I have the following problem in group theory from a previous exam in a college course which stumped me:
Let $ G $ be a group such that $ G/Z(G) $ is abelian. We are asked to show that for any fixed odd $m$ the set $G^m=\{g^m | g \in G\}$ is a normal subgroup of $G$
The fact that $ G/Z(G) $ is abelian implies that the commutators (all elements of the form $xyx^{-1}y^{-1}$ for $ x,y\in G $) are in the center of $ G $ meaning they commute with all elements of $ G $. This is all I could work out with the problem. The normality of the set $ G^m $ is trivial as is showing it's closed to inverses, but how do I show it is closed under products? I thought about constructing a group homomorphism whose kernel is the set $ G^m $ but cannot figure out what such a homomorphism might look like? Why odd not even? This is where I am stuck and cannot work out a solution at all, so I obviously cannot show it is a subgroup. I would appreciate any help.
Hint: in a nilpotent group with class at most $2$ we have
$$(xy)^m = x^my^m [y,x]^{m(m-1)/2}.$$ Also, the map $[\cdot,x]$ from $G$ to $G'$ is a group homomorphism for every $x \in G$.