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 $[a,b]=e_G$.

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.
