Let $G$ be a group. Consider $\gamma_1(G):=G$ and $\gamma_{i+1}(G):=[\gamma_i,G]$, we denote $G_{\text{ab}}$ the abelianization of $G$. Then $\gamma_{i+1}/\gamma_{i+2}(G)$ is an epimorphic image of $G_{\text{ab}}\otimes\Big(\gamma_i(G)/\gamma_{i+1}(G)\Big)$ by the map that sends $(x\gamma_2(G),y\gamma_{i+1}(G))$ to $[x,y]\gamma_{i+2}(G)$. In other word $\gamma_{i}(G)/\gamma_{i+1}(G)$ is an epimorphic image of $\otimes ^{i+1}G_{\text{ab}}$. Now I have to prove the theorem:
Let $\mathcal{P}$ be a property preserved by tensor product, epimorphic image and extension. If $G$ is nilpotent and $G_{\text{ab}}$ satisfies $\mathcal{P}$ then $G$ satisfies $\mathcal{P}$.
I have no idea how to use the extension and nilpotent property, some ideas?
Start by considering the descending chain $G \supseteq \gamma_2(G) \supseteq ... \supseteq \gamma_i(G) \supseteq \gamma_{i+1}(G) \supseteq ... \supseteq \gamma_{c+1}(G)=1$, where $c$ is the nilpotency class of $G$.
Now, look at the quotients: $G/\gamma_2(G)=G_{ab} \in \mathcal{P}$, and $\gamma_2 (G) / \gamma_3 (G)$ is an epimorphic image of $G_{ab} \otimes G_{ab}$, hence $\gamma_2(G)/\gamma_3(G) \in \mathcal{P}$. Proceeding in this way, $\gamma_i(G) / \gamma_{i+1}(G) \in \mathcal{P}$, since it is an epimorphic image of $G_{ab} \otimes ... \otimes G_{ab}$. This means that all the factors satisfy $\mathcal{P}$, but we also know that it is preserved by extensions: $G/ \gamma_2(G) \in \mathcal{P}, \gamma_2(G) / \gamma_3(G) \in \mathcal{P}$ implies that $G/\gamma_3(G)\in \mathcal{P}$, and going down again on our chain, we finally reach that $G/\gamma_{c+1}=G \in \mathcal{P}$, as we wanted.