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.

  • $\begingroup$ 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? $\endgroup$ Commented Dec 13, 2019 at 15:02
  • 1
    $\begingroup$ 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. $\endgroup$
    – freakish
    Commented Dec 13, 2019 at 16:01
  • $\begingroup$ oh I'm sorry, I should make the question clear. $\endgroup$
    – youknowwho
    Commented Dec 15, 2019 at 0:51

1 Answer 1


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 (1), as any pro-nilpotent group, a pro-cyclic group is the product of its Sylows.

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.


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