Question:
Let $G$ be a finite group acting faithfully and transitively on a set $A$. Assume there exists $N\trianglelefteq G$ such that $N$ acts regularly on $A$. Prove $G_\alpha$ embeds as a subgroup of ${\rm Aut}(N)$, where $\alpha\in A$.
Thoughts:
As $N$ acts regularly on $A$, we have that $G=G_\alpha N$ and $G_\alpha\cap N=1$ for all $\alpha\in A$. Since $G$ acts faithfully and transitively on $A$, we know that there is a bijection, $\phi$, from $G$ to the symmetric group. Since $N\trianglelefteq G$, we have that $G/N$ acts faithfully on $A$ where $(gN)\cdot\alpha=g\cdot\alpha$. From here, I am not sure if I should try and do something with the orbit-stabilizer theorem, or if there is a better approach to sort of tie everything together.
Any help is, as always, greatly appreciated! Thank you.
EDIT: I've seen the solution provided by Jun Koizumi, but I am more interested to see how one could prove the statement given the definition of regular action rather than passing onto the stronger statement.
I will prove a slightly stronger statement:
Theorem. Let $G$ be a group acting faithfully on a set $A$. Assume that there exists a normal subgroup $N\unlhd G$ such that $N$ acts transitively on $A$. Then for each $\alpha \in A$, its stabilizer $G_\alpha$ embeds as a subgroup of $\operatorname{Aut}(N)$.
Proof. Define a homomorphism $\varphi\colon G\to \operatorname{Aut}(N)$ by $$ \varphi(g)=(h\mapsto ghg^{-1}). $$ Let $\alpha\in A$. I will prove that $\varphi|_{G_\alpha}$ is injective. Suppose that $g\in G_\alpha$ is in the kernel of $\varphi$, i.e. $h=ghg^{-1}$ holds for any $h\in N$. Then we have $$ h\cdot \alpha = g\cdot (h\cdot (g^{-1}\cdot \alpha)) = g\cdot(h\cdot \alpha), $$ so $g$ fixes $h\cdot \alpha$ as well. Since the action of $N$ on $A$ is transitive, we get $g\cdot \beta=\beta$ for all $\beta\in A$. Since the action of $G$ on $A$ is faithful, we get $g=1$.