Let $G$ be a group in which, for some integer $n > 1$, $(ab)^n = a^nb^n$ for all $a, b ∈ G$. Show that (a) $G(n) = \{x^n | x ∈ G\}$ is a normal subgroup of $G$. (b) $G(n- 1) = \{x^{(n- 1)} | x ∈ G\}$ is a normal subgroup of $G$.
I have been trying this for a long time but not been able to make any progress. This should be a basic question, but I got no clue... Please help!
$G(i)$ is normal for all $i$, so it remains to check that they are closed under multiplication.
If $x^n, y^n \in G(n)$, then $x^n y^n = (xy)^n \in G(n)$, so indeed $G(n)$ is a subgroup.
For $x^{n-1}, y^{n-1} \in G(n-1)$, $x^{n-1} y^{n-1} = x^{-1} x^n y^n y^{-1} = x^{-1} (xy)^n y^{-1} = (yx)^{n-1} \in G(n-1)$.