An exercise asks to prove that: If $d > 2$, then the projective general linear group $PGL_d(F)$ of dimension $d$ over a field $F$ is 2-transitive but not 3-transitive on the set of points of the projective geometry $PG_{d-1}(F)$. My question is on why the $d > 2$ assumption is necessary.
First, here is a proof of the assertion. The projective points are the 1-dimensional subspaces of $F^d$ and $GL_d(F)$ acts on this set of projective points $\Lambda$. Given a nonzero vector $u=(u_1,\ldots,u_d)$ in $F^d$, let $[u_1,\ldots,u_d]$ denote the projective point containing $u$. The action $\phi: GL_d(F) \rightarrow Sym(\Lambda)$ has kernel the scalar matrices $Z$ and so $PGL_d(F):=GL_d(F)/Z$ acts on $\Lambda$ faithfully. Define $PGL_d(F)$ to be the image $\phi(GL_d(F))$. We show this image is 2-transitive and not 3-transitive.
(i) Let $u,w,x,y \in F^d - \{0\}$ be such that $[u_1,\ldots,u_d] \ne [w_1,\ldots,w_d]$ and $[x_1,\ldots,x_d] \ne [y_1,\ldots,y_d]$. Then $u$ and $w$ are linearly independent, as are $x$ and $y$. Thus, the column vectors $u,w$ can be extended to a matrix $U$ whose columns form a basis for $F^d$, and similarly $x,y$ can be extended to a matrix $X$. Thus, $AU=X$ has a solution $A=XU^{-1}$ in $GL_d(F)$. Hence, $GL_d(F)$ is 2-transitive on the set of projective points $\Lambda$.
(ii) Pick any $u,v,x,y \in F^d - \{0\}$ such that $v \ne cu, y \ne cx, \forall c \in F$. Let $w=u+v$ and $z \ne x+y$. (For eg, in $\mathbb{R}^2 - \{0\}$, take $u,v,w$ to be $e_1,e_2,e_1 + e_2$, and take $x,y$ arbitarily, but take $z = x+y+y \ne x+y$). Then $\nexists A \in GL_d(F)$ that takes $(u,v,w)$ to $(x,y,z)$ because any such $A$ would be linear, and so since $w=u+v$, $A$ must map $w$ to $x+y$, whereas $z \ne x+y$. (So, to prove $GL_d(F)$ is not 3-transitive, take the 3 distinct projective points to be such that the third is a linear combination (in fact a sum) of the first two, but in the image, the 3 distinct projective points are chosen so that the third is not a sum of the first two.) Hence $PGL_d(F)$ is not 3-transitive.
I was looking at where the $d > 2$ condition is used in the proof. Note that in (ii), to be able to choose $z$ to be a nonzero vector other than $x,y$ or $x+y$, we require that $|F^d - \{0 \}| \ge 4$, and this is guaranteed if $d > 2$ (or if $d=2$ but $|F| \ge 3$). Only if $d=2$ AND $|F|=2$ is $|F^d - \{0\}|=3$, and in this case the choice of $w$ and $z$ is uniquely determined to be $w=u+v, z=x+y$, and so $GL_2(2)$ (and hence $PGL_2(2)$) is forced to be 3-transitive also. Now this argument implies that $PGL_2(F)$ is 3-transitive iff $|F| = 2$, but this result contradicts an earlier result I've seen (proved using linear fractional mappings) that $PGL_2(F)$ is sharply 3-transitive (even if $|F| \ge 3$). What am I missing?
Let $\langle \vec{u_1} \rangle = \{ c\vec{u_1}:c \in F \} , \langle \vec{u_2} \rangle, \langle \vec{u_3} \rangle$ be three distinct points in $$PG_{d-1}(F) = \{ \{ c \vec{v} : c \in F \} : \vec{0} \neq \vec{v} \in F^d \}$$ Let $\langle \vec{v_1} \rangle, \langle \vec{v_2} \rangle, \langle \vec{v_3} \rangle$ be three distinct points in $PG_{d-1}(F)$.
Suppose that $\vec{u_3} = \alpha_1 \vec{u_1} + \alpha_2 \vec{u_2}$ and $\vec{v_3} = \beta_1 \vec{v_1} + \beta_2 \vec{v_2}$. If $\alpha_i \neq 0$, then $\langle \alpha_i \vec{u_i} \rangle = \langle \vec{u_i} \rangle$. We cannot have $\alpha_1 = \alpha_2 =0$ lest $\vec{u_3} = \vec{0}$. We cannot have $\alpha_i = 0$ lest $\langle \vec{u_3} \rangle = \langle \vec{u}_{3-i}\rangle$. Hence $\langle \alpha_i \vec{u_i} \rangle = \langle \vec{u_i} \rangle$ and we might as well choose our original $\vec{u}_i$ so that $\alpha_i = 1$.
The same argument lets us choose $\beta_i=1$.
Now both $\{ \vec{u}_1, \vec{u}_2 \}$ and $\{ \vec{v}_1, \vec{v}_2 \}$ are linearly independent (by definition of distinct points in $PG_{d-1}(F)$), so we can find some matrix $A$ in $\operatorname{GL}_d(F)$ with $A\vec{u_i} = \vec{v_i}$ for $i=1,2$. Since $\vec{u_3} = \vec{u_1} + \vec{u_2}$ and $\vec{v_3} = \vec{v_1} + \vec{v_2}$ we get $A\vec{u_3} = A\vec{v_3}$. Hence $\bar A \in \operatorname{PGL}_d(F)$ takes the triples of points $\langle \vec{u_1} \rangle, \langle \vec{u_2} \rangle, \langle \vec{u_3} \rangle$ to the triple of points $\langle \vec{v_1} \rangle, \langle \vec{v_2} \rangle, \langle \vec{v_3} \rangle$.
If $d=2$, then every triple of points is of this form and $\operatorname{PGL}_d(F)$ is 3-transitive in its natural action.
If $d>2$, then replace $\vec{v}_3$ with a vector such that $\{ \vec{v_1}, \vec{v_2}, \vec{v_3} \}$ is linearly independent. As before we can ensure $\alpha_i=1$, but now $\beta_i$ do not exist. We still get that $$A(c \vec{u_3}) = A(c\vec{u_1}+c\vec{u_2}) = cA(\vec{u_1}) + cA(\vec{u_2}) = c\vec{v}_1 + c\vec{v_2},$$ but now this vector is not of the form $d \vec{v_3}$ for any $d \in F$ (unless $c=d=0$) since $\{ \vec{v_1}, \vec{v_2}, \vec{v_3} \}$ is linearly independent.
In other words, if $A(\langle \vec{u_i} \rangle) = \langle \vec{v_i} \rangle$ for $i=1,2$, then (choosing a multiple of $\vec{v_i}$ so that $A\vec{u_i} = \vec{v_i}$ for $i=1,2$) we have $A(\langle \vec{u_3} \rangle) = \langle \vec{v_1}+\vec{v_2} \rangle \neq \langle \vec{v_3} \rangle$, because $\langle \vec{v_1} + \vec{v_2} \rangle \cap \langle \vec{v_3} \rangle = \langle \vec{0}\rangle$. Hence $\operatorname{PGL}_d(F)$ does not act 3-transitively in its natural action if $d>2$.