$G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$ Suppose that $G$ is a group that there exists no element $x \neq e$ such that $x^2=e$. Moreover, for every $a,b \in G$ we have $(ab)^2=(ba)^2$. Prove that $G$ is Abelian.
Well, I attempted to prove that $(aba^{-1}b^{-1})^2=e$ because then if I could prove it that would imply $aba^{-1}b^{-1}=e$ which is the same as $ab=ba$. In other words I wanted to show the order of $[a,b]=aba^{-1}b^{-1}$ is two for every $a$ and $b$ in $G$. I spent almost an hour trying to show that by anything that came to my mind. I tried to brute force the problem by writing any possible equation that I could come up with but I failed :/
Any ideas?
 A: Edit: Andreas Caranti have pointed out, this proof work iff the all the elements of the group have finite order. So it doesn't answer completely but
is more a partial result.
Let $x,a \in G$ then by hypothesis we have that there's $b \in G$ such that $ab=x$ then 
$$a^{-1}x^2a=a^{-1}(ab)^2a=a^{-1}ababa=baba=(ba)^2=(ab)^2=x^2$$
So $x^2 \in Z(G)$ for all $x \in G$.
So we get the quotient $G/Z(G)$ in which all the elements have order two.
If there's an $x \in G \setminus Z(G)$ then $xZ(G) \ne Z(G)$ should have order $2$, and by properties of homomorphisms of groups the order of $x$ should be divided by $2$. That's absurd since by hypothesis $G$ can't have elements of order $2$ hence it can't have either elements which have order a multiple of $2$.
A: Elementary solution to this problem. 
The problem is equivalent to that the $x^2 \in Z(G)$ for any $x\in G$, as stated in a previous response. 
Under these conditions we show that $(xyx^{-1}y^{-1})^4=e.$
$$(xyx^{-1}y^{-1})^4=(xyx^{-1}y^{-1})^2(xyx^{-1}y^{-1})(xyx^{-1}y^{-1})=$$
$$=(xyx^{-1})(xyx^{-1}y^{-1})^2(y^{-1})(xyx^{-1}y^{-1})=$$
$$= xy(x^{-1}x)(yx^{-1}y^{-1})(xyx^{-1}y^{-1})(y^{-1})(xyx^{-1}y^{-1})=$$
$$=xy^2(x^{-1}y^{-1})(xyx^{-1})(y^{-1})^2(xyx^{-1}y^{-1})=$$
$$=y^2(xx^{-1})(y^{-1}xy(y^{-1})^2)(xyx^{-1}y^{-1})=(yxy^{-1})(yx^{-1}y^{-1})=e.$$
Considering that the group no has elements of order $ 2 $ result, step by step, that $(xyx^{-1}y^{-1})^2=e$, $(xyx^{-1}y^{-1})=e$ and $xy=yx$ and therefore the group is commutative.
A: Set $b=a^{-1}x$. We have $x^2=a^{-1}x^2a$, i.e.  $ax^2=x^2a$ for all $x,a$. Since $x^2$ runs all group, then $G$ is Abelian.
Correction: This proof is valid only for a finite group. Thanks to DonAntonio.
Addendum: I am not sure that this assertion is true for infinite groups. A candidate \for a counter-example is $G=\langle a,b|a^2=b^2, (ab)^2=(ba)^2\rangle$.
A: The following solution is the same as in the duplicate, just made shorter. As in the answer of Boris Novikov, for every $x\in G$ the element $x^2$ is in the centralizer of $Z(G)$ of $G$, by using
$(ab)^2=(ba)^2$ for $b=a^{-1}x$:
$$
x^2 =(ab)^2=(ba)^2=(a^{-1}xa)^2=a^{-1}x^2a\ .
$$
Let now $s,t\in G$ be two elements. We show $st=ts$. Consider
$$a = s^{-2}t^{-2}\; stst\ .
$$
One easily shows $a^2=1$ by using $s^{\pm 2},t^{\pm 2}\in Z(G)$:
$$
\begin{aligned}
a^2 
&= s^{-4}t^{-4}\cdot stst\cdot stst\\
&= s^{-4}t^{-4}\cdot stst\cdot tsts\\
&= s^{-4}t^{-4}\cdot sts\cdot t^2\cdot sts\\
&= s^{-4}t^{-2}\cdot st\cdot s^2\cdot ts\\
&= s^{-2}t^{-2}\cdot s\cdot t^2\cdot s\\
&= s^{-2}\cdot s^2\\
&=1\ .
\end{aligned}
$$
From the assumption, $a=1$, i.e.
$$
1=s^{-2}\; stst\; t^{-2}=s^{-1}\; ts\; t^{-1}\ ,
$$
so after multiplying from left with $s$, and from right with $t$ we get
$st=ts$.
$\square$
A: $\forall x,a\in G,ax,a^{-1} \in G\Rightarrow ax^2a^{-1}=x^2\Rightarrow ax^2=x^2a$
$\forall x,y\in G$
$xyxy=yxyx\Rightarrow x^{-1}y^{-1}xy=yxy^{-1}x^{-1}$
$(xyx^{-1}y^{-1})^2=xy(x^{-1}y^{-1}xy)x^{-1}y^{-1}=xy^2xy^{-1}(x^{-1})^2y^{-1}$=$x^2y^2y^{-2}x^{-2}=e$
$order(xyx^{-1}y^{-1})\neq 2\Rightarrow xyx^{-1}y^{-1}=e$
$xy=yx$
