which generators of SU(n) generate the elements of its center Zn? It is easy to show the center of SU(n) is $\mathbb{Z}_n$ from its definition, with the elements being $e^{i2\pi m/n}$. However, for $N>2$, I didn't see anywhere giving the explicit generators of them, i.e., some coefficients $\omega_j$ so that $e^{i\omega_jT^j}=e^{i2\pi/n}$, where $T^j$ are the generators in the fundamental representation. Is there a general expression of such coefficients $\omega_j$?
 A: Center of SU(N) in the fundamental representation
In the fundamental representation, the Lie algebra elements are $N\times N$ traceless Hermitian matrices. 
Use $e_{ij} $ to represent the matrix where only the element on ith row and jth column is nonzero, for any unitary matrix
\begin{equation}
U = \exp( i \sum_{i=1}^N x_i e_{ii} +  \sum_{i<j}  z_{ij} e_{ij}  + \sum_{i > j } \bar{z}_{ij}  e_{ij} ) \quad x_i \in \mathbb{R}, \sum_{i=1}^N x_i = 0 \,\, z_{ij} \in \mathbb{C}
\end{equation}
Elements in the center commute with any the other elements in the group, i.e. if $c \in Z(\text{SU}(N))$, then
\begin{equation}
U c  = c  U \implies c = Uc U^{\dagger} \quad \forall U \in \text{SU(N)}
\end{equation}
Now take $\partial_{z_{ij}}$ derivative on this expression and then evaluate the value at $U = I$, we have
\begin{equation}
0 = i e_{ij} c - ic e_{ij}  \implies [ e_{ij}, c] = 0
\end{equation}
The $\partial_{\bar{z}_{ij}}$ derivative extend the range of $i,j$ to all pairs of $i\ne j$. The computation of the commutator for one pair of $i,j$ shows that the ith row, jth row, ith column and jth column of $c$ are zero except for the diagonal element. Moreover the two diagonal elements are the same. Iterating over all pairs of $i,j$ then implies $c = \omega I$, where $\omega$ is a constant. 
Being a special unitary group element, c has $\det c = \omega^N = 1$, so $\omega$ is the root of unity. We have therefore constructed the center of the special unitary group:
\begin{equation}
Z( \text{SU(N)} ) = \left\{ \exp\left( 2\pi i \frac{m}{N} \right) I | m = 1, 2, \cdots, N \right \} = \mathbb{Z}_N
\end{equation}
and its element can be generated by taking 
\begin{equation}
x_i = \left\lbrace
      \begin{aligned}
        &2\pi \frac{m}{N} &  \quad  1 \le i \le N-1 \\
        &-2\pi \frac{m(N-1)}{N} & \quad  i = N\
      \end{aligned} \right. 
\end{equation}
