I am working on the following task:
Let $*$ be a transitive group action of $G$ on $Ω$.
Show: $*$ is faithful if and only if for all $\omega \in Ω$ the
stabiliser $G_ω$ doesn´t have any non-trivial normal subgroup of $G$.
My idea:
Let $*$ be faithful. Assume there exists $\omega \in Ω$ and $N$ non-trivial normal subgroup of $G$ so that $N\subset G_{\omega}$ . Then there exists $n\in N, n\neq e$. It follows $n * \omega=\omega $. Because $*$ is transitive, there exists $g\in G$ so that $g * \omega = \alpha, \alpha \in Ω$ . It follows $$n\alpha=n(g\omega)=gg^{-1}n(g\omega)=g(g^{-1}ng)\omega=gmw=gw=\alpha$$ with $m\in N\subset G_{\omega }$.
is this correct?
I am stucked at the other direction…
Now for all $\omega \in Ω$ the
stabiliser $G_ω$ doesn´t have any non-trivial normal subgroup of $G$.
Assume there exists $g\in G, g\neq e$ so that $g \alpha=\alpha$ for all $\alpha \in Ω$.
Thank you for any help!
Best Answer
The kernel of the action is the intersection of all the stabilizers, which can't be nontrivial, because otherwise it would be a nontrivial subgroup of (all) the stabilizers, normal in $G$. So it is trivial and hence the action is faithful.