Prove that $K = \cap_{g \in G} g^{-1}G_xg$

Question

Relevant definitions:

$$G_x = \{g \in G \mid \rho_g(x) = x\}$$

Question:

Let $\rho: G \to S(X): g \mapsto \rho_g$ be a transitive group action. Let $K := \ker \rho$. Then $K = \cap_{g \in G} g^{-1}G_xg$

My attempt (which I know is wrong, since I didn't use transitivity!):

$\boxed{\supset}$

Let $k \in \cap_{g \in G} g^{-1}G_xg$. Then, for all $g \in G$, there exists $l \in G_x$ such that $k = g^{-1}lg$, such that $\rho_k = \rho_{g^{-1}lg} = 1_X$, meaning that $k \in K$

$\boxed{\subset}$

Let $k \in K$ and $g \in G$. Then, consider the group element $h := gkg^{-1}$. Then, we have $\rho_h(x) = x$, such that $h \in G_x$. Now, it follows that $k = g^{-1}hk \in g^{-1}G_xg$ for any $g \in G$, as desired.

Can someone point out where my mistake is? I am quite sure I made a mistake since I never used transitivity.

Answer 1

You can't conclude that $\rho_k = 1_X$ just from what you've written. However, it follows from the fact the $k = g^{-1} l g$ that $k \in G_{\rho_{g^{-1}}(x)}$. Can you see how to use transitivity now?

The second part looks ok except for a typo (I guess you meant $k = g^{-1}h g \in g^{-1}G_xg$ instead of $k = g^{-1}h k \in g^{-1}G_xg$).