transitive actions are primitive if and only if stabilizers are maximal subgroups

Theorem 1.

If $G$ is transitive on the set $A$, then $G$ is primitive on $A$ if and only if for each $a\mathrm{\in }A$, $G_a$ is a maximal subgroup of $G$. Here $G_a\mathrm{=}{\mathrm{Stab}}_G\mathit{}\mathrm{(}a\mathrm{)}$ is the stabilizer of $a\mathrm{\in }A$.

Proof.

First claim that if $G$ is transitive on $A$ and $B\subset A$ is a block (http://planetmath.org/BlockSystem) with $a\in B$, then $G_B=\{\sigma \in G\mid \sigma (B)=B\}$ is a subgroup of $G$ containing $G_a$. It is obvious that $G_B$ is a subgroup, since

$$\sigma \in G_B\Rightarrow \sigma (B)=B\Rightarrow {\sigma }^{-1}(\sigma (B))={\sigma }^{-1}(B)\Rightarrow B={\sigma }^{-1}(B)\Rightarrow {\sigma }^{-1}\in G_B$$

$$\sigma ,\tau \in G_B\Rightarrow (\sigma \tau )(B)=\sigma (\tau (B))=\sigma (B)=B\Rightarrow \sigma \tau \in G_B$$

But also, if $\sigma \in G_a$ for $a\in B$, then $\sigma (a)=a$, so $\sigma (B)\cap B\ne \mathrm{\varnothing }$ and thus $\sigma (B)=B$ since $B$ is a block system and thus $\sigma \in G_B$. This proves the claim.

To prove the theorem, note that for each $a\in A$, there is by the claim a $1-1$ correspondence between containing $a$ and subgroups of $G$ containing $G_a$. Thus, $G$ is primitive on $A$ if and only if all blocks are either of size $1$ or equal to $A$, if and only if any group containing $G_a$ is either $G_a$ itself or $G$, if and only if for all $a\in A$, $G_a$ is maximal in $G$.∎