Let $E\_{ij}(x)\in \mathrm{Mat}\_{7\times7}(\overline{\mathbb{F}}\_2)$ be the matrix with zeros everywhere, except for the value $x$ at $ij$. Set $$a(x)=1+E\_{12}(x)+E\_{34}(x)+E\_{56}(x),\quad b(y)=1+E\_{23}(y)+E\_{45}(y)+E\_{67}(y),$$ for two indeterminates $x,y$. Set $$G\_8=\langle a(x), b(y) \mid x, y\in\overline{\mathbb{F}}\_2\rangle.$$ Is $G\_8$ Zariski closed and connected?


For easier visualization, here are the generating matrices:


$$
a(x)=\begin{pmatrix} 
1&x& & & & & \\
 &1&0& & & & \\
 & &1&x& & & \\
 & & &1&0& & \\
 & & & &1&x& \\
 & & & & &1&0\\
 & & & & & &1
\end{pmatrix} 
,\
b(y)=\begin{pmatrix} 
1&0& & & & & \\
 &1&y& & & & \\
 & &1&0& & & \\
 & & &1&y& & \\
 & & & &1&0& \\
 & & & & &1&y\\
 & & & & & &1
\end{pmatrix} 
$$


To put things on a more general footing, given such a description of a subgroup of an algebraic group, how does one show that it is connected?



Is this unipotent group, over characteristic 2, connected?
Good news and bad news here: This group is connected and Zariski closed for nice general reasons. However, I assume that Dror considered this group because $a(1)$ and $b(1)$ generate a subgroup isomorphic to the dihedral group of order $16$, and are thus relevant to [my question](https://mathoverflow.net/questions/262745). It turns out that the group generated by $a(x)$ and $b(y)$ has dimension at least $9$. The issue is that "the $\mathbb{F}\_2$ points of the group generated by $X$" can be much larger than "the group generated by the $\mathbb{F}\_2$ points of $X$".


**Good news:** See Section 2.h of Milne, *Algebraic Groups* for the following results. Let $G$ be an algebraic group scheme over $k$, let $X$ be a geometrically connected, geometrically reduced, subvariety of $G$ over $k$ and let $\phi: X \to G$ be a map defined over $k$ such that $\phi(X)$ is closed under inversion.


Then there is a smallest algebraic sub-group-scheme $H$ of $G$ such that $\phi$ factors through $H$, and this group is connected. Define $\phi\_n : X^n \to G$ by $(x\_1, \ldots, x\_n) \mapsto \phi(x\_1) \cdots \phi(x\_n)$, then $\phi^n : X^n \to H$ is dominant for $n$ sufficiently large.


Furthermore (this part isn't in Milne, but seems right to me), let $n$ be large enough that $\phi^n : X^n \to H$ is dominant. Then I claim that $\phi^{2n} : X(k^{\mathrm{alg}})^{2n} \to H(k^{\mathrm{alg}})$ is surjective. Proof: Let $U$ be an nonempty open set within the image of $\phi^n$. Since $H$ is irreducible, for any $h \in H(k^{\mathrm{alg}})$, the open sets $U$ and $U^{-1} h$ have nonempty intersection, so we can find some point $u\_1 \in H(k^{\mathrm{alg}})$ which is both in $U(k^{\mathrm{alg}})$ and of the form $u\_2^{-1} h$ for $u\_2 \in U(k^{\mathrm{alg}})$, and thus $h = u\_1 u\_2$.


So we can literally think of $H$ as the subgroup of $G$ generated by $X$, as long as we work with points in the algebraic closure.


**Bad news:** Let $G$ be the group of $7 \times 7$ upper triangular matrices with $1$'s on the diagonal. Define
$$G\_k = \{ g \in G : g\_{ij}=0 \ \mbox{for} \ i < j < i+2^k \}.$$
So $G = G\_0 \supset G\_1 \supset G\_2 \supset G\_3 = \{ \mathrm{Id} \}$.
Each $G\_k$ is normal in $G$, and each $G\_k/G\_{k+1}$ is a product of copies of $\mathbb{G}\_a$. Let $H\_k = H \cap G\_k$. I will verify that $\dim H\_0/H\_1 = 2$, $\dim H\_1/H\_2 = 3$ and $\dim H\_2/H\_3 \geq 2$. So the group $H$ has dimension at least $7$. Details follow. Note that we always have $g\_{ij} = g\_{(i+2)(j+2)}$ for any $g \in H$.


**Computation of $H\_0/H\_1$** Every product in $H$ is of the form
$$\begin{bmatrix}
1 & x & \ast & \ast & \ast & \ast & \ast \\
0 & 1 & y & \ast & \ast & \ast & \ast \\
0 & 0 & 1 & x & \ast & \ast & \ast \\
0 & 0 & 0 & 1 & y & \ast & \ast \\
0 & 0 & 0 & 0 & 1 & x & \ast \\
0 & 0 & 0 & 0 & 0 & 1 & y \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}$$
and every $(x,y)$ occurs, so $H\_0/H\_1 \cong \mathbb{G}\_a^2$.


**Computation of $H\_1/H\_2$** Let $h=\prod\_j \left( a(x\_j) b(y\_j) \right) \in H\_1$. Then $\sum x\_i = \sum y\_j=0$. Then $h\_{13} = h\_{35} = h\_{57} = \sum\_{i<j} x\_i y\_j$ and $h\_{24} = h\_{46} = \sum\_{i \geq j} x\_i y\_j$. We moreover have $h\_{13} + h\_{24} = \sum\_{i,j} x\_i y\_j = \left( \sum\_i x\_i \right) \left( \sum\_j y\_j \right) = 0$, so $h\_{13} = h\_{24}$. So we have $h\_{13} = h\_{35} = h\_{57} = h\_{24} = h\_{46}$. Also, $h\_{14} = h\_{36}$ and $h\_{25} = h\_{47}$. So all matrices of $H\_1$ are of the form
$$\begin{bmatrix}
1 & 0 & p & q & \ast & \ast & \ast \\
0 & 1 & 0 & p & r & \ast & \ast \\
0 & 0 & 1 & 0 & p & q & \ast \\
0 & 0 & 0 & 1 & 0 & p & r \\
0 & 0 & 0 & 0 & 1 & 0 & p \\
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}$$
and $H\_1/H\_2$ embeds into $\mathbb{G}\_a^3$. We compute that $a(x) b(y) a(x) b(y)$ has $(p,q,r) = (xy, x^2 y, x y^2)$ and every vector in $(\mathbb{F}\_2^{\mathrm{alg}})^3$ is a sum of vectors of the form $(xy, x^2 y, x y^2)$, so $H\_1/H\_2 \cong \mathbb{G}\_a^3$. 


**Computation of $H\_2/H\_3$** At this point, I already know that $\dim H \geq 5$, so it is not a $4$-dimensional unipotent group. I wasn't able to fully compute $H\_2$. For $h \in H\_2$, we definitely have $h\_{15}=h\_{37}$, so $\dim H\_2 \leq 5$. Every element of $H\_2$ that I have found obeys the additional equation $h\_{15} = h\_{26}$. I can find enough elements to generate the $4$-dimensional subspace cut out by these equations. So $H\_2$ is a subgroup of $\mathbb{G}\_a^5$ which contains at least $\mathbb{G}\_a^4$.


