# 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: $x$ can be written as a power of $x^3$, and $y$ can be written as a power of $y^3$. Can you see why $x$ and $y$ must then commute? – Tobias Kildetoft Mar 18 '14 at 8:13
• Elements in $\langle x \rangle$ are the same as in $\langle x^3 \rangle$ in some order. => $x = x^{3m}$ for some m => $x^{9m}y^{9m} = y^{9m}x^{9m} ~ \forall ~x,y \in G.$ Uhm, Can i do something from here? – MathMan Mar 18 '14 at 8:24
• I am not quite sure where the $9$ came from. Note that if $z$ and $w$ commute, then so do any powers of $z$ and $w$ (here we have $z = x^3$ and $w=y^3$). – Tobias Kildetoft Mar 18 '14 at 8:26
• That 9 came from substituting $x=x^{3m}$ back in the equation. So, we have $x=x^{3m}$ and $y=y^{3n} ~ \forall x,y ~ \in ~H$ => $xy =x^{3m}y^{3n}$ and $yx =y^{3n} x^{3m}$ – MathMan Mar 18 '14 at 8:31
• note that you cannot be sure that you can use the same exponent for $x$ and $y$ (in going from the third power to the first power). But if you replace one of those $m$'s by an $m'$ then note that those $x^{3m}y^{3m'}$ can be rearranged since those are third powers of suitable elements. – Tobias Kildetoft Mar 18 '14 at 8:34

Hint: If $k:=o(x)$ and $3$ are relatively prime, there exist $p,q$ such that $kp+3q=1$, hence $$x=x^{kp+3q}= \left( x^q \right)^3.$$
Hint: Consider the group $\langle x^3, y^3 \rangle$. This subgroup is abelian and also contain $x$ and $y$. So they commute.
• Please could you explain why $\langle x^3, y^3\rangle$ contains $x$ and $y$? – a student Dec 8 '15 at 6:21
• @student. Order of $x$ say $m$ is relatively prime to 3. So you can fine integer $r$ and $s$ such that $mr+3s=1$. Clearly $x^{3s}\in \langle x^3, y^3\rangle$. But $x^{3s}=x^{1-mr}=x$. So $x\in \langle x^3, y^3\rangle$. – D. N. Dec 8 '15 at 9:29
$1$. $(|x|,3)=1 \implies a|x|+b\cdot 3=1 \implies x=x^{b\cdot3}=(x^b)^3 \implies$ 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.