Let $A \unlhd G$ and $o(x) = 3$ for each $x \notin A$, then $[B, B^x] = 1$ for abelian subgoups $B \le A$. 
Let $A$ be a normal subgroup of $G$. If each element from $G \setminus A$ has order $3$, then $[B, B^x] = 1$ for every abelian subgroup $B \le A$ and $x \in G \setminus A$.

Any hints for this exercise? I have totally no idea how to exploit the fact that $x$ has order three (guess this might be crucial here...). All I see is that $B^x \le A$ by normality of $A$ and that we must show $[b, \hat b^x] = 1$ for $b, \hat b \in B$. 

For $b \in B$, $1 = (xb)^3 = x^3b^{x^2}b^xb=b^{x^{-1}}b^xb$, so $b^{-1}=b^{x^{-1}}b^x$ (1).
For some other element $c \in B$, conjugating (1) by $c$ (which commutes with $b$) gives $b^{-1} = b^{x^{-1}c}b^{xc}$.
But we could also replace $x$ by $c^{-1}x$ in (1), giving $b^{-1} = b^{x^{-1}c}b^{c^{-1}x} = b^{x^{-1}c}b^{x}$, so $b^{xc}=b^{x}$,
and hence $c$ commutes with $b^{x}$.
