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].

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