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Questions tagged [abelian-groups]

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Characteristically simple locally compact abelian groups

Say a topological group $G$ is topologically characteristically simple if there does not exist a closed subgroup $1 < K < G$ such that $K$ is invariant under all automorphisms of $G$ (here `...
Colin Reid's user avatar
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Direct limits of a matrix and its transpose

Let $A \in M_n(\mathbb Z)$ and $A^T$ denote the transpose of $A$. Define the direct limits $$H_1 = \mathrm{colim} (\mathbb Z^n \xrightarrow{A} \mathbb Z^n \xrightarrow{A} \mathbb Z^n \xrightarrow{A} \...
Toke Nørgård-Sørensen's user avatar
5 votes
1 answer
432 views

Is the annihilator of the intersection of two subgroups of a (countable) discrete abelian group generated by the annihilators of the two subgroups?

Let $G$ be a (countable) discrete abelian group and denote by $\hat{G}$ its Pontryagin dual, i.e. the compact abelian group of group homomorphisms $\chi:G \longrightarrow \mathbb{T}$. Recall that, for ...
Nico Stammeier's user avatar
5 votes
2 answers
219 views

Torsionless not separable abelian groups

A torsionless abelian group $A$ is one for which any element $a\neq 0$ can be sent to a nonzero element of $Z$ by some homomorphism $A\rightarrow Z$ (integers). Equivalently, $A$ can be embedded as a ...
GMark's user avatar
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0 answers
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Can an infinite abelian $p$-group be tall and thin?

Does there exist an abelian $p$-group $A$ with countable Ulm invariants and uncountable height? Here by height, I mean the minimal ordinal $\rho$ such that $p^\rho A$ is divisible [1]. For an ordinal ...
Tim Campion's user avatar
5 votes
0 answers
202 views

Invariant measure on coset space and integrable functions

Let $G$ be a locally compact abelian group, and $H$ a closed subgroup. Let $C_c(G)$ be the space of continuous, compactly supported complex valued functions on $G$. General theory of Haar measure ...
D_S's user avatar
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194 views

Which rings are the endomorphisms ring of some abelian groups?

Which rings are (isomorphic to) the endomorphisms ring of some abelian group? Is there any necessary and sufficient condition?
Sara.T's user avatar
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Subgroups and quotients of an abelian pro-finite group

It is well known that every subgroup $H$ of a finite abelian group $G$ is isomorphic to a quotient of $G$. I'm wondering whether there is a counterpart for profinite groups. For example is it true ...
user106317's user avatar
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0 answers
171 views

Result of Larsen and Lunts on rationality of power series with coefficients in a free abelian group

Let $G$ be a free abelian group (not necessarily finitely-generated) and $F$ be the fraction field of the group ring of $G$. Let $\Theta$ be the set of power series in $F[[t]]$ such that each nonzero ...
Yuzhou Gu's user avatar
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2 answers
449 views

Number of torsion-free abelian groups

Let $\mathfrak{c}$ be the cardinality of the continuum. How much Choice, if any, is needed to prove that there are $2^{\mathfrak{c}}$ distinct (mutually nonisomorphic) torsion-free abelian groups of ...
Bruce Blackadar's user avatar
4 votes
2 answers
553 views

Non-torsion part of the abelianisation of congruence subgroups

I've posted this question on math.stackexchange, but haven't gotten any responses so I'm trying here instead. Let $A = F_q[T]$ be the ring of polynomials in one variable with coefficients in a finite ...
Liam Baker's user avatar
4 votes
4 answers
630 views

A question about the additive group of a finitely generated integral domain

Let $R$ be an integral domain of characteristic 0 finitely generated as a ring over $\mathbb{Z}$. Can the quotient group $(R,+)/(\mathbb{Z},+)$ contain a divisible element? By a "divisible element" I ...
Sidney Raffer's user avatar
4 votes
1 answer
426 views

Is $\mathbb Z$ prime in the class of abelian groups?

Let $B$, $C$, and $D$ be abelian groups such that $\mathbb Z\times B$ is isomorphic to $C\times D$. Is there a group $E$ such that $C$ or $D$ is isomorphic to $\mathbb Z\times E$? Reference: page 263 ...
Tri's user avatar
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507 views

Co-finite type abelian groups

Suppose $B$ is an abelian group such that for every integer $n\ge 1$, the $n$-torsion subgroup $B[n]$ is finite. Let $B_{\rm tor} = \varinjlim_{n\ge 1} B[n]$ be the torsion subgroup of $B$. Is it ...
user avatar
4 votes
2 answers
227 views

Maximal subgroups of finite abelian $2$-groups

Suppose $G$ is a finite abelian $2$-group, and $S$ is a subset of $G$, $\langle S\rangle=G$,$S^{-1}=S$,$e\notin S$. How to determine whether there exists a maximal subgroup $M$ of $G$, such that $S$ ...
lunch zheng's user avatar
4 votes
2 answers
1k views

Cyclic subgroups of finite abelian groups

I learned from MO Subgroups of a finite abelian group that the problem of enumerating subgroups (not up to isomorphism) of finite abelian groups is a difficult one. Are there simple formulas if one ...
user22518's user avatar
4 votes
1 answer
242 views

Quadratic refinements of a bilinear form on finite abelian groups

$\DeclareMathOperator\Hom{Hom}$Let $A$ be a finite abelian group and $\text{Sym}(A)$ the (abelian) group of symmetric bilinear forms over $A$ valued in $\mathbb{R}/\mathbb{Z}$. A quadratic function on ...
Andrea Antinucci's user avatar
4 votes
1 answer
588 views

Why does tensor product in Ab(V) require colimits in V?

In Tom Leinster's book on operads, he gives Ab(V), the category of abelian groups in a symmetric monoidal category V, as an example of a multicategory that doesn't arise from a monoidal category, ...
ziggurism's user avatar
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4 votes
1 answer
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Cardinality of the set of elements of fixed order.

Let us consider the group $G:=\mathbb{Z}_N^a$ (the product of the cyclic group with $N$ elements with itself $a$ times). Suppose we are given a number $m$ that divides $N$. I would like to know how ...
Math's user avatar
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Why does the category of abelian groups satisfy the axiom AB6?

In his Tohoku article, Grothendieck says that the category $\mathbf{Ab}$ of abelian groups satisfies the axiom AB6, namely "All small colimits exist in $\mathbf{Ab}$. Moreover for any index ...
rtwo's user avatar
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1 answer
498 views

Short exact sequence $0\to \mathbb Z\to A \to \mathbb R \to 0$

Does every short exact sequence $0\to \mathbb Z\to A \to \mathbb R \to 0$ split in the category of Abelian groups?
Richard P.'s user avatar
4 votes
2 answers
323 views

Cancellation theorem for lattices

By a lattice, we mean a finitely generated, free $\mathbb{Z}$-module together with a symmetric bilinear form. Typical examples are the hyperbolic lattices $U$ and the root lattices $A_{n}, D_{n}, E_{n}...
M Koerner's user avatar
4 votes
1 answer
2k views

Minimal generation for finite abelian groups

Let $G$ be a finite abelian group. I know of two ways of writing it as a direct sum of cyclic groups: 1) With orders $d_1, d_2, \ldots, d_k$ in such a way that $d_i|d_{i+1}$, 2) With orders that are ...
Calc's user avatar
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4 votes
1 answer
216 views

"Universal" abelian p-groups

Let $p$ be a prime number. I am interested in the abelian groups $G$ with the following property: (U) every finite abelian $p$-group $A$ admits a monomorphism $A\hookrightarrow G$. In other words, ...
Yuri Bilu's user avatar
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4 votes
1 answer
211 views

Nonempty intersection of cosets of finite-index subgroups

$\DeclareMathOperator\lcm{lcm}$This question is crossposted from MSE. Let $H_1,\dots,H_{n+2}$ be cosets of finite-index subgroups of $\mathbb{Z}^n$ and suppose for all $i=1,\dots,n+2$, $\bigcap_{j\neq ...
Saúl RM's user avatar
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4 votes
1 answer
405 views

A question on bi-character of finite abelian group

Setting: $G$ is a finite abelian group and any bicharacter on $G$, where a bi-character on $G$ is a map $b:G \times G \to \mathbb{Q}/\mathbb{Z}$ such that $$b(x+y,z)=b(x,z)+b(y,z),b(x,z+y)=b(x,z)+b(x,...
enjuikuo's user avatar
4 votes
1 answer
406 views

The action of the unitary divisors group on the set of divisors and odd perfect numbers

Let $n$ be a natural number. Let $U_n = \{d \in \mathbb{N}\mid d\mid n \text{ and } \gcd(d,n/d)=1 \}$ be the set of unitary divisors, $D_n$ be the set of divisors and $S_n=\{d \in \mathbb{N}\mid d^2 \...
user avatar
4 votes
1 answer
221 views

Sum of divisors and unitary divisors as the eigenvalue and the spectral norm of some addition matrix?

Let $n$ be a natural number and $D_n$ be the set of divisors. We can make this set to a ring by observing that each divisor $d$ has $$0 \le v_p(d) \le v_p(n)$$ Hence we can add two divisors $d,e$ by ...
user avatar
4 votes
1 answer
383 views

Finite dimensional compact abelian group that is not a product of connected and a totally disconnected

Let $G$ be a compact abelian group. A compact abelian group is said to have dimension $n$ if $\dim_\mathbb{Q} \mathbb{Q}\otimes \hat G = n$. Equivalently one can show that this holds if $G$ is ...
TopGroups's user avatar
4 votes
1 answer
277 views

Is there a good notion of kernels of quadratic forms on abelian groups?

Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the ...
Bipolar Minds's user avatar
4 votes
2 answers
715 views

The center of a(n endomorphism) ring is a PID

Let $A$ be a torsion-free Abelian group, and let $\mbox{End}(A)$ be its endomorphism ring. Denote by $R\doteq\mathbf{Z}(\mbox{End}(A))$ the center of $\mbox{End}(A)$. What would be necessary and/or ...
Bedovlat's user avatar
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4 votes
1 answer
182 views

How far is a countably infinite reduced abelian $p$-group from being an infinite direct sum?

Question Let $G$ be a countably infinite reduced abelian $p$-group. Is it always possible to write it has an infinite direct sums of non-trivial groups? If it is not true, how far is $G$ from being an ...
PHL's user avatar
  • 343
4 votes
0 answers
212 views

When does a short exact sequence of abelian groups with $B\cong A\oplus C$ split?

$\hspace{20pt}$Duplicate on stackexchange. This question, in a way, extends this one. The question is what are some sufficient conditions on the abelian group $B$ so that if $B\cong A\oplus C$ and a ...
cnikbesku's user avatar
  • 171
4 votes
0 answers
63 views

Possible questions about the Tate-Shafarevich subgroup of a Galois hypercohomology group?

$\newcommand{\wt}{\widetilde}$ Let $n=1,2$. There are infinite torsion abelian groups $H^1$, $H^2$ killed by some natural number $m$. There are finite subgroups $$ {\rm Sha}^1 \subset H^1,\quad ...
Mikhail Borovoi's user avatar
4 votes
0 answers
113 views

Duality for finite quotient groups of finitely generated free abelian groups

$\newcommand{\Z}{{\Bbb Z}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\Hom}{{\rm Hom}} $ The following lemma is certainly known. Lemma (well-known). Let $B$ be a lattice (that is, a finitely generated ...
Mikhail Borovoi's user avatar
4 votes
0 answers
153 views

Image of $\rm{lim}^1$ functor

In category of abelian groups, we know that — values of $\rm{lim}^1$ on countable systems are precisely cotorsion groups — values of $\rm{lim}^1$ on systems of finitely generated groups are of the ...
Denis T's user avatar
  • 4,600
3 votes
2 answers
902 views

Definable subsets of the integers as an abelian subgroup?

Consider the integers as a first-order structure in the language {0,+,-} of abelian groups. I suspect that the collection of definable subsets (without parameters) of this structure is an algebra ...
zetavolt's user avatar
  • 133
3 votes
2 answers
305 views

Are $H^3(A,U(1))$ and $\operatorname{Ext}^1(A,A^\vee)$ isomorphic for $A$ finite Abelian?

Motivated by three-dimensional Dijkgraaf-Witten TQFTs for finite Abelian groups $A$, that are classified by $H^3(A,\mathbb{R}/\mathbb{Z})$, it seems natural that this group is (naturally) isomorphic ...
Andrea Antinucci's user avatar
3 votes
1 answer
1k views

Automorphisms of an infinite direct product of abelian groups

Let $G = \prod_p \mathbb{Z}/p\mathbb{Z}$, where $p$ ranges over all primes, considered as an abelian group. What does $\text{Aut}(G)$ (or even $\text{End}(G)$) look like? I know that that if we take $...
Iian Smythe's user avatar
  • 3,115
3 votes
1 answer
1k views

Isomorphic Abelian Group [closed]

How many different non-isomorphic Abelian groups of order n are possible ??
Mahesh Gupta's user avatar
3 votes
2 answers
831 views

Divisible torsion $\mathbb{Z}$-modules

I am trying to prove that for any divisible torsion $\mathbb{Z}$-module $V$, this map $$f:\mathbb{Q}/\mathbb{Z}\otimes_E\text{Hom}(\mathbb{Q}/\mathbb{Z},V)\longrightarrow V\mbox{ defined by } f((q+\...
Fat's user avatar
  • 33
3 votes
1 answer
474 views

Structures of subgroups of a finite abelian p-group

$\newcommand\la{\langle}\newcommand\ra{\rangle}$Let $G=\mathbb{Z}/p^{i_1}\times\cdots\times\mathbb{Z}/p^{i_r}$ with $i_1\leq\ldots\leq i_r$ be a finite abelian $p$-group. Then there can be many ...
YJ Kim's user avatar
  • 321
3 votes
1 answer
153 views

On decomposition of finite Abelian groups

It is easy to see that for any finite Abelian group $G$ and any numbers $a,b$ with $|G|=ab$ there exist a subgroup $A\subset G$ and a subset $B\subset G$ such that $|A|=a$, $|B|=b$ and $G=A+B$, where $...
Taras Banakh's user avatar
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
3 votes
1 answer
336 views

What are the almost periodic functions on the complex plane?

The almost periodic functions on the real line can be characterized as uniform limits of trigonometric functions. I was wondering whether a similar definition exists on the complex plane (a locally ...
Merry's user avatar
  • 173
3 votes
1 answer
202 views

Intermediate lattices $C\mathbb{Z}^n \subseteq \Lambda \subseteq \mathbb{Z}^n$

Let $C \in \mathfrak{gl}(\mathbb{Z},n)$ be a symmetric full rank integer valued matrix (in my case it is the symmetric part of a Cartan matrix). Let $\Lambda \subseteq \mathbb{Z}^n$ be a full rank ...
Bipolar Minds's user avatar
3 votes
2 answers
318 views

Character kernels in the lattice of subgroups of a finite abelian group

I am looking for any efforts that have been made to characterize the character kernels (equivalently, the subgroups yielding cyclic quotients) inside the lattice of subgroups of a finite abelian group....
benblumsmith's user avatar
  • 2,851
3 votes
1 answer
125 views

Uncountable divisible groups and the existence of order-preserving isomorphisms of their subsets

Let $(G,+,0,<)$ be an ordered divisible group of uncountable dimension. Consider the subset $G^{<0}$ of $G$. Question: Are $G$ and $G^{<0}$ isomorphic as ordered sets? Does there exist an ...
N.W.'s user avatar
  • 33
3 votes
1 answer
171 views

Compatible total orderings of the group $\mathbb{Z}^\mathbb{N}$

Given the additive group of the module $\mathbb{Z}^\mathbb{N}$ and a total ordering of the group that is compatible with addition and where $\chi_{\{n\}} > 0$ for all $n \in \mathbb{N}$, can we say ...
Anschel Schaffer-Cohen's user avatar
3 votes
1 answer
169 views

Maximal zero-sum free sequences of $C_3^n$

I am working on the Davenport constant for groups, $D(G)$, which is the minimal number $d$ such that every sequence or multiset of $d$ elements of the group $G$ always contains some non-empty zero-sum ...
Mikel Martinez Puente's user avatar