I'm stuck somewhere in the following problem in Isaacs Finite Group Theory [4D.4], I would appreciate if you could help:
Problem: Let $A$ act via automorphisms on $G$ , where $G$ is a $2$-group and $A$ has odd order. Show that if $A$ fixes every element $x$ in $G$ such that $x^4=1,$ then the action of $A$ on $G$ is trivial.
My attempt:
Case 1:In the case $G$ is abelian, the result follows from Fitting's theorem, $G$ = $C_G(A)$ $\times$ $[G,A].$ Since $C_G(A)$ and $[G,A]$ has trivial intersection and $C_G(A)$ contains each element $x$ in $G$ such that $x^4=1,$ $[G,A]$ cannot have an element of order $2.$ But $[G,A]$ is a $2$-group and every nontrivial $2$-group has an element of order $2$. Therefore, it must be $[G,A ]= 1,$ and the claim follows.
Case 2: Let's use induction on $\vert G \vert.$ If $[G,A]$ is a proper subgroup of $G,$ then by inductive hypothesis $[G,A,A]=1.$ In a coprime action, it must be $[G,A,A]=[G,A]$, the result follows. Hence, we can assume that $[G,A]=G.$ How can I get the desired result from this point?
Thanks.