Does the center of a perfect group not contain all elements of prime order? 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?
 A: Short version: if $p$ is odd and all elements of order $p$ are central in $G$, then $G$ has a normal $p$-complement, i.e., a normal $p'$-subgroup $K$ such that $|G:K|$ is a power of $p$. This follows from Theorem 5.3.10 from Gorenstein, which states that if $p$ is odd and a $p'$-automorphism of a $p$-group $P$ acts trivially on $\Omega_1(P)$ then it is the identity.
Thus, if $G$ has this property for any odd prime then $G$ is not perfect, because it has a $p$-quotient.
Original post follows, which says things about $p=2$, and I leave here for posterity, and for noting that I completely forgot about that theorem from Gorenstein's book.

I originally thought you meant every prime dividing $|G|$. This question is a lot harder than I first thought.
Notice that the property that all elements of prime order being central is inherited by subgroups. In particular, if $H$ is a normal subgroup of $G$ then the soluble residual of $H$, i.e., the last term in the derived series for $H$, satisfies your conditions.
Let's start with $p=2$, and let $G$ be a counterexample to your claim. Bob Griess proved in the 1978 paper Finite groups whose involutions lie in the center that if all involutions of $G$ lie in the centre of $G$ and $O_{2'}(G)=1$ then the soluble residual of $G$ is a direct product of $\mathrm{SL}_2(q)$s, or a central extension of $A_7$.
Let $H$ be the normal subgroup $O_{2'}(G)X$, where $X$ is one of the direct factors, and let $H_1=H^{(\infty)}$ be its soluble residual. Then $H_1$ is also a
counterexample, and is non-trivial as it has a simple composition factor. Thus $H_1=G$, and we may assume that $X=G/O_{2'}(G)=\mathrm{SL}_2(q)$ or $A_7$.
Now I have to leave, but my current plan is to choose a prime $p$ such that the Sylow $p$-subgroup of $X$ is cyclic, quotient out by $O_{p'}(G)$, and show that the Sylow $p$-subgroup doesn't have the property. I will return! Unless someone else solves it first.
EDIT!!! I should have read Bob's paper more. Remark: if $p$ is an odd prime and $G$ has this property for elements of order $p$ then $G$ is $p$-nilpotent.
A: 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.
A: Suppose $G$ is not trivial. Since $G'=G$, then $G$ is not abelian and $Z(G)$ is a proper subset of $G$. Pick an element $g$ of $G-Z(G)$ and consider the subgroup $H=\langle g\rangle$. If every element of $H$ with prime order is in $Z(G)$, then $H\leq Z(G)$, which is false since $g\notin Z(G)$. Therefore, at least one of those elements is not in $Z(G)$.
EDIT:
From the comments below I realized that "If every element of $H$ with prime order is in $Z(G)$, then $H\leq Z(G)$" is false. The proof works only if the order of $g$ is a product of distinct primes... which is not the general case.
