# 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 $G_a ## 1 Answer Let$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\$.

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 Perfect, thanks. – Carl Dec 9 '11 at 19:55