Let $G$ be a group that acts transitive on $X$. Show that if $N$ is a normal subgroup in $G$ and for a $x\in X$ holds $N\subseteq G_x$. Then lies $N$ in the kernel of the group action.
[This is one part of a task in a textbook, so that $G$ acts transitive might not be needed.]
My proof goes like this:
A group action $G\times X\to X$ can be translated into a group homomorphism $\varphi: G\to\operatorname{Sym}(X), g\mapsto \tau_g$ (left translation, so $\tau_g(x)=gx$).
We want to show $N\subseteq\ker\varphi=\{g\in G: \tau_g(x)=x\quad\text{for every}\,\,x\in X\}=\{g\in G: gx=x\quad\text{for every}\,\, x\in X\}$. Here I am unsure if it is actually correct to add 'for every $x$ in $X$". I thought so, because $g\in\ker\varphi$ if $\tau_g$ is the identity. So it has to hold $gx=x$ for every $x\in X$.
We have that for some $y\in X$ it is $N\subseteq G_y$.
Let $n\in N$. Since $n\in G_y$ we have $ny=y$. Also $G$ acts transitive. So $Gx=\{gx: x\in X\}=X$. Which means for every $x\in X$ we have a $g_x\in G$ with $g_xx=x$.
A proof of the statment should now go something like this:
$nx=ng_xx\stackrel{\text{N normal}}{=}g_xnx\stackrel{?}{=}g_xx=x$
Where I do not know how to justify this last equality from the assumption (if possible) or neither do I know if I really need that $G$ acts transitive here.
Do you mind checking on my thoughts, and correct them if necessary?
Thanks in advance.
Best Answer
Transitivity is crucial here. For example, consider action of $G$ on $G\cup\{p\}$ which is the translation on $G$ but fixes $p$. Then $G_p=G$ contains all its normal subgroup,s but obviously a nontrivial normal subgroup cannot fix anything else.
I'll leave you to prove $G_{gx}=gG_xg^{-1}$. Then $N=gNg^{-1}\subseteq G_{gx}$ for all $g\in G$, so $N\subseteq\bigcap_{g\in G}G_{gx}=\ker(G\to S_X)$ by transitivity.