Prove that $[N_{G}(G_{\alpha}) : G_{\alpha}] = |\mathrm{fix}(G_{\alpha})|$ when the action of $G$ is transitive.

Question

(For an attempt, see the answers.)

If $G$ acts transitively on $\Omega$, and $\alpha \in \Omega$, show that $$ [N_{G}(G_{\alpha}):G_{\alpha}] = |\mathrm{fix}(G_{\alpha})| $$ where $\mathrm{fix}(G_{\alpha}) = \{\beta \in \Omega \mid \beta^{G_{\alpha}} = \beta\}$.

Recall: that an action is transitive means that for any two different points $\alpha$, $\beta$ there is $x \in G$ such that $\alpha^{x} = \beta$.

Answer 1

Consider $\Gamma:=\{G_{\alpha}x\mid x\in N_{G}(G_{\alpha})\}$. Then we can define \begin{align*} \varphi:&\Gamma\rightarrow \operatorname{fix}(G_{\alpha})\\ &G_{\alpha}x\mapsto \alpha^{G_{\alpha}x}=\alpha^{x} \end{align*} First we check that $\varphi$ is well define: $$ (\alpha^{x})^{G_{\alpha}}=\alpha^{xG_{\alpha}} =\alpha^{G_{\alpha}x}=\alpha^{x} $$ hence $\alpha^{x}\in \operatorname{fix}(G_{\alpha})$. By $$ \alpha^{x}=\alpha^{y}\iff xy^{-1}\in G_{\alpha}\iff G_{\alpha}x=G_{\alpha}y $$ $\varphi$ is a injection. To show that $\varphi$ is a surjection. Consider $\forall \beta\in\operatorname{fix}(G_{\alpha})$, there is $x\in G$, such that $\beta=\alpha^{x}$. Since \begin{align*} &\beta^{G_{\alpha}}=\beta\\ \Rightarrow&\alpha^{xG_{\alpha}}=\alpha^{x}\\ \Rightarrow&xG_{\alpha}x^{-1}\leq G_{\alpha}\\ \Rightarrow&x\in N_{G}(G_{\alpha}) \end{align*} so $\varphi(G_{\alpha}x)=\beta$, $\varphi$ is a surjection hence a bijection.

Answer 2

Here's the way that comes to my mind. Am I correct in assuming you are looking at Ex. 1.6.3 in Dixon & Mortimer's book? Let me know if I use any facts that you are not familiar with.

We want to show that $N_G(G_\alpha)$ acts on $\Omega$ in such a way that $\mbox{fix}(G_\alpha)$ is an orbit under this action, or equivalently, $N_G(G_\alpha)$ acts transitively on $\mbox{fix}(G_\alpha)$. If this is true, then by the Orbit-Stabilizer Theorem, we have that $$|N_G(G_\alpha)| = |\mbox{fix}(G_\alpha)|\cdot |(N_G(G_\alpha))_\alpha| = |\mbox{fix}(G_\alpha)|\cdot |G_\alpha|$$ so the result shall follow as soon as we prove transitivity.

You first want to show that $\mbox{fix}(G_\alpha)$ is actually a union of orbits under the action of $N_G(G_\alpha)$. This follows because of the formula $g^{-1}G_{\Delta}g = G_{\Delta^g}$ where $\Delta \subseteq \Omega$.

To see that the action is transitive, let $\alpha, \beta \in \mbox{fix}(G_\alpha)$, then, by the transitivity of $G$, there is $g \in G$ such that $\alpha^g = \beta$. Observe that $g^{-1}G_{\alpha}g =G_{\alpha^g} = G_\beta$. Convince yourself that $G_\alpha = G_\beta$ since $\alpha, \beta \in \mbox{fix}(G_\alpha)$ and $|G_\alpha| = |G_\beta|$. Now, $g \in N_G(G_\alpha)$, and since $\beta$ was chosen arbitrarily, it follows that $N_G(G_\alpha)$ is transitive on $\mbox{fix}(G_\alpha)$.