Direct computation yields

∎

By Theorem 1, the elements from $[X,Y]$ or $[Y,X]$ are products of commutators of the form $[x,y]$ or $[y,x]$ with $x\in X$ and $y\in Y$. ∎

This is mainly a brute-force calculation. We can easily calculate the first factor $y^{{-1}}[x,y^{{-1}},z]y$ explicitly using theorem 1:

Let $h_{1}:=x^{{-1}}y^{{-1}}xz^{{-1}}x^{{-1}}$, the “first half” of $y^{{-1}}[x,y^{{-1}},z]y$. Let $h_{2}$ be the element obtained from $h_{1}$ by the cyclic shift $S\colon x\mapsto y\mapsto z\mapsto x$, and $h_{3}$ be the element obtained from $h_{2}$ by $S$. We have

which gives us

and, by applying $S$ twice

In total, we have

∎

The group $[Z,X,Y]$ is generated by all elements of the form $[z,x^{{-1}},y]$ with $x\in X$, $y\in Y$ and $z\in Z$. Since $N$ is normal, $y^{{-1}}[x,y^{{-1}},z]y$ and $x^{{-1}}[z,x^{{-1}},y]x$ are elements of $N$. The Hall–Witt identity then implies that $x^{{-1}}[z,x^{{-1}},y]x$ is an element of $N$ as well. Again, since $N$ is normal, $[z,x^{{-1}},y]\in N$ which concludes the proof. ∎

By expanding:

The other identities are proved similarly. ∎