Let $A$ be a normal subgroup of $G$. Suppose that every element in $G\,\backslash\, A$ has order $\bf 3$. Then $[B,B^x]=1$ for all Abelian subgroups $B\leq A$ and $x\in G\,\backslash\, A$.
I have been told that my task must have something to do with Chapter VI, On the isomorphism of a Group with Itself, para 66. of the famous book—“Burnside, W.: Theory of Groups of Finite Order, 2nd edn., Cambridge 1911; Dover Publications, New York 1955”. [ It’s a trick about order $3$, which was mentioned in the comments and Derek Holt’s answer below. ]
Although we‘ve made many attempts indeed and have made a breakthrough (Derek Holt’s answer), yet we haven’t been able to figure out how to use the normality of $A$ and abelianity of $ B$, on which I’m still struggling...
It would be greatly appreciated if you are kind enough to provide a reasonable answer!
PS: It’s exercise 1.5.6 of the book The Theory of Finite Groups, An Introduction. Berlin: Springer, 2004.