Suppose we know that $K$ is the kernel of $G_{\alpha}$ acting on $\Omega-\alpha$. Is $K$ precisely the kernel if $G$ on $\Omega$.
I feel I have a proof. Could anyone check whether my proof is correct please?
Suppose $k\in K$, then $k$ maps $\beta$ to itself for any $\beta$ in $G_{\alpha}$, then $k$ fixes everything in $\Omega$, so is in the kernel of $G$ acting $\Omega$.
Conversely, if $k$ is in the kernel of $G$ acting on $\Omega$, then $g \in G_{\alpha}$ and fixes everything in $\Omega-\alpha$, so is in $K$. Hence the claim.
Is this correct?
Yeah, this is correct. Clearly if K${}_{\text{1}}$, then K${}_{\text{1}} \leq$ G($\alpha$). And K= K${}_{\text{1}} \cap$ G($\alpha$) = K${}_{\text{1}}$.