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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?

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In the example you describe, we have ${\rm PSL}(d,q) \le G \le {\rm P \Gamma L}(d,q)$, and the point stabilizer has the structure $N \rtimes X$ with $N$ elementary abelian of order $q^{d-1}$, and ${\rm SL}(d-1,q) \le X \le {\rm \Gamma L}(d-1,q)$, with the natural action of $X$ on $N$, which is faithful and irreducible.

It is not hard to see that any subgroup of index 2 in $N \rtimes X$ is of the form $N \rtimes Y$ with $|X:Y| = 2$, and $Y$ is still faithful and irreducible on $N$. So $N \rtimes Y$ has trivial centre. In most cases, this follows from the fact that $N \rtimes {\rm SL}(d-1,q)$ is perfect and hence must be contained in any subgroup of index 2 in $N \rtimes X$. This is true except when either (i) $d=2$, or (ii) $d=3$ and $q=2$ or 3, and you can prove it directly in those cases.

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    Okay, this is starting to make sense to me now. Meanwhile, I found your subgroup $N$ of order $q^{d-1}$, which clears up the other problem I was having, and so the rest seems to be falling into place. Thank you again for your help.2012-03-18