On p-groups having maximal class Let's suppose P is p-group of maximal class and let N be a normal subgroup of P whose index in P is greater than or equal to $p^2$. How can we show that N turns out to be one of the terms of the lower central series for P?
Considering the lower central series for P, it's clear that all the successive quotients are of order p, except for the top-most quotient which is of order $p^2$. It seems clear that N cannot be properly squeezed in the lower central series, but then I'm not sure how to show N is actually one of the terms.
 A: We know that for $p$-groups, the length of upper and lower central series is same. Thus, if $|G|=p^n$, $n\geq 3$, and if $G$ is of maximal class then $G$ has upper central series 
$$1 < Z_1(G) <Z_2(G) <\cdots < Z_{n-2}(G) < G$$
where $|Z_i(G)|=p^{i}$. Since, the lower central series also has same length, it is 
$$G >\gamma_2(G) >\gamma_3(G) >\cdots > \gamma_{n-1}(G)>1$$
where $[G\colon \gamma_2(G)]=p^2$ and other successive indices are $p$ as you said. Therefore $\gamma_{n-1}(G)$ is a normal subgroup of order $p$. 
In $p$-group, normal subgroups intersect center non-trivially, hence $Z(G)=Z_1(G)$ and $\gamma_{n-1}(G)$ intersect non-trivially, and since both are of order $p$, they must be equal. 
Proceeding almost as before, we can see that $Z_2(G)=\gamma_{n-2}(G)$ ( factor out $G$ by $Z_1(G)=\gamma_{n-1}(G)$, apply above procedure).
Thus for $p$-group of maximal class, upper and lower central series coincide.
Coming to your question: let $N$ be a normal subgroup of $G$. If $|N|=p$, then $N\cap Z_1(G)\neq 1$, and since both have same order $N=Z_1(G)$. 
If $|N|>p$, then again, $N\cap Z_1(G)\neq 1$ implies $Z_1(G)<N$. Let $\overline{G}=G/Z_1(G)$, and $\overline{N}=N/Z_1(G)$. By induction on $|G|$, this image of $N$ should be one of the term of lower/upper central series of $\overline{G}$, hence $N$ should be one of the terms of upper/lower central series of $G$.
