A finite union of infinite cyclic subgroups of a group $G$ is never a group. $\mathbf{Question}$: 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. 
 A: 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.
A: 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 $m=1$). 
