I am attempting to understand the details of a part of this answer to a question on Math Overflow. The assertion is that a point stabiliser in a doubly transitive group of prime degree cannot have a subgroup of index two with nontrivial centre. I've dealt with most of the cases, but am having trouble with this last one.
Let $G$ be an almost simple permutation group of prime degree $p$ with socle $\operatorname{PSL}(d,q)$ (that is, $\operatorname{PSL}(d,q)\leq G\leq\operatorname{Aut}(\operatorname{PSL}(d,q))$), acting on $1$-dimensional subspaces of $\mathbb{F}_{q}^{d}$, the $d$-dimensional vector space over the field $\mathbb{F}_{q}$ with $q$ elements. (Or, on $(d-1)$-dimensional subspaces.) We think of this as a permutation group in which the points are lines (or hyperplanes). We are assuming that the degree $\frac{(q^{d} - 1)}{(q - 1)}$ is the prime $p$, but $q$ may be a power of a different prime. I want to show that a subgroup of index $2$ in a point stabiliser in $G$ has trivial centre.
Can someone please explain why this is true in this case?