Permutation Group Embedded into Holomorph of its Regular Normal Subgroup

Theorem

If $G⩽\mathrm{Sym}(\Omega )$ and $G$ has regular normal subgroup $X$, then $G$ on $\Omega$ is permutationally isomorphic to a subgroup $G_R$ of $\mathrm{Hol}(X)$ on $X$ via an natural isomorphism $\theta :G\to G_R$ and a bijection $\beta :{\alpha }^x\mapsto x$ (for some fixed $\alpha \in \Omega$).

\begin{proof} Since $X$ is a regular normal subgroup, $\Omega$ can be identified as $X$. Since $X⊲G$, we have $X_R⊲G_R$ and so $G_R⩽N_{\mathrm{Sym}(\Omega )}(X_R)=\mathrm{Hol}(X)$. Then it is easy to verify it is a permutationally isomorphic. \end{proof}