Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $\operatorname{Stab}_G(a), a\in S$, is a maximal subgroup of $G$.

Question

I am self-studying "Classical Groups and Geometric Algebra" by Larry C. Grove. This is the 2nd question of the exercises of the 0th Chapter.

Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $\operatorname{Stab}_G(a), a\in S$, is a maximal subgroup of $G$.

Unfortunately, I can't even start.

1. $G$ acts transitively on $S$ $\iff$ $\operatorname{Orb}_G(a)=S, \forall a\in S$. So, we can reach any element of $S$ applying $G$ to any element of $S$.
2. If the action is primitive there is no block like $B\subseteq S$, with $|B|\ge 2, B\neq S$, such that for each $x \in G$ either $xB=B$ or $xB\cap B = \emptyset$.
3. $\operatorname{Stab}_G(a)$ is a maximal subgroup if there is no subgroup containing $\operatorname{Stab}_G(a)$ other than $\operatorname{Stab}_G(a)$ itself and $G$.

Answer 1

($\implies$)

Let $H$ is a proper subgroup of $G$ containing $Stab_G(a)$ and $B=\{ha|h\in H\}$ is a subset of $S$. Assume that $xB\cap B \neq \emptyset$ for some $x \in G$. There exist $h,h'\in H$ such that $xha=h'a$. Then, $h'^{-1}xha=a\implies h'^{-1}xh\in Stab_G(a)\implies h'^{-1}xh\in H\implies h'(h'^{-1}xh)h^{-1}=x\in H\implies xB=B\implies B\text{ is a block.}$

$B$ is a block means $B=\{a\}$ or $B=S$ because $G$ acts primitively on $S$.

• $B=\{ha|h\in H\}=\{a\} \implies ha=a, \ \forall h\in H \implies H=Stab_G(a)$.
• $B=\{ha|h\in H\}=S \implies H\text{ acts transitively on }B.$ We also have $G$ acts transitively on $B$. So there exist $g\in G$ and $\bar h\in H$ such that $ga=\bar ha$. Then, $\bar{h}^{-1}ga=a\implies \bar{h}^{-1}g\in H\implies G\subseteq H\implies H=G$.

($\impliedby$)

Let $G$ acts on $S$ imprimitively such that there exists some block $B\subseteq S$ with $a\in B$. Then, $Stab_G(B)=\{g:gb=b, \ g\in G, \ and \ \forall b\in B\}\supset\{g:ga=a, \ g\in G \}=Stab_G(a)$. Let $\alpha, \ \beta\in Stab_G(B)$, and $b\in B$. Then $(\alpha\beta)\cdot b=\alpha\cdot(\beta\cdot b)=\alpha\cdot b=b\implies \alpha\beta\in Stab_G(B).$ Moreover, $\alpha^{-1}\cdot b=\alpha^{-1}\cdot(\alpha\cdot b)=(\alpha^{-1}\alpha)\cdot b=b\implies \alpha^{-1}\in Stab_G(B).$ Thus, $Stab_G(B)$ is a proper subgroup containing $Stab_G(a)$.