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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 sh …

I just stumbled across the answer to this in Fuchs' 2015 book on Abelian Groups. The papers Hill, Paul, Two problems of Fuchs concerning tor and hom, J. Algebra 19, 379-383 (1971). ZBL0228.20027. and …

Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions. …

The answer to Question 1 is no. Let $A=\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ and let $B$ be the subgroup generated by $(2,1)$. Since $B$ is cyclic of order $4$, if it were contained in a …

There's a construction of a rank two (and therefore not locally cyclic) abelian group with endomorphism ring $\mathbb{Z}$, and therefore automorphism group cyclic of order 2, in "On the cancellation …

$\mathbb{Z}$ 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 …

Here’s a quick homological proof. Suppose $F$ is finite and $H$ torsion free. Then $F\cong\text{Hom}(F,\mathbb{Q}/\mathbb{Z})$, so $$\text{Ext}^1(H,F)\cong\text{Ext}^1\left(H,\text{Hom}(F,\mathbb{Q}/ …

Yes, it must. And $G$ doesn't need to be countable. Let $H$ be the $p$-primary component of the torsion subgroup of $G$. Then the natural map $H/pH\to G/pG$ is injective, so $H$ also satisfies (1), an …

I don't think it's true for $G=\mathbb{F}_2^3$ and $a=b=3$. If there were such sets $A$ and $B$, they must have exactly three elements each. By applying a translation and a group automorphism, we ma …

Arturo Magidin's answer is absolutely correct, but there's an earlier counterexample than Corner's. This question is Kaplansky's first "test problem" in his 1954 book on Infinite Abelian Groups, and …

The Baer-Specker group $B$, the direct product of countably many copies of $\mathbb{Z}$, is semi-free but not free. It is semi-free, because for any nonzero element $x\in B$ there is some projection $ …

You can choose integers $m_1,m_2,n_1,n_2$ so that $m_1$ and $n_1$ are coprime to $p$ and $m_2$ and $n_2$ are coprime to $q$, and such that $n_1m_2b(a_1,b_2)=0=n_2m_1b(a_2,b_1)$. Then $$b(n_1a_1+n_2a_2 …

For the first question, $$\bigoplus_{p\text{ prime}}\mathbb{Z}/p\mathbb{Z}$$ is a counterexample.

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 relati …

"Groups with a small number of automorphisms" by H. de Vries, A. B. de Miranda (Math Zeitschrift (1957/58) Volume 68, Issue 1, pp 450-464) link gives examples with cyclic automorphism groups of order …