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?


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 $1, 2, 2, 4, 6, 12$.

  • $\begingroup$ 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? $\endgroup$ – StefanH Feb 1 '16 at 21:31
  • $\begingroup$ 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 ${\rm GL}(2,3)$). $\endgroup$ – Derek Holt Feb 1 '16 at 22:13
  • $\begingroup$ But why you suppose that $A \cong C_2 \times C_3$ and not $A \cong C_6$? $\endgroup$ – StefanH Feb 1 '16 at 22:38
  • $\begingroup$ That question doesn't make much sense. $\endgroup$ – Derek Holt Feb 1 '16 at 23:11
  • $\begingroup$ Yes, this question was stupid; sorry. $\endgroup$ – StefanH Feb 2 '16 at 9:25

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