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The lower $p$-central $p$-series of a finite $p$-group $G$ is defined as follows: $$G=P_1(G)>P_2(G)>P_3(G)>\cdots P_m(G)=1,$$ where $P_i(G)$ is the smallest normal subgroup of $G$ contained in $P_{i-1}(G)$ such that in $G/P_i(G)$, the term $P_{i-1}(G)/P_i(G)$ is central as well as elementary abelian (is this correct?)

Now analogously define upper $p$-central $p$-series pf $G$ as follows:

$$1=Q_0(G)<Q_1(G)<Q_2(G) < \cdots < Q_k(G)=G,$$ where modulo $Q_i(G)$, the subgroup $Q_{i+1}(G)$ becomes largest elementary abelian central subgroup (i.e. $Q_{i+1}(G)/Q_i(G)$ is largest elementary abelian central subgroup in $G/Q_i(G)$.)

The two series are defined in a similar manner as we define the standard upper and lower central series of any group. Since the upper and lower central series have same length, my natural question is the following:

Does the upper and lower $p$-central $p$-series of an y$p$-group have same length?

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    $\begingroup$ Yes, and the proof is analogous. The lower series is the fastest descending such series and the upper series is the fastest ascending in the sense that any such series has its terms contained in the upper series and containing the lower series. $\endgroup$ – Derek Holt Nov 9 '16 at 10:03

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