central series of $\frac{G}{Z_2(G)}$

Question

Let $G$ be finite p-group I am trying to make central series for $\frac{G}{Z_2(G)}$ and more inportant what is nilpotency class of $\frac{G}{Z_2(G)}$
since $\frac{Z_{i+1(G)}}{Z_i(G)}=Z(\frac{G}{Z_i(G)})$
I wonder if this can be used to make a connection between central series of $\frac{G}{Z_2(G)}$ and central series of G
for $Z'_1(G)$ it is easy $\frac{Z'_1(G)}{1}=Z(\frac{\frac{G}{Z_2(G)}}{1})=\frac{Z_3(G)}{Z_2(G)}$
for other $Z'_i$ I can't get the any result

Answer 1

If $G$ has nilpotence class $c$ then the ascending central series looks like

$$Z_0<Z_1<\cdots<Z_c $$

with $Z_0(G)=1$, $Z_c(G)=G$ and each $Z_{i+1}$ is $G$'s center mod $Z_i$. If we apply the projection map $G\to G/Z_i(G)$ (for some fixed $i$) to the above series, we get

$$\overline{Z_i}<\overline{Z_{i+1}}<\cdots<\overline{Z_c}$$

for the ascending central series of $G/Z_i$ (overlines denote images under projection). To prove this, you must prove that $\overline{Z_{j+1}}$ is the center of $\overline{G}$ mod $\overline{Z_j}$.

Thus the nilpotence class of $G/Z_i(G)$ equals $c-i$.