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I would like help understanding why the statement in bold in the following proof is true.

$\textbf{Thm.}$. A non-trivial normal subgroup of a 3-transitive permutation group of odd degree greater than 3 is 3-transitive.

$\textbf{Proof.}$ Let G be a 3-transitive group of odd degree n>3 acting on a set X and let N be a non-trivial normal subgroup of G. Then N is 2-transitive by a theorem of Jordan; and $N_{[x,y]}$ (the stabiliser of the unordered pair) has $k$ orbits of length $l$ on X\{x,y}, where $kl$=n-2, so that $l$ is odd. Now $N_{(x,a)}$ (the stabiliser of the ordered pair) is a normal subgroup of $N_{[x,y]}$ with index 2, so each $N_{[x,y]}$-orbit splits into at most two $N_{(x,a)}$-orbits of the same size. Since $l$ is odd, no splitting can occur, and these two subgroups have the same orbits on X\{x,y}.

Let P be a Sylow 2-Subgroup of $N_{[x,y]}$. Then P has at least k fixed points, one in each orbit of $N_{[x,y]}$. $\textbf{Since |P| is greater that the 2-part of a 2-point stabiliser, it follows that k=1}$, that is, that N is 3-transitive.

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We have that $N_{(x,y)}$ is normal in $N_{[x,y]}$ of index $2$.

Since $N$ is $2$-transitive, the orders of these groups do not depend on the choice of $x$ or $y$. In particular, a $2$-Sylow $P_{[x,y]}$ of $N_{[x,y]}$ is not contained in $N_{(w,z)}$ for any alternative pair $w,z$, and has order greater than the corresponding $2$-Sylow $P_{[w,z]}$.

OK, so assume that $P_{[x,y]}$ has $k \ge 2$ orbits. Then $P_{[x,y]}$ has $k \ge 2$ fixed points. So there are $2$ points $w$ and $z$ such that every element of $P_{[x,y]}$ fixes both $w$ and $z$. So there is an inclusion

$$P_{[x,y]} \subset P_{(w,z)}$$

and this is a contradiction as above. If $k = 1$, then $P_{[x,y]}$ would necessarily fix only $k = 1$ point $z$ and $P_{[x,y]} \subset P_{z}$ is no contradiction.

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  • $\begingroup$ In the notation I introduced, N_{[x,y]} is the setwise stabiliser, and N_{(x,y)} is the pointwise stabiliser, and the latter is a normal subgroup of the former, not the other way round. What I don't understand is why it follows that there is only one orbit. $\endgroup$ – Sveti Ivan Rilski Apr 3 at 14:06
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    $\begingroup$ @ivansvetirilski If there are $k \ge 2$ orbits, then $P_{[x,y]}$ fixes $2$ points. That is where $k \ge 2$ is used. $\endgroup$ – user760870 Apr 3 at 17:00

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