Question about upper central series of a group

Question

I am trying to figure out nilpotent groups, and it says that a group $G$ is nilpotent if there is a non-negative integers $c$ such that $Z^c(G) = G$.

Now an upper central series is something like $Z^0(G) \leq Z^1(G) \leq Z^2(G) \leq \ldots$ where $Z^0(G) = 1$ and $Z^1(G) = Z(G)$. But it doesn't say what $Z^2(G)$ is, it just says that $Z^2(G) / Z^1(G) = Z(G / Z^1 G)$. Since $G / Z^1(G)$ is a group whose elements are sets, $Z^2(G) / Z^1(G)$ will also be a set whose elements are sets. How do you solve for $Z^2(G)$, if you want to check if $Z^2(G) = G$?

Answer 1

Read this link it may help you https://en.wikipedia.org/wiki/Central_series

Answer 2

Consider the group $Z(G/Z(G))$. Then it is a subgroup of the factor group $G/Z(G)$, say $H/Z(G)$. Therefore $H$ is a subgroup of $G$ containing $Z(G)$. Now we write $H = Z_2(G)$. For example, consider the dihedral group $G = D_8$. You can check that $|Z(G)|=2$. Now $G/Z(G)$ is a group of order $4$ and in particular is abelian. Therefore $$G/Z(G) = Z(G/Z(G)) = Z_2(G) /Z(G).$$ Hence we have $G = Z_2(G)$ and $G$ is of class 2.