Primitive Permutation Group with Subdegree 3

This is exercise 8B.7 in Isaacs's Finite Group Theory. Let $G$ be a primitive permutation group on $\Omega$, and let $H$ be the stabilizer of $\alpha\in\Omega$. Suppose $H$ has an orbit of size $3$ on $\Omega-\alpha$ (say on the 3 points $\beta$,$\gamma$,$\delta$), and define $D$ to be the stabilizer of $\beta$ in $H$ (so that $D$ is the pointwise stabilizer of $\lbrace\alpha,\beta\rbrace$).

I 'm trying to show $D$ is a 2-group; the hint in FGT is to let $K=core_H(D)$ and show $O^2(D)=O^2(K)$. I have done this. Now I need to show $O^2(K)=1$; most likely I need to show $O^2(K)$ is normal in $G$. So I need some $g\in G-H$ normalizing $O^2(K)$. Probably I can find some $g\in G$ permuting $\lbrace\alpha,\beta,\gamma,\delta\rbrace$, so that $K^g=K$.

So does anyone know how to finish the exercise?

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You have proved that $O^2(D) \lhd H = G_\alpha$, where $D = G_{\alpha,\beta}$. Now $|G_\alpha:D|=3$ and hence also $|G_\beta:D|=3$, so the orbit of $G_\beta$ containing $\alpha$ also has length 3. So we also have $O^2(D) \lhd G_\beta$. But now $N_G(O^2(D)$ properly contains $H$, so either we contradict primitivity of $G$, or we get $O^2(D)$ normal in $G$, and since $D$ has fixed points, this implies $O^2(D)=1$.
that $|G_\alpha|$ divides $3.2^4$.
Yes, thank you, that is very simple. I believe I was so wrapped up in seeing these things as permutation groups, I didn't stop to think of $D$ as an abstract group, with a well-defined $O^2$. – user641 Oct 21 '11 at 14:19