I have the next problem:
Let $G$ an abelian and finitely generated group. Suppose that $a\in G$ has infinity order. Prove that: $G$ isn't cyclic if and only if there exists $b\in G\setminus\{e_G\}$ such as $\langle a\rangle\cap \langle b\rangle = \{e_G\}$.
The left implication is simple: We have $\langle a\rangle \langle b\rangle \cong \langle a\rangle\times \langle b\rangle$ because $G$ is abelian and $\langle a\rangle\cap \langle b\rangle = \{e_G\}$. If $\langle a\rangle\times \langle b\rangle$ is cyclic then there exists $(m,n)\in \mathbb{Z}\times\mathbb{Z}$, $m,n\neq 0$ such as $\langle a\rangle\times \langle b\rangle = \langle (a^m,b^n)\rangle$, then $m=1$ because $\text{ord}(a) = \infty$, so there exists $j$ such as $(a,b^{n+1}) = (a,b^n)^j = (a^j,b^{nj})\Rightarrow j=1\Rightarrow b^{n+1} = b^n\Rightarrow b=e_G$, contradiction. So $\langle a\rangle\times \langle b\rangle$ isn't cyclic, then $\langle a\rangle \langle b\rangle$ isn't cyclic and finally $G$ isn't cyclic.
For the right implication, I don't have any idea about how to start. I will be very grateful if you can help me.