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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 ...

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$\mathbb{Z_{24}}$ is the only cyclic one - all other groups are non-abelian except $\mathbb{Z_{12}}\times \mathbb{Z_{2}}$, which you noted is not cyclic. Among the remaining ones, $\mathbb{Z_{12}}\times \mathbb{Z_{2}}$ is the only abelian one, so this is out too. If you have already seen the center of a group, you may proceed as follows. Among the ...

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