Let $G$ be a finite group acting on some set $\Omega$ with $|\Omega|$ even. Let $S \in \mbox{Syl}_2(G)$. Further let $Q \le S$ such that $|S : Q| = 2$ and suppose we have some $x \in S \setminus Q$. Suppose that $x$ interchanges two points $\alpha$ and $\beta$, and that $S_{\omega} = 1$ for $\omega\notin \{\alpha,\beta\}$.

Now look what $X := \langle x \rangle$ does on $\Omega\setminus\{\alpha,\beta\}$. Either $|S| = |X|$, or $|X| < |S|$ and the orbits of $X$ have size $|X|$. In the first case $S$ is cyclic, and hence by Burnside's p-complement theorem we have some normal $2$-complement, and so we can find $N$ with $|G : N| = 2$. If $|X| < |S|$ then the $X$-orbits partition each $S$-orbit in an even number of $X$-orbit of $2$-power size with at least two different orbits, hence if we look at $\Omega\setminus \{\alpha,\beta\}$ we find that we got an even number of $X$-orbits of size $|X|$ (= $2$-power length). By supposition $x$ interchanges $\alpha$ and $\beta$, and hence $x$ induces an odd permutation on $\Omega$ (as the $X$-orbits are cycles who are odd permutations, but we got an even number of of them, so on $\Omega \setminus \{\alpha, \beta\}$ $x$ induces an even permutation). Hence $G$ is not contained in the alternating group on $\Omega$, and if we denote by $N$ the intersection of $G$ with the alternating group, then $|G : N| = 2$.

In either case we can find $N$ such that $|G : N| = 2$ and by construction $G = NX$.

Now my question:

Why is every element from $S \setminus Q$ not in $N$?

For this suppose $y \in S \setminus Q$. Then with $Y := \langle y \rangle$ we can do a similar analysis. If $S = Y$ then of course we must have $y \notin N$ for otherwise $N$ would contain a full Sylow $2$-subgroup and hence could not have index $2$. So suppose $S \ne Y$, with a similar analysis as for $x$ we find that $y$ induces an odd permutation on $\Omega$. So if $S \ne X$ we have that $N = G \cap A_{\Omega}$, and so $y \notin N$. But what if $S = X$, here I do not see why $y \notin Y$?

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    $\begingroup$ If $S = X$ then $S=Y$ so that case does not occur. $\endgroup$ – Derek Holt Dec 4 '15 at 12:33
  • $\begingroup$ Ah okay I see as $S$ is cyclic we would have $Y \le Q$ if $S \ne Y$, as the subgroup $Q$ is unique in a cyclic group. $\endgroup$ – StefanH Dec 4 '15 at 12:45

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