Let $$G = P_0(G) > P_1(G) > \dots > P_c(G) = 1$$ be a lower exponent $p$-series of a finite $p$-group $G$, i.e. $P_{i+1}(G) = [G, P_i(G)]P_i(G)^p$.
In the $p$-quotient algorithm, one frequently uses that the sections $P_i(G) / P_{i+1}(G)$ are elementary abelian and central in $G$. But why is this true?
Of course, $P_0(G) / P_1(G) = G/\Phi(G)$ is elementary abelian.
But how do I see that e.g. $P_1(G) / P_2(G)$ is elementary abelian? I tried to use the fact that $(G/P_2(G))/(P_1(G)/P_2(G)) \cong G/P_1(G) \cong C_p^n$ for some $n$, but I could not obtain the result from there.
Since $P_{i+1}$ contains $[P_i,P_i]$ (which is contained in $[G,P_i]$), the quotient $P_i/P_{i+1}$ is necessarily abelian.
Since $P_{i+1}$ contains $P_i^p$, the quotient $P_i/P_{i+1}$ is of exponent $p$.
So the quotient is abelian of exponent $p$, i.e., elementary abelian.