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Results tagged with abelian-groups
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user 22989
For questions about groups whose elements commute.
4
votes
Accepted
Almost free group without the Specker group as a subgroup
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 …
- 35.2k
6
votes
Accepted
A question about freeness of a certain class of abelian groups
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 $ …
- 35.2k
5
votes
Accepted
$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$
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 …
30
votes
If $A$, $B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for a...
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 …
- 35.2k
1
vote
Accepted
Cotorsion-freeness in uncountable products of abelian groups
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 …
- 35.2k
4
votes
Number of orbits for abelian group actions
You could take $G=\mathbb{Z}\times\mathbb{Z}$, $X=G$ with the regular action, and $Y$ the disjoint union of two copies of $X$, and let $\{G_i\}$ be the family of cyclic subgroups of $G$.
- 35.2k
11
votes
Accepted
Do these properties of a countable abelian group guarantee a Prüfer subgroup?
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 …
- 35.2k
5
votes
A question on bi-character of finite abelian group
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
$$b(n_1a_1+n_2a_2 …
- 35.2k
20
votes
Accepted
Classification of subgroups of finitely generated abelian groups
The answer to Question 1 is no.
Let $A=\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
and let $B$ be the subgroup generated by $(2,1)$.
Since $B$ is cyclic of order $4$, if it were contained in a …
- 35.2k
13
votes
Accepted
Finite-by-torsion-free abelian groups (or compact abelian groups with finitely many components)
Here’s a quick homological proof.
Suppose $F$ is finite and $H$ torsion free. Then $F\cong\text{Hom}(F,\mathbb{Q}/\mathbb{Z})$, so
$$\text{Ext}^1(H,F)\cong\text{Ext}^1\left(H,\text{Hom}(F,\mathbb{Q}/ …
- 35.2k
8
votes
Accepted
On decomposition of finite Abelian groups
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 …
- 35.2k
5
votes
Co-finite type abelian groups
For the first question,
$$\bigoplus_{p\text{ prime}}\mathbb{Z}/p\mathbb{Z}$$
is a counterexample.
- 35.2k
4
votes
Epimorphisms $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ are split
I'll start by describing the notation that I'll use.
I'll think of elements of $\mathbb{Z}^\mathbb{N}$ as infinite column vectors
$${\bf x}=\begin{pmatrix}
x_0\\x_1\\x_2\\\vdots
\end{pmatrix}$$
of in …
- 35.2k
6
votes
Two abelian groups, each being direct factor of the other
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 …
- 35.2k
2
votes
Accepted
On describing a sort of "well-behaved" subgroups of a free abelian group
It is proved in
Joel M. Cohen and Herman Gluck, MR 254028 Stacked bases for modules over principal ideal domains, J. Algebra 14 (1970), 493--505,
that the answer is yes. They credit Kaplansky for as …
- 35.2k