I am trying to find the cyclic Sylow $l$-subgroups of $A = \Bbb Z_{p^3q} \times \Bbb Z_{p^2} \times\Bbb Z_{q^2r}$, where $p,q,r$ are prime numbers.
I have found the elementary divisor composition of $A$ as $\Bbb Z_{p^3} \times \Bbb Z_q \times\Bbb Z_{p^2} \times\Bbb Z_{q^2} \times Z_r$.
From what I have written in my notes, it seems that I can use the elementary divisor composition to find isomorphisms of the Sylow $l$-subgroups for $l = p, q, r$, and use those to determine which are cyclic -- but I'm not sure how to do that.
My gut feeling is that the Sylow $r$-subgroup is cyclic. But I can't really explain why that is.
The order of $A$ is $|A|=p^{5}q^{3}r$. So the $p,q,r$-sylow subgroups of $A$ is of orders $p^{5},q^{3},r$, respectively. First, "recall" that, in general, the $s$-sylow subgroups of a group $G$ are pairwise isomorphic, where $s\mid|G|$ prime. hence, the question is to find for which prime $l\in\{p,q,r\}$ there exists a cyclic $l$-sylow subgroup. Now, your guess is correct, because the subgroup $P_{r}$ of A generated by $(0,0,q^{2})$ is $r$-sylow (why?).
Second, lets denote $B=\mathbb{Z}_{p^{3}}\times\mathbb{Z}_{p^{2}}$ and $C=\mathbb{Z}_{q}\times\mathbb{Z}_{q^{2}}$, which are both non-cyclic groups. you may verify that the maps are $\varphi_{p}:B\rightarrow A$, given by $(1,0)\mapsto(q,0,0)$, $(0,1)\mapsto(0,1,0)$, and $\varphi_{q}:C\rightarrow A$, given by $(1,0)\mapsto(p^{3},0,0)$, $(0,1)\mapsto(0,0,r)$, are injective. It remains to note that $Im(\varphi_{p})$,$Im(\varphi_{q})$ are $p,q$-sylow subgroups of $A$, respectively.
Alternatively, you can use your decomposition to find out, in a more “direct” way, that there exist a $p,q$-sylow subgroups of $A$ which are, respectively, isomorphic to $\mathbb{Z}_{p^{3}}\times\mathbb{Z}_{p^{2}}$ and $\mathbb{Z}_{q}\times\mathbb{Z}_{q^{2}}$.