How "big" can the center of a finite perfect group be?

Solution 1:

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.

Solution 2:

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 $|G/Z(G)| = p^{2n+1}(p^2-1)/2$.