TITLE: With the exception of $\mathbb{Z}$, every infinite abelian group contains a subgroup isomorphic to $\mathbb{Z}^2$?
QUESTION [5 upvotes]: With the exception of $\mathbb{Z}$, every infinite abelian group
  contains a subgroup isomorphic to $\mathbb{Z}^2$.

Is this statement true? I don't have much experience working with non-finite groups so any advice would be appreciated, thanks!

REPLY [8 votes]: This is false. You could consider for a (non trvial) finite abelian group $G$ the group $\mathbb{Z} \times G $. 
But there are also infinite abelian groups that do not contain a copy of $\mathbb{Z}$ even. Take for example the product of infinitely many copies of $\mathbb{Z}/2 \mathbb{Z}$.
Also note that $(\mathbb{Q},+)$ does not contain a copy of $\mathbb{Z}^2$ (yet $(\mathbb{R},+)$ does).

REPLY [5 votes]: No, consider $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ : it's infinite, not isomorphic to $\mathbb{Z}$ (it has torsion) and does not contain $\mathbb{Z}^2$.