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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'm supervising an undergraduate research project. Among other things, I've got the student to look at this paper of Gene Kopp and John Wiltshire-Gordon. This question arose from a missing complex con …

I think it follows from Theorem 1.1 of "Subgroups of Infinite Symmetric Groups" by Macpherson and Neumann (J. London Math. Soc. (1990) s2-42 (1): 64-84) that there is no minimal generating set of $S(\ …

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 …

For any prime $p$, the endomorphism monoid of $\mathbb{Z}[\frac{1}{p}]$ is a commutative monoid with zero whose submonoid of nonzero elements is the direct sum of a cyclic group of order two (generate …

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 …

No. There are non-abelian groups $G$ for which all subgroups are normal, such as the quaternion group of order 8. So the intersection of all normalizers is just $G$.

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 …

Here's a slightly different answer, less group-theoretic and more representation-theoretic than Geoff's. Rephrasing your question in terms of $\mathbb{F}_pG$-modules, you are asking about a faithful …

Yes. In fact, $V$ has an infinite dimensional semisimple quotient, which is decomposable since any simple $\mathbb{F}_pG$-module is a quotient of $\mathbb{F}_pG$ and so is finite-dimensional. Let $V' …

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}/ …

It may well be a lot easier than this, but it follows from the answer to the weaker question Can a group be a finite union of (left) cosets of infinite-index subgroups? that it's not possible.