Normal subgroup and singleton orbits

Question

Here is the exercise 20.11 from Groups and Symmetry from Armstrong :

Let $H$ be a subgroup of $G$ and write $X$ for the set of left cosets of $H$ in $G$. We have the action : $$ g(xH) = gxH$$ Show that $H$ is a normal subgroup iff every orbit of the induced action of $H$ on $X$ contains just one point.

$\Rightarrow h(xH)=hxH=hHx=xH$ so it is clear that $G(xH)=\lbrace xH \rbrace $

For the other way I don't know how to do.

Answer 1

You have already done one implication. The other one is as follows:

For any $h \in H$ and $x \in G$, you have $h(xH)=xH$, so $h \in xHx^{-1}$. So $H = xHx^{-1}$ for any $x$ which means that $H$ is normal in $G$.

Answer 2

By the Orbit-Stabilizer Theorem:

\begin{alignat}{1} |O(xH)| &= [H:\operatorname{Stab}_H(xH)],\space\forall x \in G \\ \end{alignat}

where

\begin{alignat}{1} \operatorname{Stab}_H(xH) &= \{h\in H\mid hxH=xH\} \\ &= \{h\in G\mid hxH=xH\wedge h\in H\} \\ &= H\cap xHx^{-1}, \space\forall x\in G \end{alignat}

Therefore,

\begin{alignat}{1} &|O(xH)|=1,\space\forall x\in G \iff \\ &H\cap xHx^{-1}=H,\space\forall x\in G \iff \\ &H\subseteq xHx^{-1},\space\forall x\in G \iff \\ &H\unlhd G \end{alignat}