Prove that a certain Group is Abelian Let $G$ be a group with following two properties:

*

*for all $a, b \in G$ we have $(ab)^2=(ba)^2$

*every element $a \in G$ of order $2$, ie $a^2=e_G$, is already the neutral element $a=e_G$
question: is this group already Abelian?
what I tried: we can derive a nice identity of commuators $[a,b]:=aba^{-1}b^{-1}$:
$$  aba^{-1}b^{-1} = b^{-1}a^{-1}ababa^{-1}b^{-1}= b^{-1}a^{-1}(ab)^2a^{-1}b^{-1}=  b^{-1}a^{-1}(ba)^2a^{-1}b^{-1} = b^{-1}a^{-1}ba $$
Therefore $[a,b]=[b^{-1},a^{-1}]$ and we have to show that $[a,b]=e_G$.
 A: As $\left(ab\right)^2 = \left(ba\right)^2$, then $e_G = b^{-1}a^{-1}b^{-1}a^{-1}baba$, for all $a,b \in G$.
Now
\begin{align}
\left[a,b\right]^2 &= \left(aba^{-1}b^{-1}\right)^2 \\
        &= aba^{-1}b^{-1}aba^{-1}b^{-1} \\
        &= ab\left(b^{-1}a^{-1}b^{-1}a^{-1}baba\right)a^{-1}b^{-1}ab\left(b^{-1}a^{-1}b^{-1}a^{-1}baba\right)a^{-1}b^{-1} \\
  &= \left(b^{-1}a^{-1}ba\right)\left(b^{-1}a^{-1}ba\right) \\
        &= \left(b^{-1}a^{-1}ba\right)^2,
\end{align}
and
$$
\left[a,b\right]^2 = \left(\left(ab\right)\left(a^{-1}b^{-1}\right) \right)^2 = \left(a^{-1}b^{-1}ab\right)^2.
$$
Then
\begin{align}
\left[a,b\right]^4 &= \left(a^{-1}b^{-1}ab\right)^2 \left(b^{-1}a^{-1}ba\right)^2 \\
& = \left(a^{-1}b^{-1}ab\right)\left(a^{-1}b^{-1}ab\right)\left(b^{-1}a^{-1}ba\right)\left(b^{-1}a^{-1}ba\right)\\
& = e_G.
\end{align}
By the second property $\left[a,b\right]^2 = e_G$, therefore $\left[a,b\right] = e_G$, then G is Abelian.
A: Proof that $G$ in abelian: Given condition implies that $(x^{-1}\cdot yx)^2=(yx \cdot x^{-1})^2$, i.e. $x^{-1}y^2x=y^2$.
This being true for all $x$ implies that  all squares are in center of $G$. Then,
$$
(ab)^2=abab= ab\cdot (ba\cdot  a^{-1}b^{-1})\cdot ab=ab^2a [a,b]=a^2b^2[a,b]
$$
(last equality uses - squares are in center).
Changing role of $b,a$, we get
$$(ba)^2=b^2a^2[b,a].$$
Since $(ab)^2=(ba)^2$, we get
$$
a^2b^2[a,b]=b^2a^2[b,a].
$$
Again, squares are in center, we can do cancellation of squares here to get
$$
[a,b]=[b,a]=[a,b]^{-1} \,\,\,\,\Rightarrow \,\,\,\,  [a,b]^2=1 \,\,\,\, (Q.E.D.)
$$
