Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)?




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Edited to add: As no answers are forthcoming, does anyone know what happens if we allow arbitrary modules in place of abelian groups?



$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$
This is not a complete answer, but a construction that might give an answer.


I'll start by constructing a ring with several objects (a.k.a. preadditive
category) $\mathcal{C}$ by generators and relations, using similar ideas as in
Leavitt's ring with $R\not\cong R\oplus R\cong R\oplus R\oplus R$.


The objects $X\_{1}, X\_{2},\dots$ are indexed by the positive integers. For each
$m\leq n$ there are generators $\alpha\_{n,m},\beta\_{n,m}: X\_{n}\to X\_{m}$ and
$\gamma\_{n,m},\delta\_{n,m}:X\_{m}\to X\_{n}$.


The relations are designed to make
$$X\_{n}\oplus X\_{n}\cong X\_{1}\oplus X\_{2}\oplus\cdots\oplus X\_{n}$$
for all $n$ once we close under finite direct sums. For example, for $n=3$
impose the relations that make the matrices
$$
\begin{pmatrix}
 \alpha\_{3,1}&\beta\_{3,1}\\\alpha\_{3,2}&\beta\_{3,2}\\\alpha\_{3,3}&\beta\_{3,3}
\end{pmatrix}
\text{ and }
\begin{pmatrix}
 \gamma\_{3,1}&\gamma\_{3,2}&\gamma\_{3,3}\\\delta\_{3,1}&\delta\_{3,2}&\delta\_{3,3}
\end{pmatrix}
$$
mutually inverse.


Now form a ring by adjoining a unit:
$$R=\mathbb{Z}\oplus\bigoplus\_{m,n}\operatorname{Hom}\_{\mathcal{C}}(X\_{m},X\_{n}).$$
For each $n$ let $e\_{n}\in R$ be the idempotent corresponding to the
identity endomorphism of $X\_{n}$, let $P\_{n}=e\_{n}R$ be the corresponding
projective right $R$-module, and let $P=\bigoplus\_{n}P\_{n}$.


Then $P\_{n}\oplus P\_{n}\cong P\_{1}\oplus P\_{2}\oplus\cdots\oplus P\_{n}$ for each
$n$, so
\begin{align}
 P\oplus P&\cong (P\_{1}\oplus P\_{1})\oplus(P\_{2}\oplus P\_{2})\oplus(P\_{3}\oplus
 P\_{3})\oplus\cdots\\
 &\cong P\_{1}\oplus (P\_{1}\oplus P\_{2})\oplus(P\_{1}\oplus P\_{2}\oplus
 P\_{3})\oplus\cdots\\
 &\cong P^{(\omega)}
\end{align}


**Question 1.** Is $P\cong P\oplus P$?


Probably not, as there seems no obvious way to produce such an isomorphism.


If the answer to Question 1 is ``no'', then this answers the supplementary
question in the OP about modules.


As for abelian groups:


Clearly $R$ is countable.


**Question 2.** Is $R$ torsion free?


Probably, as there seems no obvious way to produce a torsion element.


**Question 3.** Is $R$ reduced?


Probably, as there seems no obvious way to produce a divisible element.


If the answers to Questions 2 and 3 are both ``yes'', then Corner's theorem
applies, and $R\cong\operatorname{End}(B)$ for some abelian group $B$.


Let $A\_{n}=e\_{n}(B)$ and $A=\bigoplus\_{n}A\_{n}$. Then
$$A\oplus A\cong A^{(\omega)}.$$


**Question 4.** Is $A\cong A\oplus A$?


I don't see any reason that it should have to be (note that $B$ is not uniquely
determined, so the answer to Question 4 might conceivably depend on the
particular choice of $B$).


