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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.