TITLE: $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$
QUESTION [90 upvotes]: Are there abelian groups $A$ with $A \cong A \oplus \mathbb{Z}^2$ and $A \not\cong A \oplus \mathbb{Z}$?

REPLY [2 votes]: Also not an answer, but the following Boolean algebra gives a similar example of William Hanf.
Let B be the two element boolean algebra, and we will use an uncountable subuniverse of a countable power of B.  Let A be the subalgebra with subuniverse $\{\sigma \in \{0,1\}^{\omega}:$ there is $n$ so that for all $k \geq n, \sigma(2k)=\sigma(2k+1)\}$.  Then A Is isomorphic to A $\oplus$ B $\oplus$ B , but not to A $\oplus$ B .
This suggests putting $Z$ in for B and seeing what comes of it.
For more details, consult Algebras, Lattices, Varieties Volume I of McKenzie, McNulty, and Taylor, chapter 5 section 5.1 exercise 9 (p. 267 in my copy).  I do not know what happens if the signature is restricted to that of Abelian groups.
Gerhard "Not Feeling That Ambitious Today" Paseman, 2016.01.02