Theorem 1   Let $x,y\in G$ , then $[x,y]^{-1}=[y,x]$ .

Proof. Direct computation yields \begin{equation*} [x,y]^{-1}=(x^{-1}y^{-1}xy)^{-1}=y^{-1}x^{-1}yx=[y,x]. \end{equation*}

Theorem 2   Let $X,Y$ be subsets of $G$ , then $[X,Y]=[Y,X]$ .

Proof. 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 $y\in Y$ .

Theorem 3 (Hall-Witt identity)   Let $x,y,z\in G$ , then \begin{equation*} y^{-1}[x,y^{-1},z]yz^{-1}[y,z^{-1},x]zx^{-1}[z,x^{-1},y]x=1. \end{equation*}

Proof. 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 \begin{equation*} h_2^{-1}=(y^{-1}z^{-1}yx^{-1}y^{-1})^{-1}=yxy^{-1}zy \end{equation*}which gives us \begin{equation*} y^{-1}[x,y^{-1},z]y=h_1h_2^{-1}, \end{equation*}and, by applying $S$ twice
In total, we have \begin{equation*} y^{-1}[x,y^{-1},z]yz^{-1}[y,z^{-1},x]zx^{-1}[z,x^{-1},y]x=h_1h_2^{-1}h_2h_3^{-1}h_3h_1^{-1}=1. \end{equation*}

Theorem 4 (Three subgroup lemma)   Let $N$ be a normal subgroup of $G$ . Furthermore, let $X$ , $Y$ and $Z$ be subgroups of $G$ , such that $[X,Y,Z]$ and $[Y,Z,X]$ are contained in $N$ . Then $[Z,X,Y]$ is contained in $N$ as well.

Proof. 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.

Theorem 5   For any $x, y, z \in G$ we have \begin{eqnarray*} [xy, z] & = & [x,z]^y [y,z] \\ {[}x,yz] & = & [x,z][x,y]^z \\ {[}x,y]^z & = & [x^z, y^z] \\ {[}x^z, y] & = & [x, y^{z^{-1}}] \end{eqnarray*}where $a^b$ denotes $b^{-1} a b$

Proof. By expanding: \begin{eqnarray*} [xy,z] & = & y^{-1}x^{-1} z^{-1} xyz \\ & = & y^{-1} x^{-1} z^{-1} \cdot xz \cdot z^{-1} x^{-1} \cdot xyz \\ &=& y^{-1} [x,z] \cdot y \cdot y^{-1} \cdot z^{-1} x^{-1} \cdot xyz \\ &=& [x,z]^y \cdot y^{-1} z^{-1} y z \\ &=& [x,z]^y [y,z] \end{eqnarray*}The other identities are proved similarly.