I know that $\mathbb Z_n$ is the center of the Lie group $SU(n)$. Therefore, the quotient $SU(n)/\mathbb Z_n$ forms a group. However, I am not sure whether it also forms a Lie group. I also wonder what is $\mathfrak{su}(n)/\mathbb Z_n$ at the Lie algebra level.
So far, I have only seen examples of quotient group $G/H$ in which the quotient $H$ is itself a connected Lie group; then if $H$ is a normal, closed, connected Lie subgroup, $G/H$ is a Lie group. However, I don't have intuition for what we would obtain if we quotient a Lie group by a finite group.
In general, if $G$ is a Lie group and $N$ is a normal subgroup which is closed (in the topological sense), then $G/N$ is a Lie group. If $N$ is discrete, then the projection map $\pi : G \to G/N$ is a covering map. In particular, there is an isomorphism $(d\pi)_e : T_eG \to T_e(G/N)$, so the Lie algebras of $G$ and $G/N$ are isomorphic.
Therefore $SU(n)/\mathbb{Z}_n$ is a Lie group and its Lie algebra is isomorphic to $\mathfrak{su}(n)$.