Collineation group as a subgroup of the symmetric group

TL;DR

This paper proves that a certain subgroup of the permutation group of a projective space, containing the projective group and a specific transformation, is actually the entire symmetric group when the space is infinite.

Contribution

It extends known results by showing that under specified conditions, the subgroup must be the full symmetric group for infinite projective spaces.

Findings

For finite spaces, the subgroup contains the alternating group.

For infinite spaces, the subgroup equals the entire symmetric group.

The result generalizes previous finite case findings to infinite cases.

Abstract

Let $\Psi$ be the projectivization (i.e., the set of one-dimensional vector subspaces) of a vector space of dimension $\ge 3$ over a field. Let $H$ be a closed (in the pointwise convergence topology) subgroup of the permutation group $\mathfrak{S}_{\Psi}$ of the set $\Psi$. Suppose that $H$ contains the projective group and an arbitrary self-bijection of $\Psi$ transforming a triple of collinear points to a non-collinear triple. It is well-known from \cite{KantorMcDonough} that if $\Psi$ is finite then $H$ contains the alternating subgroup $\mathfrak{A}_{\Psi}$ of $\mathfrak{S}_{\Psi}$. We show in Theorem \ref{density} below that $H=\mathfrak{S}_{\Psi}$, if $\Psi$ is infinite.