The lower exponent $p$-central series for a $p$-group $G$ is defined by $G=P_1(G) > P_2(G) > \ldots > P_c(G) = 1$, where $$P_i(G)=[P_{i-1}(G), G] P_{i-1}(G)^p.$$ If $G_i=G/P_i(G)$ and $A:G_{i+1} \to G_i : g P_{i+1}(G) \mapsto g P_i(G)$, then $\ker(A) = P_i(G)/P_{i+1}(G)$.
Notice $\ker(A) \leq G_{i+1}$. Is $\ker(A) = P_i(G)/P_{i+1}(G) \cong P_i(G_{i+1})$?