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3 votes
1 answer
153 views

A question about freeness of a certain class of abelian groups

Lets call an abelian group $G$, to be semi-free (or SF) if every nonzero subgroup of $G$ is isomorphic to $\mathbb{Z}\times H$ for some abelian group $H$. Is every semi-free group, a free group? If ...
Mostafa - Free Palestine's user avatar
28 votes
2 answers
863 views

$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$

Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)? Edited to add: As no answers are forthcoming, does anyone ...
Pace Nielsen's user avatar
  • 18.7k
0 votes
0 answers
162 views

Finite-exponent abelian groups

Let $G$ be an abelian group and $G=\bigoplus_{i=1}^t{{\Bbb{Z}}_{p_i}^{n_i}}^{(\Lambda_i)}$ where each $\Lambda_i$ is a set (at least one of $\Lambda_i$ is infinite). Since $G_{\Bbb{Z}}$ is a finite-...
Najmeh Dehghani's user avatar
6 votes
2 answers
249 views

Two abelian groups, each being direct factor of the other

Let $M$ and $N$ be two abelian groups. Suppose that $M$ is a direct factor of $N$ (i.e. there are homomorphisms $i:M\rightarrow N$ and $p:N\rightarrow M$ such that $p\circ i=id_M$) and $N$ is also a ...
pi_1's user avatar
  • 1,463
3 votes
0 answers
46 views

What is an example of an integral domain with a module that is 1-separable but not separable?

Let R be an integral domain. All modules under discussion are torsion free unital left R-modules.     An R-module is completely decomposable if it is the direct sum of rank 1 submodules.     An R-...
user108215's user avatar
7 votes
1 answer
373 views

On describing a sort of "well-behaved" subgroups of a free abelian group

I found this question when I tried to figure out what kind of subgroups of a free abelian group behave just as well as in the finitely generated case. Let $M$ be a free abelian group and $N$ a ...
Censi LI's user avatar
  • 403
13 votes
1 answer
1k views

Classification of symtrivial modules over a PID

Let us call a module $M$ over a commutative ring $R$ symtrivial if the symmetry $M \otimes M \to M \otimes M, a \otimes b \mapsto b \otimes a$ equals the identity (the same notion applies to arbitrary ...
Martin Brandenburg's user avatar