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First, just a quick comment about terminology. It's possible that there are varying conventions, but I have seen "almost free" used to mean that all subgroups of an abelian group $G$ that have cardina …

There are also abelian examples (these are called superdecomposable abelian groups). For example, by Theorem 5.1 of Corner, A. L. S., Every countable reduced torsion-free ring is an endomorphism ring, …

No. The third condition implies that $U$ is countable, and so has countably many finite subsets, and so has countably many finitely generated subgroups. But there are uncountably many finitely generat …

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 …

Any simple group $G$ only has $G$ and the trivial subgroup as retracts.

A similar question was asked on math.stackexchange a few years ago, and I posted the following answer. I've just looked again at Conlon's paper, and I'm afraid it's a bit short on explicit examples. = …

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 …

In fact, more is true. Every direct product of cotorsion-free abelian groups is cotorsion-free. This is clear because the cotorsion-free abelian groups are precisely those that have no nonzero homomor …

In Alperin, J. L., Large abelian subgroups of p-groups, Trans. Am. Math. Soc. 117, 10-20 (1965). ZBL0132.27204, the second part of Theorem 1 gives a group of order $2^{50}$ with no abelian subgroups o …

Assuming that by "sub-algebra" you mean "unital sub-algebra": Every group algebra has a one-dimensional module (the trivial module), so any subalgebra has a one-dimensional module. But many finite-dim …

In a very short paper put on the arXiv recently, García, Margolis and del Río give examples of nonisomorphic finite $2$-groups $G$ and $H$ with $\mathbb{F}_2G\cong\mathbb{F}_2H$, thus solving the modu …

It seems that the $p>2$ part of this was proved in Kulakoff, A., Über die Anzahl der eigentlichen Untergruppen und der Elemente von gegebener Ordnung in $p$-Gruppen., Math. Ann. 104, 778-793 (1931). Z …

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 …

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 …

Representations of $B$ (or at least an equivalent category) are studied in the literature under the name of "cohomological Mackey functors". Theorem 1.1 of Bouc, Serge; Stancu, Radu; Webb, Peter, On t …