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.
This is not a complete answer, but I hope it helps.
Let $b \in A$ and $x \in G \setminus A$. We claim that $b$ commutes with $b^x = x^{-1}bx$.
Let $u = bx^{-1}bx$ and $v = x^{-1}bxb$. We want to show that $u=v$.
Now, since $b\in A,\,x\not\in A\Rightarrow bx^{-1}\not\in A\Rightarrow(bx^{-1})^3=1$, we have $(bx^{-1})^2u = bx$.
Similarly, using $x^{-2}=x$ and $(bx)^3=1$, we get $(bx^{-1})^2v = bx^{-1}bxbxb = bx^{-1}(bx)^3x^{-1} = bx^{-2} = bx$.
So $(bx^{-1})^2u = (bx^{-1})^2v$ and hence $u=v$, as claimed.
Added March 20: As Zach Teitler has pointed out, I had answered this question here. So I am afraid it is a duplicate.