TITLE: A finite union of infinite cyclic subgroups of a group $G$ is never a group.
QUESTION [5 upvotes]: $\mathbf{Question}$: 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.

REPLY [4 votes]: 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 $m=1$).