Finite group's center Let $N$ be a normal subgroup of group $G$. There exists a prime $p$ such that $G/O^p(N)$ is simple group. Is there any element in the center of group $G$?
 A: Yes, there can be non-identity elements of the center of $G$.
Let $G=N$ be cyclic of order $p$. Then $O^p(N) = \langle n \in N : \gcd(p,|n|)=1 \rangle = 1$ and $G/O^p(N)$ is simple of order $p$.
For an example with $G \neq N$, take $G$ perfect of order 120, $N=Z(G)$ of order 2, and $p=3$, then $O^p(N) = N$ and $G/N$ is simple of order 60.

How can $G/O^p(N)$ be simple if $O^p(N) \neq N$? There is only one way: for $N=G$ and $G/O^p(G)$ be cyclic of order $p$. Then of course this can happen: take $G=N=S_5 \times C_3$ of order 360, and $p=2$, then $O^p(N) = A_5 \times C_3$ is order 180, and the quotient is simple of order 2.
The p-residual
If $N$ is a finite group and $\mathcal{X}$ is a class of finite groups closed under isomorphism, quotients, subgroups, and finite direct products, then $O^{\mathcal{X}}(N) = \bigcap\{ M \unlhd N : N/M \in \mathcal{X} \}$ is the unique normal subgroup of $N$ such that $N/M \in \mathcal{X}$ if and only if $M \geq O^{\mathcal{X}}(N)$. If $\mathcal{X}$ is the class of abelian groups, then $O^{\mathcal{X}}(N) = [N,N]$ is the derived subgroup. If $\mathcal{X}$ is the class of $p$-groups, then $O^{\mathcal{X}}(N) = O^p(N)$ is the subgroup in the question.
One can also view $O^p(N)$ as the subgroup generated by the Sylow $q$-subgroups for $q \neq p$, as Cauchy's theorem describes the only obstruction to being a $p$-group is having non-identity elements of order a power of a distinct prime $q$.
Obviously $O^p(N)$ is a characteristic subgroup of $N$ and contains all elements of order coprime to $p$. In particular, it can be a very large subgroup, so quotienting out by it can lose all sorts of information, including the center. Requiring $G/O^p(N)$ to be simple merely requires $N$ and $O^p(N)$ to be “large”, so it makes it very easy for them to contain central elements (which are then lost during the quotient).
