How "big" can the center of a finite perfect group be? A perfect group is a group where the derived (commutator) subgroup $G'$ of $G$ equals $G$. $G'$ measures the "non-abelian-ness" of $G$, in a sense. This suggests that "many" elements of perfect $G$ don't commute, though "many" others must commute in any finite group, given that any element of a finite group generates a cyclic subgroup.
It occurred to me that the center $Z(G)$ of $G$ might provide some guidance: e.g. the center (elements that commute with everything) of a perfect group must be "small", perhaps as small as trivial (e.g. the center of perfect $A_5$ is trivial). But some Googling suggests that the center of a perfect group need not be trivial (Gr$\ddot{u}$n's lemma: $G/Z(G)$ is centerless, but $Z(G)$ may be nontrivial). I think I'm interpreting this correctly, but I could be wrong!

Question: Does $G$ being a finite perfect group say anything about the structure of $Z(G)$? In particular, is there a lower bound on its index $[G:Z(G)]$ in terms of $|G|$?

 A: There are two nice bounds,
Let's $|G:Z(G)|=n$ and $G=G'$
1.) There is a minimal generating set $S$ for  $G$ such that $|S|\leq n^2$
2.) For all $g\in G$, $g^n=e$.
First arguments proof comes from the idea that $[x,y]=[z_1u,z_1s]$ for $z_1,z_2\in Z(G)$. By manupulating, we will get $[x,y]=[u,s]$. Hence every commutator can be written as a commutator of elemets which is taken from left coset representator of $Z(G)$. We will get $n^2$ diffrent commutator.
Second arguments proof comes from transfer theory and it is nontrivial fact.
A: There are such bounds, but $|Z(G)|$ can be bigger than you might expect.
The Schur Multiplier of a $p$-group $P$ of order $p^n$ has order at most $p^{n(n-1)/2}$ (see here), which is about $|P|^{\log |P|}$. For a perfect group $G$, $Z(G)$ is a quotient of the Schur Multiplier of $G/Z(G)$, and the order of this is at most the product of the orders of the Schur Multipliers of its Sylow subgroups, so we get $|Z(G)| \le n^{\log_2 n}$, where $n = |G/Z(G)|$, which is an improvement on the bound given by mesel's answer.
Some improvements on this may be possible, but there are examples in which $|Z(G)| = n^{O(\log n)}$. Specifically, let $p \ge 5$ be prime, and let $V$ be the natural module for ${\rm SL}(2,p)$. Then there is a perfect group $G$ with $Z(G)$ elementary abelian of order $p^{n(n+1)/2}$, and $G/Z(G)$ an extension of $n$ copies of $V$ by ${\rm SL}(2,p)$. So $|G/Z(G)| = p^{2n+1}(p^2-1)/2$.
