Let $G$ be a group, $\lambda:G\to S_G$ s.t. $\lambda(g)(x)=gx$, ${\rm Hol}(G)=\lambda(G){\rm Aut}(G)$. Prove ${\rm Hol}(G)=G\rtimes{\rm Aut}(G)$.

Question

Let $G$ be a group.

Denote $\lambda: G \to{\rm Sym}(G)$ such that $\lambda (g)(x)=gx$.

Denote ${\rm Hol}(G)=\lambda(G){\rm Aut}(G)$ such that the action of ${\rm Aut}(G)$ on $G$ is the regular action.

Prove ${\rm Hol}(G)=G \rtimes {\rm Aut}(G)$.

Using the theorem

If $G=KH , H \triangleleft G , K\leq G, K\cap H=\{e\},$ then $G =H \rtimes K $.

I know ${\rm Aut}(G)\leq {\rm Hol}(G)$.

I have to prove $G \triangleleft Hol(G) $.

Let $g'\in G, g\in{\rm Hol}(G) , f\in{\rm Aut}(G),$ hence $g^{-1}g'g=\lambda(g^{-1})fg'\lambda(g)f\in G$.

Is my approach correct?

Thanks!

Answer 1

Since $\varphi\lambda_a=\lambda_{\varphi(a)}\varphi$, we have that $\lambda(G)$ and $\operatorname{Aut}(G)$ commute, and hence $\lambda(G)\operatorname{Aut}(G)\le\operatorname{Sym}(G)$. Moreover, $(\lambda_b\varphi)\lambda_a(\lambda_b\varphi)^{-1}=$ $\lambda_b(\varphi\lambda_a\varphi^{-1})\lambda_b^{-1}=$ $\lambda_b\lambda_{\varphi(a)}\lambda_{b^{-1}}=$ $\lambda_{b\varphi(a)b^{-1}}\in\lambda(G)$, whence $\lambda(G)\unlhd\lambda(G)\operatorname{Aut}(G)$. Finally, $\lambda_a\in\operatorname{Aut}(G)$ if and only if $a=e_G$ if and only if $\lambda_a=Id_G$, whence $\lambda(G)\cap\operatorname{Aut}(G)=\{Id_G\}$. Now use the theorem and note that $G\cong\lambda(G)$.