 at 16:27

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