# Projectivizing a group: how to go from AGL(n,K) to PGL(n+1,K)


Given $G_{\infty,0}$, how do I find $G$?

I want to generalize the example of $X=\PGL(n+1,K)$ acting on $\mathbb{P}^n(K)$, the set of 1-dimensional subspaces of $K^{n+1}$. The stabilizer $X_\infty$ of a subspace are the image of those matrices of the form $\begin{bmatrix} A & v \\ 0 & c \end{bmatrix}$ where $A \in \GL(n,K)$, $v \in K^n$, and $c \in K^\times$, but since we mod out by scalar matrices we can replace $A$ by $\tfrac1c A$ and $c$ by $1$ to get the stabilizer is the (isomorphic) image of $X_\infty = \AGL(n,K)$, the set of all matrices of the form $\begin{bmatrix} A & v \\0 & 1 \end{bmatrix}$ where $A \in X_{\infty,0} = \GL(n,K)$, $v \in K^n$. Note $X_{\infty,0}$ is the stabilizer of a point in $X_{\infty}$.

I'd like to shrink the group $G_{\infty,0} \leq X_{\infty,0}$ that $A$ is chosen from in the point stabilizer. $A$ will still act irreducibly, but will be smaller. I want $G_{\infty}$ to keep the same structure as an affine group (I think that happens automatically, but am not positive; certainly once $G_{\infty}$ is affine, the point stabilizer of $0$ is of the form described).

For instance, if $G_{\infty,0} = \operatorname{SL}(n,K)$, then what is $G$? I think $G=\operatorname{PSL}(n+1,K)$ only rarely works.

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If I have done the calculation correctly this time, then the point stabilizer of ${\rm PSL}(n+1,K)$ is equal to ${\rm ASL}(n,K)$ if and only if $a^{n+1}=1$ for all $0 \ne a \in K$, which is true if and only if $K$ is finite of order $q$ and $n+1|q-1$. – Derek Holt Mar 4 '14 at 9:16