## The commutator generated subgroup operator is commutative

Let $G$ be a group. Prove that if $x,y \in G$ then $[y,x] = [x,y]^{-1}$. Deduce that for any subsets $A,B \subseteq G$, $[A,B] = [B,A]$. (Recall that $[A,B]$ is the subgroup of $G$ generated by the commutators $[a,b]$.)

Note that $[x,y][y,x] = (x^{-1}y^{-1}xy)(y^{-1}x^{-1}yx) = 1$. By the uniqueness of inverses, $[x,y]^{-1} = [y,x]$.

Now $[A,B] = \langle S \rangle$, where $S = \{ [a,b] \ |\ a \in A, b \in B \}$. Now $S^{-1} = \{ [a,b]^{-1} \ |\ a \in A, b \in B \}$ $= \{ [b,a] \ |\ a \in A, b \in B \}$, so that $[A,B] = \langle S \rangle$ $= \langle S^{-1} \rangle$ $= [B,A]$.