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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)?

Notice that $\Bbb Z$ is not cancellable, 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 question, such a group $A$ would possibly provide an answer to that 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. 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.

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    $\begingroup$ I feel like I am lost in a maze of twisty little questions, all alike which all ask whether there exists foo which is isomorphic to foobarbar but not to foobar or something confusingly similar. It's one of the drawbacks of stackexchange that there isn't really a way to provide a summary of all of these questions at once. $\endgroup$ – Gro-Tsen Feb 5 '17 at 18:25
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$\mathbb{Z}$ 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 $A\cong A\oplus\mathbb{Z}$.

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  • $\begingroup$ Thank you very much! I didn't search enough, since it was already asked there (where the references of the papers are also given). $\endgroup$ – Watson Feb 5 '17 at 11:24

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