Suppose $G$ is a group and $x^3y^3 = y^3x^3 ~ \forall ~x,y \in G.$ Let $H = \{x \in G |~~ |x| $ is relatively prime to 3$\}$. Suppose $G$ is a group and $x^3y^3 = y^3x^3 ~ \forall ~x,y \in G.$ Let $H = \{x \in G |~~ |x| $ is relatively prime to 3$\}$. Prove that elements of H commute with each other  

$Attempt$: $x^3y^3 = y^3x^3 ~ \forall ~x,y \in G.$ 

And $H = \{x \in G |~~ |x| $ is relatively prime to 3$\}$

=> $\langle x \rangle = \langle x^3 \rangle ~ \forall ~x \in H$ and $O(x) = O(x^3)~ \forall ~x \in H$.

=> Elements in $\langle x \rangle$ are the same as in $\langle x^3 \rangle$ in some order.

=> $x = x^{3m}$ for some m

=> $x^{9m}y^{9m} = y^{9m}x^{9m} ~ \forall ~x,y \in G.$ 

Now i am unable to proceed further.

( Please note that this is a question from Gallian and i have studied only till cyclic groups before this question exercise.)

Thank you
Hint: If $k:=o(x)$ and $3$ are relatively prime, there exist $p,q$ such that $kp+3q=1$, hence $$x=x^{kp+3q}= \left( x^q \right)^3.$$
$1$. $(|x|,3)=1 \implies a|x|+b\cdot 3=1 \implies x=x^{b\cdot3}=(x^b)^3 \implies$ elements of $H$ commute.

$2$.$(|x|,3)=1,(|y|,3)=1 \implies (|x||y|,3)=1$. But $(|xy|)|(|x||y|)$, hence $(|xy|,3)=1 \implies$ $H$-subgroup.