Exercise 12, § 4

Let G be a group such that, for a fixed integer $n > 1$, $(xy)^n = x^n y^n$ for all $x \in G, y \in G$. If $G^{(n)}$ denotes the set of $x^n$, where $x$ runs through $G$, and $G_{(n)}$ the set of $x \in G$ such that $x^n = e$, show that $G^{(n)}$ and $G_{(n)}$ are normal subgroups of $G$; if $G$ is finite, the order of $G^{(n)}$ is equal to the index of $G_{(n)}$. Show that, for all $x, y$ in $G$, also $x^{1-n} y^{1-n} = (xy)^{1-n}$ and deduce that $x^{n-1} y^n = y^n x^{n-1}$; conclude from this that the set of elements of $G$ of the form $x^{n(n-1)}$ generates a commutative subgroup of $G$.