For a while I thought that, 'intuitively,' $G/Z(G)$ where $G$ is nonabelian with nontrivial center would have trivial center - that is, it would be 'totally nonabelian.' I realize now that this is not necessarily true, as in cases where $G/Z(G)$ has order $p$ or $p^2$ where $p$ is prime.
I also realize that what fed this idea and made it feel intuitive was the use of phrases about "modding out" or "quotienting out" normal subgroups. To me, this phrasing makes it sound like we're sort of 'removing' that subgroup. And indeed we are, in some cases. $\mathbb{Z}_6$ has elements of orders 1, 2, 3, and 6. Taking $\mathbb{Z}_6 / \{0,3\}$ (that is, 'quotienting out' a subgroup of order 2) we obtain $\mathbb{Z}_3$, and have effectively 'removed' the elements of order 2, since $\mathbb{Z}_3$ has only elements of orders 1 and 3.
So my question is: are there specific conditions under which $G/Z(G)$ (as described above) has trivial center?
More generally, can we say anything about when taking the quotient by a normal subgroup actually 'removes' the elements that characterize that subgroup?