Let us consider $ SU(2)$ irrep. with a $1\times k$ Young tableaux. Then the center group of such linear representation depends on the value of $k$ in the following sense. If $k\in$odd, then the center is nontrivial, namely $Z_2$, while the center is trivial when $k\in$even.
I wonder whether it can be generalized to $SU(N)$ irrep. with a general Young tableaux consisting of $b$ blocks. What I can guess or expect is the result should be that the center is $Z_{N/(b,N)}$ where $(b,N)$ is the greatest common divsor of $b$ and $N$. Is this statement correct? If so, how to prove it?
The matrix representing the center generator of $SU(N)$ can be calculated within some (larger) irreducible representation since the matrix is proportional to the identity matrix. More specifically, any Young-tableaux representation with $b$ blocks is contained in the $b$-fold tensor product of fundamental representations, namely \begin{eqnarray} \text{irrep}(b)\in\otimes_1^bN. \end{eqnarray} Thus we can calculated the prefactor by the right-hand side: \begin{eqnarray} Z(b)&=&\left[\exp(i2\pi/N)\right]^b\nonumber\\ &=&\exp(i2\pi b/N), \end{eqnarray} which generates $Z_{N/(b,N)}$.