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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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Thank you very much for this. But there are some things I don't understand, beginning with why $G$ is contained in $\operatorname{P{\Gamma}L}(d,q)$ (for $d > 2$). Could not the transpose-conjugate automorphism ($a\mapsto (a^{-1})^{T}$) belong to $G$? –  James Mar 16 '12 at 0:36
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No, for $d>2$, the transpose-conjugate automorphisms interchanges $k$ and $(d-k)$-dimensional subspaces of ${\mathbb F}_q^d$ and, in particular, it does not permute the set of 1-dimensional subspaces. When $d=2$, it does permute them, but it induces an inner automorphism of ${\rm PSL}(2,q)$. –  Derek Holt Mar 16 '12 at 8:30
    
I think that one source of my difficulty is that I don't see how the transpose-inverse automorphism (say, $\tau$) acts on $\mathbb{F}_{q}^{d}$. It is, after all, given as an automorphism of $\operatorname{PSL}(d,q)$, and while $\operatorname{P{\Gamma}L}(d,q)$ comes with a natural action on $\mathbb{F}_{q}^{d}$, it is not so clear to me how to extend this action to include $\tau$, or why any such extension should be equivalent to the action of the given group $G$. I do see how $\tau$ interchanges the two conjugacy classes of stabilisers in $\operatorname{PSL}(d,q)$ (for $d > 2$). –  James Mar 16 '12 at 19:29
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I don't believe that it acts in any natural way on ${\mathbb F}_q^d$. In its action on ${\rm PSL}(d,q)$, it interchanges the stabilizer of a $k$-dimensional subspace with the stabilizer of a $(d-k)$-dimensional subspace, and so you get an induced action on the set of all subspaces of ${\mathbb F}_q^d$. But, for the problem in question, all you need to know is that (for $d>2$) it does not normalize the stabilizer in ${\rm PSL}(d,q)$ of a 1-dimensional subspace, and hence it does not normalize ${\rm PSL}(d,q)$ in its 2-transitive action on the set of 1-dimensional subspaces. –  Derek Holt Mar 17 '12 at 14:39
    
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. –  James Mar 18 '12 at 21:07

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