Non-identity central element in a central quotient Let $G$ be a group. Let $Z(G)$ be the center of $G$, the set of elements that commute with every element of $G$.
Then, can we say that there is some elements in $Z(G/Z(G))$ which is not $Z(G)$?
 A: We can say it if we like, but it isn't true in general. Simple examples: if $G$ is abelian then $G/Z(G)$ is trivial and so has no non-identity elements. If $Z(G)$ is trivial, then $G/Z(G)$ is isomorphic to $G$ and again has no non-identity central elements.
A: Let's rephrase your question as follows:

Can we say that there is an element in $G$ whose image is in $Z(G/Z(G))$, but which is not in $Z(G)$?

Not always. In the two extreme cases, $G=Z(G)$ and $Z(G)=\{1\}$, the answer is "no". And there are examples of groups in which $Z(G)\neq \{1\}$, but $G/Z(G)$ has no center, so that the answer again is "no". For instance, take $G=S_3\times C_2$. Then $Z(G) = \{1\}\times C_2$, $G/Z(G)\cong S_3$, so $Z(G/Z(G))$ is trivial.
But there are some cases where we can guarantee it. Specifically, if $G$ is nilpotent but not abelian, then the answer is "yes". In particular, if $G$ is a nonabelian $p$-group, then the answer is "yes": there are always nontrivial elements in $Z(G/Z(G))$. 
A: The answer to this question depends on $G$. If $G$ is abelian, then there's no chance, since $G/Z(G)=\{Z(G)\}$.
On the other hand, consider the dihedral group given by symmetries of the square, whose center consists just of the identity and rotation by $\pi$. Any nontrivial coset will be in the center of $G/Z(G)\cong\mathbb{Z}/2\times\mathbb{Z}/2$, which is abelian, so in this case the answer is yes.
