Let G be a finite group, acting on a set $\Omega$, and let p be the smallest prime number such that $p \biggr||G|$. Let $\alpha \in \Omega$. It's known that $|\operatorname{Orb}_G(\alpha)| = p$. Prove that $\operatorname{Stab}_G(\alpha) \vartriangleleft G$.
What I've done so far: if p is the smallest prime dividing G's order. then the biggest subgroup of G (which isn't G) will be of maximum order of $\frac{|G|}{p}$.
$\forall \beta \in \operatorname{Orb}_G(\alpha),\ \left|\operatorname{Stab}_G(\beta)\right| = \frac{|G|}{|\operatorname{Orb}_G(\beta)|} = \frac{|G|}{|\operatorname{Orb}_G(\alpha)|}=\frac{|G|}{p}$.
For every $g \in G:\ g(\alpha) = \alpha $ we get that $\forall s \in \operatorname{Stab}_G(\alpha),\ g^{-1}sg \in \operatorname{Stab}_G(\alpha)$, and for $\forall g \in G:\ g(\alpha) \neq \alpha$ we get for $\beta := g^{-1}(\alpha) \in \operatorname{Orb}_G(\alpha)$ that $\forall s \in \operatorname{Stab}_G(\alpha),\ g^{-1}sg \in \operatorname{Stab}_G(\beta)$.
This transformation from $\operatorname{Stab}_G(\alpha)$ to $\operatorname{Stab}_G(\beta)$ is injective, and the two subgroups are of the same order, so every member of $\operatorname{Stab}_G(\beta)$, for any $\beta \in \operatorname{Orb}_G(\alpha)$ is of this form.
I'm not sure how to continue from here. I need to show that all the stabiliziers subgroups are equal, or take a completely different approach. I haven't used the fact that this are p (potentially) different subgroups of the maximal possible size, which I believe is somehow relevant.
