# About finite covering by cyclic subgroups

A covering of a group $G$ a family $\{S_i\}_{i \in I}$ of subsets of $G$ such that $G = \displaystyle \bigcup _{i \in I} S_i$.

Why is true that: A group covered by finitely many cyclic subgroups is either cyclic or finite?

Remark: Is true that by Baer (see D. Robinson, Finiteness Conditions and Generalized Soluble Groups, p.105) that: Theorem 4.16: A group is central-by-finite if and only if it has a finite covering consisting of abelian subgroups?

• What does that remark have to do with the question? That stuff by Baer is way more general, deep and advanced that the question. Feb 13, 2015 at 16:51
• @Timbuc This result Baer gives us a condition for a group have a finite covering by Abelian subgroups. Of course, every cyclic group is abelian. So I figured that this result could help us in the pursuit of question I asked. And even that did not help not believe that could mess.
– user59969
Feb 13, 2015 at 17:29
• @Timbuc The result of Baer allows you to assume that $G$ ia central-by-finite, and then it's not hard to reduce to infinite cyclic-by-finite. But if you think that the question is easy without using Baer's theorem, then why not give a hint? Feb 13, 2015 at 17:32
• @DerekHolt Sorry, but I have not a clue in this case. Actually I was reading an article, and they mentioned that it was true that my question, but the theorem cited in my remark as being the answer. I do not understand how one can imply the other.
– user59969
Feb 13, 2015 at 17:36
• The Article is "On finite groups in which coprime commutators are covered by few cyclic groups" and the authors are: Cristina Acciarri and Pavel Shumyatsky.
– user59969
Feb 13, 2015 at 17:37

Some partial elementary results:

Assume $G$ is infinite and has a finite covering by finitely many cyclic subgroups $S_i=\langle s_i\rangle$, $i\in I$. We have to show that $G$ is cyclic.

If $S_i\subseteq S_j$ for some $i\ne j$, we may drop $S_i$. Hence we may assume wlog. that $S_i\not\subseteq S_j$ for $i\ne j$. Consequently, none of the $S_i$ is properly contained in a larger cyclic subgroup. Any conjugate of an $S_i$ is contained in some $S_{i'}$ and by the previous remark must in fact equal $S_{i'}$. Thus $G$ acts on the set $\{\,S_i:i\in I\,\}$ by conjugation.

If all $S_i$ are finite then so is their union, hence some of the $S_i$ must be infinite. Hence there exist subsets $J\subseteq I$ such that $\bigcap_{j\in J}S_j$ is nontrivial and each $S_j$, $j\in J$, is infinite. Let $J$ be a maximal such subset and let the corresponding intersection be $\langle a\rangle$.

Claim. $\bigcup_{j\in J}S_j=G$.

Proof. Assume $g\notin \bigcup_{j\in J}S_j$. Then the infinitely many $a^kg$ are not in any $S_j$ with $j\in J$. By pigeon-hole we find distinct $a^kg, a^lg$ in the same $S_i$, $i\notin J$, and then also $a^{k-l}\in S_i$. This contradicts maximality of $J$. $_\square$

So wlog. $I=J$. Since $a$ commutes with each $s_i$, the center has finite index in $G$, and each $S_I$ has finite index.

Here are some hints. I denote a cyclic group of order $n$ by $C_n$.

Step 1. First show that $C_{\infty} \times C_n$ is not covered by finitely many cyclic groups for any $n$, including $n=\infty$.

Note that the condition of being covered by finitely many cyclic ubgroups is inherited by subgroups. By the Baer result, for any such group $G$, $G/Z(G)$ is finite, and $Z(G)$ is clearly finitely generated, so by Step 1, $Z(G)$ is either finite, in which case we are done, or equal to $C_\infty$.

Now by a result of Schur, $[G,G]$ is finite, so it has trivial intersection with $Z(G)$. If $1 \ne g \in [G,G]$ then $\langle Z(G),g \rangle = Z(G) \times \langle g \rangle \cong C_\infty \times C_n$ for some $n$, contradicting Step 1. So $[G,G]=1$ and $G=Z(G) = C_\infty$.