# If $G_{\alpha} \cong S_4$ and $|\mbox{fix}(g)| \in \{0,3\}$ for $g \ne 1$. Then $G$ has transitive normal subgroup of index $2$.

Let $G$ be a transitive permutation group such that $|\mbox{fix}(g)| \in \{0,3\}$ for every nontrivial $g \in G$. Also suppose $|N_G(G_{\alpha}) : G_{\alpha}| = 1$, i.e. $G_{\alpha}$ is the only fixed point of all elements in $G_{\alpha}$ (by this assumption there exists some $g \in G$ such that $G_{\alpha} \ne G_{\alpha}^g$ and $G_{\alpha} \cap G_{\alpha}^g \ne 1$). Also note that by the assumption $|\mbox{fix}(g)| = 2$ for every nontrivial $g \in G_{\alpha}$.

Also assume $\operatorname{gcd}(|\Omega|, |G_{\alpha}|) \in \{1,3\}$ and $G_{\alpha} \cong S_4$ and that $G_{\alpha}$ has three non-regular orbits on $\Omega \setminus \{\alpha\}$ of size $12, 8$ and $6$. Consider the following argumentation:

Let $S$ be a Sylow $2$-subgroup of $G_{\alpha}$. Then $\alpha$ is the only fixed point of $S$ and therefore we have $N_G(S) \le G_{\alpha}$. This implies $N_G(S) = S$. By Grün's theorem, $G' \cap S = \langle S', S\cap S'^g : g \in G \rangle$. Now $S$ is a dihedral group of order $8$ and, if not all involutions of $S$ are conjugate in $G$, it follows that $G$ has a normal transitive subgroup $H$ of index $2$.

Why does $G$ has a normal transitive subgroup if not all involutions of $S$ are conjugate in $G$?

These are the things I comprehend: By the assumption $\operatorname{gcd}(|\Omega|, |G_{\alpha}|) \in \{1,3\}$ , every Sylow $2$-subgroup of $G_{\alpha}$ is also a Sylow $2$-subgroup of $G$. Also by looking at the orbit decompositions of $S$ on the orbits of $G_{\alpha}$, as they are divisible by $2$, $S$ could not have a single fixed point. So the case that it has two fixed point must be ruled out, again by orbit decomposition they must both lie in one orbit. Suppose it is $\Delta$ and $\alpha_1, \alpha_2$ are the fixed points $\ne \alpha$, then $S_{\alpha_1} \le G_{\alpha} \cap G_{\alpha_1}$ and $|\Delta| = |G_{\alpha} : G_{\alpha}\cap G_{\alpha_1}|$, but as $|G_{\alpha}\cap G_{\alpha_1}| \ge 8$ and $|G_{\alpha}| = 24$ we have $|\Delta| \le 3$, but the orbits are assumed to be larger. So $S$ has just $\alpha$ has its only fixed point. And $N_{G_{\alpha}}(S) = S$ as the number of Sylow $2$-subgroups is $3 = |G_{\alpha} : N_{G_{\alpha}}(S)|$ in $S_4$.

Also that every normal subgroup of index two must be transitive I see too, as the number of orbits is $|G : G_{\alpha}H|$, which is not divisible by $2$ as $G_{\alpha}$ contains a Sylow $2$-subgroup. So $2 = |G : H| = |G : G_{\alpha}H||G_{\alpha}H : H|$ implies $|G : G_{\alpha}H| = 1$.

• This follows from Grün's Theorem. If not all involutions are conjugate, then Grün's Theorem gives $|G' \cap S| \le 4$. – Derek Holt Jan 14 '16 at 17:52
• @DerekHolt This at least implies $|G : G'| \ge |S : G'\cap S| \ge 2$, so I am asking myself where $H$ comes from? Is $H = G'$ a possible candidate? – StefanH Jan 14 '16 at 18:47
• It tells us that $|G/G'|$ is even. – Derek Holt Jan 15 '16 at 9:16
• I see, as $|S : G' \cap S| = |SG' : G|$, and as $SG'$ is a group, this index divides $|G : G'|$, then we can choose $H$ as an appropriate inverse image in this quotient group. – StefanH Jan 15 '16 at 9:59

## 1 Answer

If $S = \{ 1, r, r^2, r^3, s, sr, sr^2, sr^3 \}$ is the dihedral group of order $8$, then for the involutions we have the conjugacy classes $$\{ r^2 \}, \quad \{ s, sr^2 \}, \quad \{ sr, sr^3 \}.$$ and $S' = \{ 1, r^2 \}$. Now $$\langle r^2, s \rangle = \{ 1, r^2, s, sr^2 \} \cong C_2 \times C_2$$ and $$\langle r^2, sr \rangle = \{ 1, r^2, sr^3, sr^2 \} \cong C_2 \times C_2.$$ but $\langle s, sr \rangle = S$. We have $|S \cap S'^g| = 2$ if $S'^g \le S$ and it has trivial intersection otherwise. Now if not all involutions are conjugate in $G$, then we have the following cases:

i) $S'^g = S'$ for all $g \in G$ with $S'^g \le S$. Hence $\langle S', S\cap S'^g \rangle = S'$.

ii) $S'^g \cap \{ s, sr^2 \} \ne \emptyset$ for some $g \in G$. Then $\langle S', S \cap S'^g \rangle \cong C_2 \times C_2$.

iii) $S'^g \cap \{ sr, sr^3 \} \ne\emptyset$. Similar as ii).

In all cases by Grün's theorem we have $|G' \cap S| \le 4$. Hence as $SG'$ is a subgroup we have $|G : G'| = |G : SG'||SG' : S| = |G : SG'||S : S \cap G'|$ and therefore $|G : G'|$ is divisble by $2$, as it is abelian we can choose some normal subgroup of index $2$ in $G / G'$, its inverse image has index $2$ in $G$ and is normal.

EDIT: Proof that $H$ is transitive. As $\operatorname{gcd}(G_{\alpha}, |\Omega|) \in \{1,3\}$ by assumption and $|\Omega| = |G : G_{\alpha}|$, every Sylow $2$-subgroup of $G_{\alpha}$ is also a Sylow $2$-subgroup of $G$. Now by assumption $S$ is a Sylow $2$-subgroup of $G_{\alpha}$, and by the above also of $G$. Hence $|G : G_{\alpha}H|$ is not divisble by $2$. But as $G_{\alpha}H = G_{\{\Delta\}} = \{ g \in G : \Delta^g = \Delta \}$ for $\alpha \in \Delta$ is the stabilizer of the $H$-orbit $\Delta$, we have $|G : G_{\alpha}H|$ different orbits of $H$. We have $2 = |G : H| = |G : G_{\alpha}H||G_{\alpha}H : H|$, which gives $|G : G_{\alpha}H| = 1$ and hence $H$ is transitive (this by the way gives that every subgroup of index two must be transitive).

• You also need to prove that the normal subgroup of index $2$ is transitive. – Derek Holt Jan 17 '16 at 14:29
• @DerekHolt: Thanks for your comment. I had this mentioned in the question, but maybe it is better to add it to the answer. I added a proof. By the way, if you have the time I would be glad if you look at my other recent questions, like from today: math.stackexchange.com/questions/1615488/… – StefanH Jan 17 '16 at 14:42