what's the pro-cyclic group with a given order?

cyclic groups are classified according to their orders. but for procyclic groups, (procyclic) means:

the group is a profinite group $$G$$, such that there exists $$g \in G$$, the smallest closed subgroup containing $$g$$ is $$G$$. this is different from the cyclic group case.

to classify procyclic groups I have 2 questions: (1) how to prove a procyclic group the direct product of its sylow p subgroups?

(2) why this definition is equivalent to: $$G$$ is an inverse limit of cyclic groups?

can anyone give me some hint? or is this a classical result? thanks.

• You may want to give more background. Are procyclic groups topological groups (you say they have closed subgroups)? Are they inverse limits of cyclic groups? Commented Dec 13, 2019 at 15:02
• That's a strange definition. If you put a trivial topology on a group then isn't it procyclic? In that situation it can be arbitrary. Commented Dec 13, 2019 at 16:01
• oh I'm sorry, I should make the question clear. Commented Dec 15, 2019 at 0:51

A pro-cyclic group is isomorphic to a Cartesian product $$\prod_{p\in \mathbb{P}} \frac{\mathbb{Z}_p}{p^{n(p)}\mathbb{Z}_p},$$ with $$n(p)\in \mathbb{Z}_{\ge 0}\cup \{\infty\}$$. This is a quotient of $$\widehat{\mathbb{Z}} = \prod_{p\in \mathbb{P}}$$, where for each prime $$p$$ you pick a pro-cyclic pro-$$p$$ group. So for each Steiniz number $$\prod_{p\in \mathbb{P}} p^{n(p)}$$ there is exactly one pro-cyclic group of that order.
For (2), the definition of pro-cyclic is that it is an inverse limit of finite cyclic groups. Given an inverse limit of finite cyclic groups $$G$$, a standard inverse limit argument shows that there exists $$g\in G$$ such that $$G=\overline{\langle g\rangle}$$. On the other hand, if $$G=\overline{\langle g\rangle}$$, then $$G\cong \varprojlim_{N\lhd_o G} G/N = \varprojlim_{N\lhd_o G} \langle g \rangle N/N \cong \varprojlim_{N\lhd_o G} \langle g \rangle/(N \cap \langle g \rangle)$$ and the quotient $$\langle g \rangle/(N \cap \langle g \rangle)$$ is a finite cyclic group, so it is on the form $$\prod_{p\in \mathbb{P}} C_{p^{m(N)}}$$ (with appropriate bounds on number of primes and consistency conditions). Thus, $$\varprojlim_{N\lhd_o G} \langle g \rangle/(N \cap \langle g \rangle) \cong \prod_{p\in \mathbb{P}} \varprojlim_{N\lhd_o G} C_{p^{m(N)}}$$ which is isomorphic to the first Cartesian product.