Claim: The transitive group $G$ is primitive on $A$ if and only if $G_a$ is a maximal group of $G$ for all $a \in A$, where $G_a$ is the stabilizer of $a$.
I've been trying to prove a stronger claim. Concretely, let $P$ be the set of all subgroups of $G$ containing $G_a$, and $Q$ be the set of all blocks of $A$ containing $a$. My claim is that there exists a bijective map $f: P \to Q$.
Proof: Given subgroup $G'$ containing $G_a$, let $f(G')=A = \{ \sigma(a) | \sigma \in G' \}$. $A$ is indeed a block, for if $\sigma \in G'$ then $\sigma(A) = A$ since $G'$ is subgroup. If $\sigma \notin G'$, then $\sigma(A) \cap A = \emptyset$ because otherwise we have $\sigma \sigma_1(a)=\sigma_2(a)$ for some $\sigma_1, \sigma_2 \in G'$, but that means $\sigma_2^{-1}\sigma\sigma_1 \in G_a$ which implies $\sigma \in G'$, a contradiction. Conversely, for each block $B$ containing $a$, let the set $G_B = \{\sigma \in G | \sigma(B) = B\}$. It's clear that $G_B$ is a subgroup containing $G_a$, this shows that $f$ is surjective. Suppose the map $f$ is not injective, say $G_1$ and $G_2$ are distinct subgroups of $G$ containing $G_a$ , let $\sigma \in G_1$ such that $\sigma \notin G_2 $, then since $f(G_1)=f(G_2)$ we have $\sigma(a) = \sigma_1(a)$ for some $\sigma_1 \in G_2$. But this would imply that $\sigma \sigma_1^{-1} \in G_a$ and therefore $\sigma \in G_2$, we reach a contradiction.
I think my proof is right, but I did not use the fact that $G$ acts transitively on $A$. Moreover, in the proof of the original's claim, the authors did use this condition, but the proof is totally different from mine. So I want to ask if the claim still remains true if we remove the condition that $G$ acts transitively on $A$? In other words, is my proof right?
Any help would be appreciated!
Edit: As @Derek Holt pointed out in the comment below, my stronger claim is wrong since the map $f$ is not well-defined yet, in particular, $f^{-1}(\{a,b,c\}) = f^{-1}(\{a,b\})=\{ (a\text{ } b), e\}$ shows that $f$ is not actually well-defined. Given a block $B$ containing $a$, the problem came from my proof is that $G_B = \{\sigma \in G | \sigma(B) = B\}$ is still a subgroup containing $G_a$, but there is no guarantee that $f(G_B) = B$. In fact, it's only know that $f(G_B) \subset B$ (why?). However, by adding the condition that $G$ acts transitively on $A$, my stronger claim will become true since in that case, we must have $B \subset f(G_B)$ and therefore $f(G_B) = B$ as desired. Here is the approach:
Proof: Since $\sigma (B)=B$ for all $\sigma \in G_B$, then by the fact that $a\in B$, we must have $\sigma(a) \in B$ for all $\sigma \in G_B$. This shows that $f(G_B) \subset B$. Let $b \in B$, since $G$ acts transitively on $A$ (here is where we need transitivity), then there is a $\sigma \in G$ sending $a$ to $b$ (i.e $\sigma(a)=b$). But then we have $\sigma(B)$ intersects $B$ and by definition of block, we must have $\sigma(B)=B$ which shows that $\sigma \in G_B$, then $f(G_B)$ contains $b$. Hence we have $f(G_B) = B$, completing the proof.
Best Answer
I believe that your stronger claim is false. Let $G = \langle (a,b) \rangle$ be a subgroup of order $2$ in ${\rm Sym}(A)$ with $A=\{a,b,c\}$.
Then there are two subgroups of $G$ containing $G_a$, namely $G_a$ (the trivial group) and $G$. But there are three blocks containing $A$, namely $\{ a \}$, $\{a,b\}$, and $\{a,b,c\}$. So there is no bijection between the two sets.