Autocommutator subgroup is a characteristic subgroup

Question

Autocommutator subgroup $K(G)$ of a group $G$ is defined as $$K(G):=\langle [g,‎\alpha‎]:g\in G,\; ‎\alpha ‎\in ‎Aut(G)‎\rangle ‎.,$$

where $[g,\alpha ]:=g^{-1}\alpha (g)$. Is $K(G)$ a characteristic subgroup in $G$?

Recall that a subgroup $H$ of a group $G$ is said to be charactristic in $G$ if $\alpha (H)\subseteq H$ for all $\alpha \in Aut(G)$.

My try: Assume that $[g,\alpha ]\in K(G)$ is an arbitrary generator and $\beta \in Aut (G)$. We need to show that $\beta ([g,\alpha])\in K(G)$. We have $\beta ([g,\alpha])=\beta (g^{-1})\beta (\alpha (g))$. But I couldn't show that $\beta ([g,\alpha])\in K(G)$.

Answer 1

Your strategy works with a little calculation: Let $g\in G$ and $\alpha,\beta\in \mathrm{Aut}(G)$. Then we have \begin{align} \beta([g,\alpha])&=\beta(g^{-1})\beta(\alpha(g))\\ &=\beta(g^{-1})g\cdot g^{-1}\alpha(g)\cdot\alpha(g)^{-1}\beta(\alpha(g)) \\ &=[g,\beta]^{-1}\cdot [g,\alpha]\cdot [\alpha(g),\beta]. \end{align} Hence, $\beta([g,\alpha])$ is again in $K(G)$.

Answer 2

$\DeclareMathOperator{\Aut}{Aut}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$You may use the following general result.

Let $A, B \le G$, and $$[A, B] = \Span{[a, b] : a \in A, b \in B }.$$ Then $[A, B]$ is normalised by both $A$ and $B$.

The proof is immediate from the standard identity $$ [x y, z] = [x, z]^{y} [y, z],\text{ or } [x, z]^{y} = [x y, z] [y, z]^{-1}, $$ and the similar one on the second term of the commutator.

Then note that $K(G) = [G, \Aut(G)]$, where the commutator is taken in the holomorph of $G$, that is, in the natural semidirect product $G \rtimes \Aut(G)$.

Answer 3

I think there is an easier calculation than in the accepted answer: Call $h:=\beta(g)$. Then $$\beta([g,\alpha])=h^{-1}\beta(\alpha(\beta^{-1}(\beta(g)))=h^{-1}(\beta\alpha\beta^{-1})(h)=[\beta\alpha\beta^{-1},h]\in K(G).$$