TITLE: Perfect $p$-group with nontrivial center
QUESTION [5 upvotes]: For a $p$-group $G$, which we know has a nontrivial center, can we have $G=G'$?
Obviously not when $G$ is finite. But the question makes sense for infinite groups (here $p$-group means that each element has finite order, which is a power of some given prime number $p$).

REPLY [2 votes]: You do not need a Tarski monster. Just take the free group $\langle a,b\rangle$ impose two relations of the form $a=w, b=w'$ satisfying a small cancellation condition, say, $C'(.000001)$ where $w,w'$ are products of commutators. The factor-group $G$ is hyperbolic (small cancellation) and perfect. Apply the method of Olshanskii from "Periodic quotient groups of hyperbolic groups" to obtain an infinite quotient $H$ of $G$ of prime exponent $p\gg 1$. Then the same proof as in Ashmanov-Olshanskii (it is in Olshanskii's book "Geometry of defining relations") proves that the Schur multiplier of $H$ is the free Abelian group of countable rank. Then take the universal central extension of $H$. 
Update: If a finite exponent is not needed, a periodic 2-generated $p$-group example can be constructed as a lacunary hyperbolic group, see my answer to this question.