Are there abelian groups $A$ with $A \cong A \oplus \mathbb{Z}^2$ and $A \not\cong A \oplus \mathbb{Z}$?

11$\begingroup$ Let me ask a weaker question: is there a compact Hausdorff space $X$ such that $X$ is homeomorphic to $X\times \mathbb{T}^2$ but not homeomorphic to $X\times \mathbb{T}$? $\endgroup$– Tomasz KaniaSep 12, 2015 at 11:49

7$\begingroup$ @Hurkyl: Not necessarily; consider the case of $A=\mathbb{Z}^\mathbb{N}$ where $p$ is the unilateral shift. In general, you can conclude that $\mathbb{Z}^\infty$ is a subgroup of $A$, but not necessarily that it is a summand. $\endgroup$– Eric WofseySep 12, 2015 at 18:49

4$\begingroup$ If I remember correctly, there exist abelian groups $A$ such that $A\cong A\oplus A\oplus A$ but $A\not\cong A\oplus A$. If this is the case, then perhaps $A$ or $A\oplus \mathbb{Z}$ will provide an answer to the present question. Note that $A\oplus\mathbb{Z}\cong A\oplus \mathbb{Z}^3$. $\endgroup$– Richard StanleySep 12, 2015 at 19:13

6$\begingroup$ @RichardStanley Why should $A\oplus\mathbb{Z}\cong A\oplus\mathbb{Z}^3$? Unless both $A$ and $A\oplus\mathbb{Z}$ have the $X\cong X\oplus X\oplus X$ property. $\endgroup$– Jeremy RickardSep 13, 2015 at 5:52

5$\begingroup$ @JeremyRickard: you are right. Haste makes waste. I withdraw my previous comments. $\endgroup$– Richard StanleySep 13, 2015 at 20:15
2 Answers
Let $A$ be the additive group of bounded sequences of elements of $\mathbb{Z}[\sqrt{2}]$. Clearly $A\cong A\oplus\mathbb{Z}[\sqrt{2}]\cong A\oplus\mathbb{Z}^2$ as abelian groups, so we just need to show that $A\not\cong A\oplus\mathbb{Z}$.
Let $A_i\cong\mathbb{Z}[\sqrt 2]$ be the subgroup of $A$ consisting of sequences whose terms are all zero, apart from possibly the $i$th term.
Lemma 1. If $\varphi:A\to\mathbb{Z}$ is a group homomorphism, then the restriction of $\varphi$ to $A_i$ is zero for all but finitely many $i$.
Proof. If not, we can choose $i_0<i_1<\dots$ so that the restriction of $\varphi$ to $A_{i_k}$ is nonzero for all $k$. The intersection of $A_{i_k}$ with $\ker(\varphi)$ has rank at most one, so we can inductively choose $x_k\in A_{i_k}$ so that $\varphi(x_k)\neq0$, $\vert x_k\vert<1$, and $x_k$ is divisible by a larger power of $2$ than any of $\varphi(x_0),\dots,\varphi(x_{k1})$.
Consider the sequences in $A$ whose $i_k$th term is either $x_k$ or $0$, with all other terms zero. Since there are uncountably many such sequences, $\varphi$ must agree on two of them. Taking the difference of these two, we get a nonzero sequence in $\ker(\varphi)$ whose first nonzero term, in the $i_k$th place for some $k$, is $\pm x_k$ and with all other terms divisible by a higher power of $2$ than $\varphi(x_k)$. But this is a contradiction, since $\varphi(x_k)=\pm\varphi(y)$ where $y$ is the sequence obtained by removing the first nonzero term. $\square$
Remark. The same proof works if we replace $A$ by the group of sequences that tend to zero, or the group of sequences such that $\sum_ia_i$ is absolutely convergent, by replacing the condition $\vert x_k\vert<1$ by sharper inequalities.
Lemma 2. If $\varphi:A\to\mathbb{Z}$ is a group homomorphism whose kernel contains every $A_i$, then $\varphi=0$.
Proof. Let $A'\leq A$ be the subgroup consisting of sequences such that $\sum_ia_i$ is absolutely convergent. Suppose $\varphi(a)\neq0$ where $a\in A$ is the sequence $(a_0,a_1,\dots)$, but that every sequence with finitely many nonzero terms is in $\ker(\varphi)$. Define a homomorphism $\theta:A'\to A$ by $$\theta(y_0,y_1,y_2,\dots)=(y_0a_0,(y_0+y_1)a_1,(y_0+y_1+y_2)a_2,\dots).$$ Then if $e(k)\in A'$ is the sequence which is zero except that the $k$th term is $1$, then $\varphi\theta\left(e(k)\right)=\varphi(a)\neq0$ for every $k$, contradicting the version of Lemma 1 that applies to $A'$. $\square$
The proofs of Lemmas 1 and 2 are adapted from wellknown proofs of the corresponding facts for the "BaerSpecker group" (the group of sequences of integers). These were first proved (I think) by Specker, but the particular proofs that I've adapted are due (I think) to Sasiada and Łos respectively. There are other proofs, and from Yves de Cornulier's comments it seems that at least some of those can also be adapted for what we need.
Prop. 3. Every group homomorphism $\varphi:A\to A$ is determined by the compositions $\varphi_{ij}:A_j\to A\stackrel{\varphi}{\to}A\to A_i$, where for each $i$, all but finitely many $\varphi_{ij}$ are zero.
Proof. Since $A$ is a subgroup of a direct product of copies of $\mathbb{Z}$ in an obvious way, this follows immediately from Lemmas 1 and 2. $\square$
In other words, this means that if we think of sequences as infinite column vectors, we can represent $\varphi$ as an infinite matrix of homomorphisms $\varphi_{ij}:\mathbb{Z}[\sqrt 2]\to\mathbb{Z}[\sqrt 2]$, with finitely many nonzero entries in each row.
Lemma 4. Let $\vartheta:\mathbb{Z}[\sqrt 2]\to\mathbb{Z}[\sqrt 2]$ be a group homomorphism that is not a $\mathbb{Z}[\sqrt 2]$module homomorphism. Then for any $\epsilon>0$ and $N>0$ there is some $x\in\mathbb{Z}[\sqrt 2]$ with $\vert x\vert<\epsilon$ and $\vert\vartheta(x)\vert>N$.
Proof. Straightforward. $\square$
Lemma 5. If $\varphi:A\to A$ is a group endomorphism, and $\varphi_{ij}$ are as above, then for all but finitely many $j$, all the $\varphi_{ij}$ are $\mathbb{Z}[\sqrt 2]$module homomorphisms.
Proof. Suppose not. Then because, by Lemma 1, for each $i$ all but finitely many $\varphi_{ij}$ are zero, we can choose $(i_0,j_0),(i_1,j_1),\dots$ so that $\varphi_{i_kj_k}$ is not a $\mathbb{Z}[\sqrt 2]$module homomorphism for any $k$, and such that $\varphi_{i_kj_l}=0$ for $k<l$.
Using Lemma 4, we can construct a bounded sequence $a=(a_j)$ of elements of $\mathbb{Z}[\sqrt 2]$ inductively so that $a_j=0$ for $j\not\in\{j_0,j_1,\dots\}$ and $$\vert\varphi_{i_kj_k}(a_{j_k})+\sum_{l<k}\varphi_{i_kj_l}(a_{j_l})\vert>k.$$ But this contradicts the fact that $\varphi(a)$ is bounded. $\square$
Theorem 6. $A\not\cong A\oplus\mathbb{Z}$.
Proof. Suppose there were such an isomorphism. Then there would be an injective map $\varphi:A\to A$ with $A/\varphi(A)\cong\mathbb{Z}$. By Lemma 5, $\varphi$ is described by a matrix $(\varphi_{ij})$ with only finitely many columns containing entries that are not $\mathbb{Z}[\sqrt 2]$module homomorphisms. So for sufficiently large $n$, if $A[n]\leq A$ consists of the sequences whose first $n$ terms are zero, then the restriction of $\varphi$ to $A[n]$ is a $\mathbb{Z}[\sqrt 2]$module homomorphism, and so $A/\varphi(A[n])$ is a $\mathbb{Z}[\sqrt 2]$module. But $A/\varphi(A[n])\cong\mathbb{Z}^{2n+1}$ as an abelian group, which is impossible, since $A/\varphi(A[n])\otimes_{\mathbb{Z}}\mathbb{Q}$ is a vector space over $\mathbb{Q}(\sqrt 2)$ and so has even dimension over $\mathbb{Q}$. $\square$
It's obvious that $A\cong A\oplus A$, and if $B=A\oplus\mathbb{Z}$ then it follows that $A\cong B\oplus B$ and $B\cong B\oplus B\oplus B$, so $B$ is an example of an abelian group $B$ with $B\cong B\oplus B\oplus B\not\cong B\oplus B$, I think rather simpler to describe than other examples I know of.
Edit (7 April 2016) I've just discovered that this question was answered by Eklof and Shelah in 1985. They reference Shabbagh for asking the question. The link I've given, from Google books, only gives a few pages of the paper, and I haven't yet got hold of a full copy, but their example seems to be more complicated (at least to describe) than mine. I've also realized that my example, if you let $B=A\oplus\mathbb{Z}$, gives an example of nonisomorphic abelian groups $A$ and $B$ with $A\oplus A\cong B\oplus B$, which is one of Kaplansky's "test problems" for abelian groups in his famous 1954 book on Infinite Abelian Groups, which also seems to be simpler to describe than other examples that I know of.

14$\begingroup$ The freeness of the group of bounded sequences of integers is a hard theorem due to G. Nöbeling, Verallgemeinerung eines Satzes von Herrn Specker, Invent. Math. 6 (1968), 4155. I learned this through Google books: books.google.com/… $\endgroup$– Todd Trimble ♦Jan 1, 2016 at 18:54

3$\begingroup$ The subgroup $B$ of finitary sequences is not a direct summand, because $A/B$ has nonzero elements which are infinitely $2$divisible, but $A$ has no such elements. $\endgroup$ Jan 1, 2016 at 23:24

2$\begingroup$ Extension of $X$ by $Y$ has 2 usual meanings, the group theorists's convention being that $X$ is the kernel and the algebraic geometer's convention being that $Y$ is the kernel. In your claim that "$A$ is extension of $\mathbb{Z}^\omega$ by the group of bounded integral sequences" if I'm correct you use the second convention. Here you can define the quotient map to be $(a_n+b_n\sqrt{2})_{n\ge 0}\mapsto (b_n)_{n\ge 0}$ (the image is the set of all integral sequences, the kernel is the set of bounded integral sequences). $\endgroup$– YCorJan 6, 2016 at 23:18

3$\begingroup$ @Adam No, I mean bounded as a real number. $\endgroup$ Jan 30, 2016 at 14:16

2$\begingroup$ @YCor Tending to zero in the real topology. $\endgroup$ Jan 30, 2016 at 14:47
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

$\begingroup$ This suggests to me a (likely unoriginal) notion of directed power. Let P be a poset (wellfounded, say), and for an algebra B consider this algebra raised to the power of P. Picking the right poset may allow similar constructions where the power is isomorphic to some subpowers (using fragments of P) and not others. I leave it for others to consider. Gerhard "May Have More Energy Tomorrow" Paseman, 2016.01.02 $\endgroup$ Jan 3, 2016 at 3:30

3$\begingroup$ I suspect that some version of this would give an example of a ring $A$ such that $A\cong A\times\mathbb{Z}^2$ but $A\not\cong A\times\mathbb{Z}$, but it seems like it would take a lot more work to prove that $A\not\cong A\times\mathbb{Z}$ as a group (if it is even true). Note that in the Boolean case, $A$ is isomorphic to $A\oplus B$ as an $\mathbb{F}_2$module, though not as an $\mathbb{F}_2$algebra. $\endgroup$ Jan 3, 2016 at 8:09

$\begingroup$ So the suggestion is the abelian group $A=\{a \in \mathbb{Z}^{\mathbb{N}} : \exists n \forall k \geq n ~ ( a(2k)=a(2k+1))\}$, right? My first question: Is $A$ free? $\endgroup$ Jan 3, 2016 at 15:01

$\begingroup$ That, or variations. I haven't gone through the proof to see how it works, but it seems to me the order properties of $\omega$ are part of what pulls it through. Thus pick some poset P as in my comment, and an appropriate subalgebra of a Pth power. Gerhard "Handwaving At A High Level" Paseman, 2016.01.03 $\endgroup$ Jan 3, 2016 at 17:23

3$\begingroup$ But this group is isomorphic to $\mathbb{Z}^{\mathbb{N}} \oplus \mathbb{Z}^{\oplus \mathbb{N}}$ via $a \mapsto ((a(2k))_{k \geq 0},(a(2k+1)a(2k))_{k \geq 0})$, right? So it won't work. $\endgroup$ Jan 3, 2016 at 17:24