If $G = G_{\alpha} \rtimes R$ and $R$ is non-abelian of order $27$ and exponent $3$, then $G_{\alpha} \cong C_2\times C_2, D_3$ or $D_4$ 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?
 A: 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 $1, 2, 2, 4, 6, 12$.
