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$\mathbf{Question}$: Consider an infinite group $G$. Let $\langle a_1\rangle , \langle a_2\rangle, \ldots, \langle a_m\rangle $ ($\langle a_i\rangle =C_i$) be a finite collection of infinite cyclic subgroups of $G$ such that $C_i \not\subset C_j$ for $i \neq j$. Then $\displaystyle\bigcup_{i=1}^mC_i$ is never a subgroup of $G$. ($m>1$)

$\mathbf{A \ restricted \ version:}$ Consider an infinite commutative group $G$. Let $<a_1>, <a_2>, ..., <a_m>$ ($<a_i>=C_i$) be a finite collection of infinite cyclic subgroups of $G$ such that $C_j \not\subset \displaystyle\bigcup_{I \setminus \{j\}}C_i$ , $I=\{1,2,3,...,m\}$. Then $\displaystyle\bigcup_{i=1}^mC_i$ is never a subgroup of $G$. ($m>1$)

I don't know how to proceed. Any help would be appreciated.

Subhasis Biswas
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    The statement is certainly wrong as stated. I mean, what happens if $m=1$? – Arturo Magidin Jun 22 '19 at 19:30
  • @ArturoMagidin $m>1$. It is certainly not of much interest when $m=1$. But yes, it must be mentioned that $m>1$ – Subhasis Biswas Jun 22 '19 at 22:34
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    Then fix it! We aren’t responsible for reading your mind of figuring out what conditions you should have put into your query. If you don’t put them, then they aren’t there. So right now, you’re wrong. If you don’t want to be wrong, then fix it in the body of the problem, not flippantly in the comments. – Arturo Magidin Jun 22 '19 at 23:19
  • Ok. So is there a thing like $\cup_{i=1}^1$? The reason I did not mention it, because I have never seen the consideration anywhere. – Subhasis Biswas Jun 22 '19 at 23:24
  • @ArturoMagidin I will try to keep this in mind – Subhasis Biswas Jun 22 '19 at 23:38
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    I do not see on what grounds you claim to be able to omit $C_1$ from the union. – Arturo Magidin Jun 23 '19 at 00:47
  • The proposed proof does not explicitly use the assumption that each $a_i$ has infinite order, which is necessary given the example involving $\mathbb{Z}_2\times\mathbb{Z}_2$. – Gabe Conant Jun 23 '19 at 02:20
  • @GabeConant Would the fact "every infinite cyclic group has only two generators" be of any help? – Subhasis Biswas Jun 23 '19 at 05:17

2 Answers2

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This is not a full proof, but it excludes large classes of groups and is way too long to fit into a comment.

I'm going to prove that no nontrivial group $G$ is the union of finitely many infinite cyclic subgroups, unless $G=\Bbb Z$ or $G$ is a noncommutative group with one end.

Suppose that $G=\bigcup_{i=1}^n \langle c_i\rangle$ where $n>1$ and $\langle c_i\rangle\simeq \Bbb Z$ for every $i$, in particular every element in $G$ has infinite order and $G$ is finitely generated. By a famous result of Freudenthal and Hopf $G$ has either $1,2$ or infinitely many ends, we look at this cases separately.

If $G$ has a single end and $G$ is abelian then $G\simeq\Bbb Z^m$, for $m>1$, by the structure theorem for finitely generated abelian groups, and it's easy to show that $\Bbb Z^m$ is not the union of finitely many cyclic subgroups, by explicitely constructing an element not contained in such an union.

If $G$ has a single end and $G$ isn't abelian I can't conclude the proof, and it's the only missing case.

If $G$ has two ends then another result of Freudenthal and Hopf shows that $G$ must be virtually $\Bbb Z$ and it's known that a torsion free virtually $\Bbb Z$ group is either trivial or $\Bbb Z$ (see for example here for a reference), which concludes the proof in this case (since it's easy to show that $\Bbb Z$ isn't the union of more than one infinite cyclic subgroup).

If $G$ has more than two ends then it must have infinitely many. Pick a ball in the Cayley graph whose complement has infinitely many connected components. The subspace $\langle c_i\rangle$ of the Cayley graph (meaning the vertices $c_i^n$ for $n\in\Bbb Z$ and the edges joining $c_i^n$ to $c_i^{n+1}$) is connected for every $i$, so it is clear that finitely many such subspaces can't cover the complement of the chosen ball, hence they can't cover the vertices of the Cayley graph either and thus $G$ is not the union of the $\langle c_i\rangle$, a contradiction.

Alessandro Codenotti
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  • I'm still a little confused on your infinite ends case, though. By "ball", do you mean a (word)-metric ball of finite radius? Because those don't ever have to have infinitely many connected complements, just arbitrarily many. Consider a free group. Even then, you're also disconnecting the cyclic groups, so it's not clear why the number of connected components of each cyclic group can't also grow without bound. – Charlie Cunningham Jun 23 '19 at 14:41
  • Ah right, I agree that this doesn't work as written, what I want to say is that I'm looking at the Cayley graph wrt the generating set ${c_i}$, so that every $\langle c_i\rangle$ is a connected subspace of the Cayley graph and it can cover at most two ends. To see that pick a ball $B_n$ and fix $i$. There is a minimum $z\in \Bbb Z$ with $c_i^z\in B_n$ (since $B_n$ is finite). Since $\langle c_i\rangle$ is connected the subgraph induced by ${c_i^k\mid k<z}$ covers at most one end. There is also a biggest $l$ such that $c_i^k\not\in B_n$ for $k>l$. The subgraph induced by – Alessandro Codenotti Jun 23 '19 at 14:57
  • ${c_i^k\mid k>l}$ covers at most one end, again by the connectedness of $\langle c_i\rangle$. $\langle c_i\rangle\setminus B_n$ could have more connected components, but they are all finite except those two, so they can't cover any end. (hopefully it works now) – Alessandro Codenotti Jun 23 '19 at 14:57
  • When I say "by connectedness of $\langle c_i\rangle$ I really mean "because taking the complement of a ball always splits it into two infinite connected components and a bunch we don't care about". The point is that $\langle c_i\rangle$ can't jump between two connected components of the complement of a ball in the Cayley graph without going through that ball again. This is getting messy to write, it might be easier to argue that $\langle c_i\rangle$ is not ramifying... – Alessandro Codenotti Jun 23 '19 at 15:02
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    Here's how I would phrase it: since there are infinitely many ends, consider 2n+1 of them. Each end is a distinct sequence of nested, unbounded, connected components of the complements of $B(e,k)$, so there must be a finite r such that each of these ends is a different unbounded component of $X = \Gamma \setminus B(e,r)$. But as argued, each $X \cap \langle c_i \rangle$ can have only 2 unbounded components, so there can be at most 2n unbounded components of X, which is a contradiction. – Charlie Cunningham Jun 23 '19 at 17:46
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Suppose $G$ is a group and $G=\bigcup_{i=1}^m C_i$ where each $C_i$ is an infinite cyclic subgroup. Note that $G$ must be torsion-free. By B. H. Neumann’s Lemma, some $C_i$ has finite index. So $G$ is torsion-free and virtually cyclic (i.e., has a cyclic subgroup of finite index). It follows that $G$ is cyclic (see this answer).

Now it’s easy to show that that we can’t have $C_i$ and $C_j$ mutually incomparable when $i\neq j$ (unless $m=1$).

Gabe Conant
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