If G is a primitive group action on A, then $G_a$ is a maximal subgroup of G

If G is a primitive group action on A, then $G_a$ is a maximal subgroup of G

A block is a subset $B \subseteq A$ such that for any $\sigma \in G$, either $\sigma (B) = B$ or $\sigma(B) \cap B = \emptyset$. A transitive group action is called primitive if the only blocks in $A$ are single elements $a \in A$ or $A$ itself.

I'm having trouble showing that the stabilizer of an element $G_a$ is maximal in $G$. Its fairly easy to show that $G_a \le G_B$, but moving beyond that point is hard. I googled and found a site saying that there is a one to one correspendance between blocks countaining $a$ and subgroups containing $G_a$, but I don't see why this is true.

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If $G_a<H<G$, can you show that $Ha$ is a block? –  Thomas Andrews Dec 9 '11 at 19:42

Let $G_a\subseteq H\subseteq G$ be a subgroup of $G$; let $B=\{ ha\mid h\in H\}$. I claim that $B$ is a block.
Indeed, assume that $x\in B\cap\sigma(B)$. Then there exist $h,h'\in H$ such that $ha = \sigma h'a$. Therefore, $h^{-1}\sigma h' a = a$, so $h^{-1}\sigma h'\in G_a$. Since $G_a\subseteq H$, then $h(h^{-1}\sigma h')h'^{-1}\in H$, so $\sigma\in H$. Therefore, $\sigma(B) = B$, since $B$ is invariant under the action of $H$.
Since $B$ is a block, by the primitivity we know that either $B=\{a\}$ or $B=A$. If $B=\{a\}$, then $ha=a$ for all $h\in H$, so $H\subseteq G_a$, hence $H=G_a$.
If $B=A$, then for every $g\in G$ there exists $h\in H$ such that $ga = ha$ (by transitivity of the action); hence $h^{-1}ga = a$, so $h^{-1}g\in G_a\subseteq H$; hence $g\in H$. This proves that $G\subseteq H$, so $H=G$.
In summary, if the action is primitive, $a\in A$, and $H$ is a subgroup of $G$ with $G_a\subseteq H\subseteq G$, then $G_a=H$ or $H=G$. That is, $G_a$ is a maximal subgroup of $G$.