My question is about this nice answer by David Speyer here.
In the answer, he sets up having a subgroup $G\le SL_3(\mathbb{F}_5)$ that fixes the conic $x^2+y^2+z^2=0$. This quadratic is homogeneous, so it has a well-defined zero set in the projective plane over $\mathbb{F}_5$. If I'm interpreting that answer correctly , it's then claimed that the restriction of $G$ to this projectivized conic gives an action on the projective line over $\mathbb{F}_5$, and thus a homomorphism from $G$ to $PGL_2(\mathbb{F}_5)$.
I understand how to get the projective line from the conic. But why does the induced action of $G$ on this conic land it in $PGL_2(\mathbb{F}_5)$? What structure of the conic is $G$ preserving that only $PGL_2(\mathbb{F}_5)$ preserves? It's not like $PGL_2(\mathbb{F}_5)$ is the full symmetry group of the projective line, so I can't see how we end up in that group.
My ultimate goal is to quickly see that we have a homomorphism from $G$ to $PGL_2(\mathbb{F}_5)$. I do know one way to show it, which is to look at the image of $G$ in $SL_3(\mathbb{F}_5)$, and show each element of $G$ when it ends up in $PGL_3(\mathbb{F}_5)$ is a fractional linear transformation on the zero set of $x^2+y^2+z^2=0$. But this seems like a lot, and I'm hoping the brevity of David Speyer's answer suggests there's an easier way to proceed.
As an example of the type of ease I was hoping for, the same answer of David Speyer looks at $G$ in $SL_3(\mathbb{F}_4)$, where it fixes the plane $x+y+z=0$. Well, a linear group that fixes a linear subspace acts linearly on that subspace, so clearly $G$ can be regarded as a subset of $GL_2(\mathbb{F}_4)$.
