Is the group of translations of an affine plane always commutative?

Question

$\DeclareMathOperator\Dil{Dil}\DeclareMathOperator\Trans{Trans}\DeclareMathOperator\Col{Col}$An affine plane is a set of points $X$ endowed with a family $\mathcal L$ of subsets of $X$, called lines, such that the following axioms hold:

1. for every distinct points $x,y\in X$ there exists a unique line $L\in\mathcal L$, containing $x,y$;

2. every line $L\in\mathcal L$ contains at least two distinct points;

3. $X$ is not a line;

4. for every line $L\in\mathcal L$ and point $x\in X\setminus L$ there exists a unique line $L_x\in\mathcal L$ such that $x\in L_x$ and $L\cap L_x=\varnothing$.

A bijection $F:X\to X$ of an affine plane $X$ is called

$\bullet$ a collineation of $X$ if for every line $L\in\mathcal L$ the sets $F[L]$ and $F^{-1}[L]$ are lines;

$\bullet$ a dilation if $F$ is a collineation such that for every line $L\in\mathcal L$ we have $F[L]=L$ or $F[L]\cap L=\varnothing$;

$\bullet$ a translation if $F$ is a dilation such that either $\forall x\in X\;F(x)=x$ or $\forall x\in X\;F(x)\ne x$.

Dilations and translations form normal subgroups $\Dil(X)$ and $\Trans(X)$ in the group of collineations $\Col(X)$ of an affine plane $X$.

Question. Is the group of translations $\Trans(X)$ commutative for every affine plane $X$?

Answer 1

Let $G$ be an arbitrary countably-infinite group. I claim that there is an affine plane $X$ such that $\operatorname{Trans}(X)$ contains a subgroup isomorphic to $G$.

By a back-and-forth argument one can show the existence of a function $f:\mathbb{N}\times\mathbb{N}\to G$ which satisfy the following conditions:

1. For any $i_1,i_2\in\mathbb N$ such that $i_1\neq i_2$, the function $\mathbb N\to G$ given by $j\mapsto f(i_1,j)^{-1}f(i_2,j)$ is a bijection
2. For any $j_1,j_2\in\mathbb N$ such that $j_1\neq j_2$, the function $\mathbb N\to G$ given by $i\mapsto f(i,j_1)f(i,j_2)^{-1}$ is a bijection

Now, the points in our affine plane will be $\left\{\left(g, i\right): g\in G, i\in\mathbb N\right\} = G\times \mathbb N$, and the lines would be:

1. $\left\{\left(g, i_0\right):g\in G\right\}$ for a fixed $i_0\in\mathbb N$
2. $\left\{\left(g_0f(i,j_0),i\right):i\in\mathbb N\right\}$ for fixed $g_0\in G$, $j_0\in \mathbb N$

One can see that, given the conditions on $f$, this is an affine plane, and for any $h\in G$ the map $\left(g,i\right)\mapsto\left(hg,i\right)$ is a translation which is nontrivial when $h\neq 1$.

Addition:

If we also assume that $G$ is not commutative then for any choice of $f$ we actually have that $\operatorname{Trans}(X)$ is isomorphic to $G$. It follows from the following lemma: for any affine plane $X$, if $u,v\in\operatorname{Trans}(X)$, $A\in X$, and $A, u(A), v(A)$ are not collinear then $u(v(A)) = v(u(A))$.

To use this lemma for our example, note that a translation of $X$ which does not come from $G$ must move $\left(1,1\right)\in X$ to some $\left(g,i_0\right)\in X$ with $i_0\neq 1$. Therefore the lemma implies that any $u\in\operatorname{Trans}(X)\setminus G$ must commute with any element of $G$. In particular, if $g,h\in G$ then both $u$ and $h\circ u$ must commute with $g$ which implies that $g,h$ are commuting.

To prove the lemma (assuming that $u,v$ are nontrivial), note that the line $\left[A,u(A)\right]$ is parallel both to the line $\left[v(A),v(u(A))\right]$ (because $v$ is a dilation) and also to the line $\left[v(A),u(v(A))\right]$ (because any line of the form $\left[P,u(P)\right]$ is conserved by $u$ (follows from being a dilation) and if two such lines would intersect then the intersection point would be fixed by $u$, contrary to $u$ being a nontrivial translation). Therefore it follows that $v(A), u(v(A)), v(u(A))$ are collinear. Similarly we have that $u(A), u(v(A)), v(u(A))$ are collinear. If $u(v(A))\neq v(u(A))$ then it follows that $u(A), v(A), u(v(A)), v(u(A))$ are collinear, and from above $A$ must also be on that line, so we get that $A, u(A), v(A)$ are collinear.