M.Isaacs’ Algebra a graduate course page 119 :
(Schur). Let $|G:Z(G)|=m<∞$. Then the map $g↦g^m$ is a homomorphism from G into Z(G).
Proof. In fact, we will show that this map is the transfer $v:G⟶Z(G)$. By the transfer evaluation lemma, we have for g∈G that
$v(g)=π(g)=∏_{tϵT_0}tg^{n_t}t^{-1}$,
Where $tg^{n_t} t^{-1}∈Z(G)$. It follows that $tg^{n_t} t^{-1}=g^{n_t}$ and
$v(g)=∏_{tϵT_0}g^{n_t} =g^{∑n_t} =g^m$,
As required.
More can be said when the hypotheses of this theorem are satisfied. If T is a transversal for Z(G) in G, an easy calculation shows that every commutator in G actually has the form $[s , t]$ for elements s , t ∈ T.
And the question is how to prove the last paragraph?
Assume $T$ is a right transversal for $Z(G)$ in $G$. Take $x$ and $y$ from $G$, then $x=zg$ and $y=z^\ast h$
for some $z,z^\ast$ from $Z(G)$ and $g,h$ from $T$. Then $$[x,y]=[zg,zh]=(zg)^{-1}(zh)^{-1}(zg)(zh)= g^{-1}h^{-1}gh=[g,h].$$