$G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$

Question

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?

Answer 1

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 $G=\langle a,b|a^2=b^2, (ab)^2=(ba)^2\rangle$.

Answer 2

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 $2$.

Answer 3

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.

Answer 4

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 $st=ts$.

$\square$

Answer 5

$\forall x,a\in G,ax,a^{-1} \in G\Rightarrow ax^2a^{-1}=x^2\Rightarrow ax^2=x^2a$

$\forall x,y\in G$

$xyxy=yxyx\Rightarrow x^{-1}y^{-1}xy=yxy^{-1}x^{-1}$

$(xyx^{-1}y^{-1})^2=xy(x^{-1}y^{-1}xy)x^{-1}y^{-1}=xy^2xy^{-1}(x^{-1})^2y^{-1}$=$x^2y^2y^{-2}x^{-2}=e$

$order(xyx^{-1}y^{-1})\neq 2\Rightarrow xyx^{-1}y^{-1}=e$

$xy=yx$