It's a fundamental result that if $G$ is a connected Lie group then the universal covering map $\widetilde{G} \to G$ has kernel a central subgroup of $\widetilde{G}$ isomorphic to $\pi_1(G)$, giving a canonical short exact sequence
$$1 \to \pi_1(G) \to \widetilde{G} \to G \to 1$$
exhibiting $\widetilde{G}$ as a central extension of $G$ by $\pi_1(G)$. The upshot is that to exhibit a (finitely generated) abelian group $\pi$ as the fundamental group of a Lie group it suffices to find a simply connected Lie group in which $\pi$ sits as a central subgroup.
Now we just observe that, for example, the center of $SU(n)$ is $\mathbb{Z}/n\mathbb{Z}$, so central quotients of $SU(n)$ (the maximal one of which is $PSU(n) \cong SU(n)/Z(SU(n))$ which satisfies $\pi_1(PSU(n)) \cong \mathbb{Z}/n\mathbb{Z}$) have $p$-torsion in their fundamental group for suitable $n$. $SU(n)$ can also be replaced by $SL_n(\mathbb{C})$ if you prefer algebraic groups to compact groups.