This question is a sequel to [Pointless groups](https://mathoverflow.net/questions/424444/pointless-groups), to which @DanielLitt produced an elegant and easy-to-understand [counter-example](https://mathoverflow.net/a/424453), and [Pointless groups II](https://mathoverflow.net/questions/424456/pointless-groups-ii), where @R.vanDobbendeBruyn [pointed out](https://mathoverflow.net/questions/424456/pointless-groups-ii?noredirect=1#comment1091035_424456) that my implicit claim about Zariski density of rational points of groups over infinite fields is only guaranteed for *reductive* groups (or perfect fields), and gave a [counterexample](https://mathoverflow.net/a/424462) (still elegant, though less easy for me to understand than a split torus over $\mathbb F\_2$!) for an infinite but imperfect field.


At the suggestion of [@R.vanDobbendeBruyn](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091038_424456) and [@YCor](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091103_424456), I am engaging in the extreme bad manners of once again modifying my search for counterexamples to rule out the counterexamples already provided. The revised question is @DanielLitt's [wording](https://mathoverflow.net/questions/424444/pointless-groups#comment1091016_424453).


So, 3rd time's a charm, hopefully … fix a *finite* field $k$. Then "group" means "smooth, connected, affine algebraic group scheme of finite type over $k$".


If $G$ is a (smooth, connected) group, and $G(k)$ is trivial, then does that imply that $k = \mathbb F\_2$ and $G$ is a split torus?



Pointless groups III
Yes.


Let $|k| = q$. Let $T$ be an $n$-dimensional torus defined over $k$ and let $\overline{T}$ be the base change of $T$ to $\overline{k}$. Then the character group $\text{Hom}(\overline{T}, \mathbb{G}\_m)$ is isomorphic to $\mathbb{Z}^n$ and the Frobenius acts on $\text{Hom}(\overline{T}, \mathbb{G}\_m)$ by a matrix $F$. If $T$ splits over $\mathbb{F}\_{q^n}$, then $F$ obeys $F^n = q^n \text{Id}$. The isomorphism class of the torus $T$ is determined by the $\text{GL}\_n(\mathbb{Z})$ conjugacy class of $F$, and we have $\# T(\mathbb{F}\_q) = \det(F - \text{Id})$. So we are reduced to the following linear algebra problem:



> 
> If $F$ is an integer matrix obeying $F^n = q^n \text{Id}$ and $\det(F - \text{Id}) = 1$, can we deduce that $q=2$ and $F = 2 \text{Id}$?
> 
> 
> 


The answer is "yes". The condition that $F^n = q^n \text{Id}$ means that the eigenvalues of $F$ are of the form $q \zeta$ for $\zeta$ a root of unity. So we have $\det(F - \text{Id}) = \prod\_i (q \zeta\_i - 1) = 1$ where the $\zeta\_i$ are various roots of unity.


But $|q \zeta - 1| \geq 1$ for any prime power $q$ and any root of unity $\zeta$, and strictly greater unless $q=2$ and $\zeta=1$, so the only solution is $q=2$ and $F = \text{Id}$. $\square$


