# Wagner's theorem about 3-transitive groups of odd degree.

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.

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.
• @ivansvetirilski If there are $k \ge 2$ orbits, then $P_{[x,y]}$ fixes $2$ points. That is where $k \ge 2$ is used. – user760870 Apr 3 at 17:00