Let $p$ be a fixed prime. A group $G$ is termed $p$-central if every element of order $p$ in $G$ lies in the center.
Having a finite $p$-group $G$ of rank $k$ (the least integer, such that every subgroup of $G$ is $k$-generated), one can prove that $G$ has a normal $p$-central subgroup of index bounded in terms of $k$. This can be proved by taking the centralizer $C$ of a maximal normal elementary abelian subgroup $A$ of $G$; the index of $C$ is clearly bounded in terms of the rank of $A$, and so in terms of $k$, and $C$ is $p$-central by a nice result of Alperin (or, in fact, of Thompson and Feit) which says that every element of order $p$ in $G$ which commutes with $A$ lies in $A$.
On the other hand, one can consider the set of all normal subgroups $N$ of $G$, for which $G/N$ is $p$-central. A nice fact is that the quotient of $G$ by the intersection $L(G)$ of all such subgroups, is still $p$-central; so $G/L(G)$ is the unique largest $p$-central quotient of $G$.
Is it true that that the order of $L(G)$ is bounded in terms of $k$ the rank of $G$?
Thanks in advance