Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a, b\in G$. Prove that $H=\{x^6\mid x \in G\}$ is a subgroup of $G$

Question

Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a, b\in G$. Prove that $H=\{x^6\mid x \in G\}$ is a subgroup of $G$

Attempt:

$e=e^6$ i.e $e \in H\neq \phi$

Let $a,b\in H$ then $a=x^6, b=y^6$ for some $x, y\in G$

$$x, y\in G \implies (xy)^3=x^3y^3 \tag 1$$

Now $ab=x^6y^6$

How to show that $ab\in H$ and $a^{-1}=x^{-6}\in H$ using $(1)$.

Well, $(x^6)^{-1} = (x^{-1})^6$, right? ... How can you rewrite $x^6y^6$ using what you know? — Ted Shifrin, Oct 03 '15 at 20:28
$(x^6)^{-1} = (x^{-1})^6$ will solve for the 2nd part. Problem is now the 1st part only. Unable to manage the 1st part. — user1942348, Oct 03 '15 at 20:29

score 3 · Accepted Answer · answered Oct 03 '15 at 20:46

From the comments, you're interested in dealing with the first part, which I'll show. For $x,y\in G$, $x(yx)^2y = (xy)^3 = x^3 y^3$, hence $(yx)^2 = x^2y^2$ by cancellation. So if $a, b\in H$, say $a = x^6$ and $b = y^6$, then $$ab = x^6y^6 = (x^3)^2(y^3)^2 = (y^3x^3)^2 = [(yx)^3]^2 = (yx)^6\in H.$$

Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a, b\in G$. Prove that $H=\{x^6\mid x \in G\}$ is a subgroup of $G$

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