Let $G$ be a torsionfree group and there exists $n\in\mathbb{N}$ such that for any $x,y \in G$ satisfy $(xy)^n=x^ny^n$. Show that $G$ is abelian.
(The question does not restrict $n$ but as Brian Moehring commented: For the question to make sense we probably need to assume $n>1$)
The question appeared in last year Israel Mathematics Olympiad for undergrads and I'm trying to solve it because I'm taking course on algebraic structures right now. Previously, at the beginning of our course in one of our exercises we had to prove a (somewhat) similar theorem:
Let $G$ be a group. We say $G$ has the property $\Phi_n$ if for any $x,y \in G$ we have $(xy)^n=x^ny^n$. Show that if $G$ has $\Phi_n,\Phi_{n+1},\Phi_{n+2}$ then $G$ is abelian.
A similar question was asked and answered here previously in In a group, does $(xy)^n=x^ny^n$ for $n\geq 3$ imply $xy=yx$?. I wanted to follow in a similar way, that is- derive some identities. These are the couple I was able to come up with: $$(xy)^n=x^ny^n\Longrightarrow x^{-1}(xy)^ny^{-1}=x^{n-1}y^{n-1}\Longrightarrow (yx)^{n-1}=x^{n-1}y^{n-1}$$
And notice that
$$(x^ny^n)^n=x^{n^{n}}y^{n^{n}}=x^{n^{2}}y^{n^{2}}$$
So in a similar fashion we can get
$$x^{n^{2}-1}y^{n^{2}-1}=(xy)^{n^{2}-1}$$
(and I also tried to decompose it back to $n^2-1=(n-1)(n+1)$ and then use the power of $n-1$ but didn't get far)
I figured since we know the group is torsionfree I need to find something to the power of $n$ that equals $e$ so that something must be $e$ but couldn't find the right thing to define. I also thought I'll end up getting something along the lines of $e=xyx^{-1}y^{-1}$ and that made me think maybe I need to use the commutator somehow. Since we know that the quotient group $G/N$ is abelian if and only if $N$ contains the commutator subgroup of $G$. If I could show the commutator is trivial and $N$ is trivial I'd get and abelian group I think? [I might be wrong about this! feel free to correct me!]. Anyhow. I didn't get far with this approach either!
Thank you very much for any help!
$$(xy)^n=x^ny^n\Longrightarrow x^{-1}(xy)^ny^{-1}=x^{n-1}y^{n-1}\Longrightarrow (yx)^{n-1}=x^{n-1}y^{n-1}$$ $$x^ny^n=(xy)^n=(xy)(xy)^{n-1}=xyy^{n-1}x^{n-1}= xy^{n}x^{n-1} $$ We now know that any $x^{n-1}$ commutes with any $y^n$. Consider $c=(x^{n-1}y^{-1}x^{1-n})y$. Then $c^n=(x^{n-1}y^{-n}x^{1-n})y^n=y^{-n}y^n=e$ and therefore $c=e$ i.e. $x^{n-1}$ is in $Z(G)$.
Now consider $d=(x^{-1}y^{-1}x)y$ and note that $y^{n-1}$ commutes with $x$. Then$$d^n=x^{-1}y^{-n}xy^n=x^{-1}y^{-1}xy=d. $$
Then $d^{n-1}=e$ and so $d=e$ and we are finished.