Projective Spaces and Groups

We aim to "fix" the following two issues: GL and SL are not 2-transitive because they can't linearly dependent vectors to linearly independent ones. SL is not simple (even though it's perfect) because it has a center. Let $V=V_n(q)$ and $n \ge 2$ throughout.

We define an equivalence relation $R$ on $V^\#$ by $vRw$ iff $v = \lambda w$ for some nonzero $\lambda \in \mathbb F_q$.

Definition We then have the projective space $\mathbb P(V)$ of projective vectors. Write $\mathbb P^{n-1}(q)$.

For a subspace $U \subseteq V$ the set of equivalence classes (projective vectors) $[U]$ (the image of $U^\#$) is a subspace of $\mathbb P(V)$ so it inherits geometric structure. The dimension of $[U]$ is the dimension of $U$ minus 1. A point is a class $[v]$ for some vector $v$, and a line is the projective class of a 2D subspace.

Given $g \in GL(V)$, $v \in V^\#$ and $\lambda \in \mathbb F_q^\#$ we have $(\lambda v)g = \lambda (vg) \in [vg]$ so we can (well) define an action by $[v]g = [vg]$. In this way $GL(V)$ and $SL(V)$ act on $\mathbb P(V)$, but not faithfully.

Lemma Let $G$ be $GL(V)$ or $SL(V)$, the kernel of the action of $G$ on $\mathbb P(V)$ is $Z(G)$.
proof: From the previous post we have seen what $Z(G)$ is: scalar multiples of the identity. If $gs = 1$ then clearly $g$ acts trivially on $\mathbb P(V)$. Conversely if $g \in GL(V)$ is in the kernel of the action then $[vg]=[v]$ for all $[v] \in \mathbb P(V)$, then for every vector $V$ we have $v g = \lambda_v v$ for some $\lambda_v \in \mathbb F^\#$ and the proof concludes in the same way as before.

Definition The projective general linear group and projective special linear group are defined by $PGL(V) = GL(V)/Z(GL(V))$ and $PSL(V) = SL(V)/Z(SL(V))$. They have faithful action on $\mathbb P(V)$. (Note: PSL might not be a subgroup of PGL anymore).

We call $[g] \in PGL$ or $PSL$ the image (???) if $g \in GL$ or $SL$ if $[v][g] = [vg]$ for $v\in \mathbb P(V)$. $PGL_n(q)$ and $PSL_n(q)$ act on $\mathbb P^{n-1}(q)$ by the order calculations in the previous post we find $|PGL_n(q)| = q^{n(n-1)/2}\prod_{i=2}^n(q^i-1)$ and $|PSL_n(q)| = \frac{q^{n(n-1)/2}}{(n,q-1)}\prod_{i=2}^n(q^i-1)$. Thus $|PGL_2(q)| = (q+1)q(q-1)$.

Proposition The permutation group $PGL_2(q)$ is sharply 3-transitive on $\mathbb P^1(q)$.
proof: For $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in GL_2(q)$ and $v \in V_2(q)^\#$, $vg = (a \lambda_1 + c \lambda_2, b \lambda_1 + d \lambda_2)$. These work out exactly as the mobius transformations when regarding $\mathbb P^1(q)$ as $\mathbb F_q \cup \{\infty\}$. We have already shown this one point extension is genrated by the mobius transforms!

Proposition Both $PGL(V)$ and $PSL(V)$ act 2-transitively on $\mathbb P(V)$.
proof: Given $([e_1],[e_2])$, $([e'_1],[e'_2])$ distinct then extend to a basis and find a map between them (why doesn't this work for all n-transitivity?)