Why is it that a permutation group $G$ on $\Omega$ with regular normal subgroup $K$ is a split extension (internal semidirect product) of $K$ and the point stabiliser $G_\alpha$ for some $\alpha\in\Omega$?
I can see the intersection of the two subgroups is trivial by regularity of $K$. I know there is an action by conjugation of the stabiliser on $K$.
But could you explain why $G$ is $K\cdot G_\alpha$ where $\cdot$ is the product of subsets of a group?
Best Answer
Take an arbitrary element $g\in G$. Since $K$ is transitive, therefore, there is an element $k\in K$ for which $\alpha^g = \alpha^k$. Hence, $\alpha^{k^{-1}g}=\alpha$, that is, $k^{-1}g\in G_\alpha$, so $g\in KG_\alpha$. Since $g$ was arbitary, it follows that $G\leq KG_\alpha$ and, as the reverse inclusion is obvious, we have $G = KG_\alpha$.