Let $G$ a finite group. If $A$ is a group of automorphisms of $G$, the subgroup $[G,A]$ is an $A$-invariant normal subgroup in $G$?

Question

By definition, $$[G,A] = \big<[g,\alpha] = g^{-1}g^\alpha; g \in G, \alpha \in A \big>.$$ For normality part, for any $x \in G$ we have $[g,\alpha]^x = (g^{-1}g^\alpha)^x = (g^x)^{-1}g^{\alpha x}$ and I don't know if it belongs to $[G,A]$.

For the A invariant part, let $\beta \in A$, $[g,\alpha]^\beta = (g^{-1}g^\alpha)^\beta = (g^\beta)^{-1}g^{\alpha \beta}$ does it belongs to $[G,A]$?

Answer 1

Take the semidirect product $H=GA$ of $G$ (normal) and $A$. Then $[G, A]=[G, H]$ which is indeed a normal subgroup of $H$ inside $G$.