Commutator subgroup and subgroup generated by square.

While reading Cours d'algèbre by D. Perrin, I found the following claim:

Proposition. If $G$ is a group, then $D(G)\subseteq G^2$, where $G^2$ is the subgroup generated by the squares of $G$.

I understand that it suffices to prove that for all $x,y\in G$, $[x,y]:=xyx^{-1}y^{-1}$ is a product of squares.

Equivalently, since $G^2$ is characteristic, hence normal, it suffices to establish that $G/G^2$ is abelian.

For now, none of my attempts have been fruitful. Any enlightenment will be greatly appreciated!

Solution 1:

Hint: prove that in a group in which every square is the identity, the group must be abelian. Then apply to $G/G^2$. Note that $g^2=1$ is equivalent to $g=g^{-1}$.