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Suppose that $G$ is a transitive permutation group and let $R$ be a regular normal subgroup which is isomorphic to the non-abelian group of order $27$ and exponent $3$. Then $G = R \rtimes G_{\alpha}$. Now suppose that $G_{\alpha}$ has an abelian subgroup $A$ of index two in $G_{\alpha}$ and $G_{\alpha}$ has three non-regular orbits, one of size $2$ and the others have size $|A|$.

Then $G_{\alpha}$ is isomorphic to $C_2 \times C_2$, or $D_3$ or $D_4$, where $D_n$ denotes the dihedral group of order $2n$.

Why is this so? I thought it has something to the with how $G_{\alpha}$ acts on $R$. The automorphism group of $R$ could be seen in this post, but this automorphism group has other dihedral subgroups, so that alone does not exclude all the other possibilities?

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StefanH
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From the information on orbit lengths, you can deduce that $|G_\alpha|$ divides $12$. so we need to rule out $|G_\alpha|=12$ and $|A|=6$.

I am only going to give a sketch proof. Suppose that $|A|=6$, so $A=\langle x,y \rangle$ with $o(x)=2$, $o(y)=3$. We know that $y$ fixes exactly $3$ points, so its centralizer in $R$ has order $3$, and we must have $C_R(y)=Z(R)$, and hence $y$ acts nontrivially on $R/Z(R)$.

Now $x$ cannot act trivially on $R/Z(R)$, and since its action centralizes that of $y$, it must act on $R/Z(R)$ by inverting every element. So $C_R(x)=Z(R)$.

Now $y$ normalizes some subgroup of $S$ of $R$ of order $9$ with $Z(R) < S$ and $x$ normalizes every subgroup of $R$ containing $Z(R)$, so $x$ also normalizes $S$. But now, using the same argument as in the action on $R/Z(R)$ it follows that $x$ inverts every element of $S$, which is a contradiction, because it centralizes $Z(R)$.

Incidentally, ${\rm Aut}(R)$ does have subgroups isomorphic to $D_6$, but their orbit lengths on $R$ are $1, 2, 2, 4, 6, 12$.

Derek Holt
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  • Why are you not considering the case of cyclic $A = \langle y \rangle$? And what you mean by "an action centralizes another action"; I just know it in context of elements or groups acting by conjugation on another group, but what should it mean in relation to two actions? – StefanH Feb 01 '16 at 21:31
  • I don't understand your question. As I said, I am considering the case $|A|=6$, and proving that it cannot occur. That at least proves that the order of $A$ has to be $2,3$ or $4$. Strictly speaking, I mean the images of the actions of $x$ and $y$ by conjugation on $R/Z(R)$ (which you could think of as elements of ${\rm GL}(2,3)$). – Derek Holt Feb 01 '16 at 22:13
  • But why you suppose that $A \cong C_2 \times C_3$ and not $A \cong C_6$? – StefanH Feb 01 '16 at 22:38
  • That question doesn't make much sense. – Derek Holt Feb 01 '16 at 23:11
  • Yes, this question was stupid; sorry. – StefanH Feb 02 '16 at 09:25