Group acting on $X$ and element of normal subgroup $H$ fixes an element of $X$ implies $H$ fixes all of $X$

Question

A group $G$ acts on a set $X$ transitively and a normal subgroup $H$ fixes a point $x_{0} \in X$, i.e. $h \cdot x_{0}=x_{0}$ for all $h \in H$. Show that $h \cdot x = x$ for all $h \in H$ and $x \in X$.

Since the action is transitive $\mathcal{O}_{x}=\{g\cdot x : g \in G,~ x \in X\}=X$ for any $x \in G$.

I've been fooling around with the fact that $g\cdot x = x_{0}$ for some $g \in G$ and using the fact that $H$ is a normal subgroup but haven't really gotten anywhere. I feel like this should follow straight from the axioms of group action and definition of normal subgroup, or am I missing something?

Answer 1

Suppose the normal subgroup $H$ fixes $x$, let $y$ be in the set and let $g$ be such that $y=gx$. Then $gHg^{-1}=H$ fixes $y$, so $H$ fixes every point.

Answer 2

This follows from the following: Let $G$ act on a set $X$,(need not be transitive) and let $Y$ be the subset consisting of all the fixed points for a normal subgroup $H\subset G$. Then $g.y\in Y$ for all $y\in Y\ g\in G$.