Holomorph is the Normalizer of $X_R$

Theorem

Let $G$ be a group. Define $G_R=\{{\rho }_g:g\in G\}⩽\mathrm{Sym}(G)$ where ${\rho }_g:x\mapsto xg$. Then we have

$$\mathrm{Hol}(X):=X_R:\mathrm{Aut}(X)=N_{\mathrm{Sym}(X)}(X_R)$$

\begin{proof} For any $\phi \in N_{\mathrm{Sym}(G)}(G_R)⩽\mathrm{Sym}(G)$, we may assume that $\phi (1)=1$. Otherwise, take ${\phi }^{\prime }=\phi \circ {\rho }_{{\phi }^{-1}(1)}$. Since $\phi \in \mathrm{Sym}(G)$, $\phi$ is bijective and so it remains to show $\phi (gh)=\phi (g)\phi (h)$ for any $g,h\in G$. Note that ${\rho }_g^{\phi }={\phi }^{-1}\circ {\rho }_g\circ \phi ={\rho }_{{\phi }^{-1}(g)}$. Then by ${\rho }_{gh}={\rho }_h\circ {\rho }_g$ we have $\phi \in \mathrm{Aut}(G)$. \end{proof}

Remark. We can readily deduce that

• $X_L=C_{\mathrm{Sym}(X)}(X_R)$.
• $X_R=C_{\mathrm{Sym}(X)}(X_L)$.
• If $X$ has a trivial center, then $⟨X_L,X_R⟩=X_L\times X_R=X_R:\mathrm{Inn}(X)$.

It is similar as Three Permutations on G Induced from G.