TITLE: Does the center of a perfect group not contain all elements of prime order?
QUESTION [7 upvotes]: Let $G$ be a finite perfect group (i.e. $G=G'$) and $Z(G)$ be its center.
I don't know whether this statement is correct:

There exists an element $x$ of prime order such that $x\notin Z(G)$.

A quick check on CFSG gives that this holds for every (quasi-)simple group. But what if $G$ is a general finite perfect group? Or is there any further descriptions on the center of perfect groups?

Another description on this question is (also I don't know if this holds):

Let $H$ be a center-less (insoluble) group (i.e. $Z(H)=1$). Then there always exists a prime divisor $p$ of $|H|$ such that the $p$-part of the Schur multiplicator of $H$ is trivial.

Is there any result on both?

REPLY [6 votes]: Your second question is much easier. Yes there exist centreless - and even perfect - finite groups $G$ such that the $p$-part of the Schur Multiplier of $G$ is nontrivial for all primes $p$ dividing $|G|$.
For example there is such a group with structure $(3^4 \times 5^3):A_5$. The primes dividing the order are $2,3,5$ and the Schur Multiplier has order $30$. You can construct lots of examples in this fashion.