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> Is there a torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z \oplus \Bbb Z$ (as groups)?
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Notice that $\Bbb Z$ is [not cancellable](https://math.stackexchange.com/questions/349826), so 
$A \oplus \Bbb Z \simeq (A \oplus \Bbb Z) \oplus \Bbb Z$ doesn't imply that $A \simeq A \oplus \Bbb Z$. Combined with [this](https://mathoverflow.net/questions/259117) question, such a group $A$ would possibly provide an answer to [that](https://math.stackexchange.com/questions/2126096) question.


An example of torsion free abelian group $A$ such that $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$ was given [there](https://mathoverflow.net/questions/218113).
The example was the additive group of bounded sequences of elements of $\mathbb{Z}[\sqrt{2}]$, i.e. 
$$A = 
\left\{
(x\_n)\_{n \geq 1} \subset \Bbb Z[\sqrt 2] \;\;:\;\;
\exists C>0,\; \forall n \geq 1,\; |x\_n| \leq C
\right\}$$
I wasn't able to adapt this example in order to answer my question.



Torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z^2$
$\mathbb{Z}$ is cancellable for *abelian* groups. This was proved in the 1950s by Walker and Cohn (independently) and is often called "Walker's cancellation theorem". The proof is only a few lines.


So if $A$ is an abelian group with $A\oplus\mathbb{Z}\cong A\oplus\mathbb{Z}^2$, then $A\cong A\oplus\mathbb{Z}$.


