Questions tagged [abelian-groups]

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7
votes
1answer
252 views

Existence of abelian group extension relative to group homomorphism

Let $f: A \to B\ $ be an abelian group homomorphism. Are there abelian groups $G,\ H,\ K$ such that $K \subseteq H \subseteq G$ and the map $$\pi \circ i: H \to G/K$$ which is the composition of ...
6
votes
4answers
2k views

Is a torsion-free abelian group finitely generated, if all of its localizations at primes $p$ are finitely generated over $\mathbb{Z}_p$?

Background: When proving that the group of $k$-isogenies $\mathrm{Hom}_k(A,B)$ between two abelian varieties is finitely generated, one first shows that the Tate map $$\mathbb{Z}_\ell\otimes_{\mathbb{...
3
votes
0answers
308 views

Homology $H_{\ast}(T, V)$

Let $A$ be a local domain. We let $T=T(A) $ be the subgroup of $\mathrm{SL}_{2}$ consisting of diagonal matrices and $V$ be the subgroup of unital matrices of $\mathrm{SL}_{2}$; i.e. $V:=\left\{\left( ...
0
votes
1answer
108 views

Fourier transform on lattice strip

I am working with a triangular lattice $L=\{n_1 a_2 + n_2 a_2 : n\in\mathbb{Z}^2 \}$ and $a_1 = \pmatrix{1 \\ 0}$ and $a_2 = \frac{1}{2} \pmatrix{-1 \\ \sqrt{3}}$, and I want to compute the Pontryagin ...
8
votes
0answers
875 views

Examples of uncountable abelian $p$-groups

Does anyone know of any interesting examples of an infinite abelian $p$-group which is uncountable? By non-interesting here I mean the direct sums of cylic and quasi-cylic groups, and totally ...
5
votes
2answers
373 views

Size of distinct sums in A

Let $G$ be an abelian group. Let $A\subset G$ be a finite set. $\sum_A$ is defined as: $$\left\{\sum_{b\in B}b \mid B\subset A\right\}$$ Is there any result similar to Freiman's Theorem for $\sum_A$? ...
67
votes
9answers
14k views

Is there a slick proof of the classification of finitely generated abelian groups?

One the proofs that I've never felt very happy with is the classification of finitely generated abelian groups (which says an abelian group is basically uniquely the sum of cyclic groups of orders $...
2
votes
1answer
169 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 ...
-2
votes
1answer
124 views

Are isomorphic quotients of abelian groups induced by automorphisms? [closed]

If I have an (abelian) group $G$ and an automorphism $\sigma: G \to G$ then for any subgroup $H$ of $G$ there is an induced isomorphism $G/H \cong G/\sigma(H)$ given by the map $gH \mapsto \sigma(g)\...
7
votes
1answer
378 views

Subgroups of the tensor product $A\otimes A$

I have this problem about subgroups of the tensor product of an abelian group $A$ with itself which arises from a complete different setting. I fell into this question studying quandles and quandle ...
5
votes
1answer
289 views

Quotient groups obtained by quotienting $G^n$ by $G^{n-1}$

Notation: For a group $G$, we write $G^n$ to denote the $n$-fold direct product of $G$ with itself. Problem set up: Consider, for a finite group $G$, and $n > 1$, the set $Q(G)_n$ of all ...
6
votes
0answers
244 views

When is every element of a coend of abelian groups contained in one of the summands?

Let $I$ be a small category and let $D : I^{\mathrm{op}} \times I \to \mathsf{Ab}$ be a functor. The coend $$\int^{i \in I} D(i,i)$$ can be constructed as the direct sum $\bigoplus_{i \in I} D(i,i)$ ...
2
votes
0answers
39 views

Cotorsion-freeness in uncountable products of abelian groups

An abelian group $A$ is cotorsion provided that whenever $A \leq G$ with $G$ abelian and $G/A$ is torsion-free, we have $G \cong A \oplus B$ for some $B \leq G$. An abelian group $A$ is cotorsion-...
1
vote
1answer
228 views

Bound for order of a group depending on number of elements of maximal order

This question has been partly answered in MSE, see here. In a paper On the Number of Elements of maximal order in a Group, it is proven that an arbitrary group $G$ with a finite number of elements of ...
8
votes
2answers
375 views

On $p$-groups with abelian automorphism group

Let $G$ be a $p$-group of order $p^{n}\geq p^{7}$ and its automorphism group is elementary abelian $p$-group. Then, it is clear that $G$ is nilpotent of class $2$. However, the converse is not true in ...
5
votes
1answer
128 views

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 `...
3
votes
0answers
125 views

The group of sequences in $G^{\mathbb{N}}$ that converge to $(0,0,0,\dots)$

Let $G$ be a discrete abelian group and $G^{\mathbb{N}}$ be the direct product (with the product topology), which consists of sequences $(a_1,a_2,a_3,\dots)$ of elements of $G$. Let $G^{\infty}$ ...
9
votes
2answers
227 views

Do these properties of a countable abelian group guarantee a Prüfer subgroup?

Suppose $(G,+)$ is a countable abelian group and $p$ is a prime number such that: The subgroup $pG$ has finite index in $G$, and For every $n \in \mathbb{N}$, $G$ contains an element of order $p^n$. ...
1
vote
1answer
202 views

How many elements of each order are there in this $p$-group? [closed]

Let $G$ be a countable Abelian $p$-group which equals a direct sum of at most countably many finite cyclic groups and at most countably many copies of the Prüfer $p$-group, where these finite cyclic ...
3
votes
0answers
176 views

Uncountable Mittag-Leffler condition?

Let $(X_\alpha)_{\alpha <\kappa}$ be an inverse system of abelian groups. If $\kappa = \omega$ (or by extension if $\kappa$ is of countable cofinality), then the Mittag-Leffler condition is a ...
8
votes
1answer
214 views

Example of an uncountable sequence of abelian groups with nonvanishing $\varprojlim^2$?

$\DeclareMathOperator{\op}{\mathrm{op}}\DeclareMathOperator{\Ab}{\mathsf{Ab}}\DeclareMathOperator{\Vect}{\mathsf{Vect}}$Question 1: What is an example of a sequence $(X_\alpha)_{\alpha<\kappa}$ of ...
5
votes
0answers
146 views

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 ...
4
votes
1answer
233 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,...
9
votes
2answers
474 views

Is there a non-degenerate quadratic form on every finite abelian group?

Let $G$ be a finite abelian group. A quadratic form on $G$ is a map $q: G \to \mathbb{C}^*$ such that $q(g) = q(g^{-1})$ and the symmetric function $b(g,h):= \frac{q(gh)}{q(g)q(h)}$ is a bicharacter, ...
6
votes
1answer
317 views

Which abelian groups are $\varprojlim^1$ groups?

Question 1: Let $\mathcal A$ be an abelian group. Does there exist an inverse system $(A^n)_{n \in \mathbb N} = (\cdots \to A^n \to A^{n-1} \to \cdots \to A^0)$ such that $\varprojlim^1 A^\bullet \...
6
votes
0answers
95 views

Cohomology of the Baer-Specker group

Let $A = \prod_{i \in \mathbb{N}} \mathbb{Z}$ be the Baer-Specker group; that is, a countably infinite product of the integers. We will consider this as a discrete abelian group. Are the higher ...
4
votes
1answer
331 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 \...
4
votes
1answer
390 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?
1
vote
1answer
137 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 ...
2
votes
0answers
56 views

Alternating $n$-homomorphism on abelian group is skew of $n$-cocycle

Let $A$ be a finitely generated abelian group. Let $c$ be a 2-cocycle on $A$, where $A$ acts trivially on $\mathbb{C}^\times$. It is well-known that the skew-map $$ c(a_1,a_2) \longmapsto \frac{c(a_1,...
10
votes
2answers
846 views

Classification of subgroups of finitely generated abelian groups

A finitely generated abelian group $A$ is isomorphic to a direct sum of cyclic groups. I am interested in an extension of this result on couples of abelian groups $(A,B),$ where $B$ is a subgroup of $...
2
votes
0answers
86 views

Who classified varieties that are commutative groups?

Who are the authors of the theorems asserting that connected varieties/manifolds which are abelian groups are isomorphic to ${\bf R}^k \times {\bf T}^n$? In the smooth setting, I guess this is due to ...
18
votes
0answers
563 views

If $A, B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?

The question is in the title. If the isomorphism $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ is natural in $G$ then this is just the Yoneda Lemma. If $A$ and $B$ are finitely generated this is also ...
7
votes
0answers
108 views

A conjecture on circular permutations of n elements in an abelian group of odd order

In 2013 I formulated the following conjecture in additive combinatorics. Conjecture. Let $G$ be an additive abelian group of odd order, and let $A$ be a subset of $G$ with $|A|=n>2$. Then, there is ...
5
votes
1answer
299 views

Can all proper sublattices of $\mathbb{Z}^n$ be generated cyclically?

Let $\Lambda \subset \mathbb{Z}^n$ be a proper sublattice (so that $\Lambda \ne \mathbb{Z}^n$). We say that $\Lambda$ is cyclically generated if there exists a matrix $M \in \text{GL}_n(\mathbb{Z})$ ...
3
votes
1answer
172 views

Can one turn finite-dimensional vector subspaces into a cancellative semigroup?

Let $V$ be a vector space over some field and let ${\rm Fin}\,V$ be the family of all finite-dimensional subspaces of $V$. Is it possible to turn ${\rm Fin}\,V$ into an commutative cancellative ...
0
votes
1answer
361 views

Abelian groups and their subgroups

It is well known that every finite abelian group is a direct product of cyclic groups. So for every $n$ every finite abelian group of exponent $n$ is a direct product of cyclic groups of order at most ...
3
votes
0answers
66 views

Hales' generalization of the stacked bases theorem (seeking a proof)

In his paper Analogues of the stacked bases theorem, published in the proceedings of a 1976 conference, A.W. Hales claimed some interesting generalizations of the stacked bases theorem for abelian ...
1
vote
1answer
103 views

Irreducible non-Abelian subgroup of $\mathrm{U}_n(\mathbb{C})$, containing diagonal matrices

Consider an irreducible non-Abelian subgroup $\mathrm{H}$ of group of unitary matrices $\mathrm{U}_n(\mathbb{C})$, that contains the subgroup of diagonal matrices. Does there exist any result ...
10
votes
1answer
1k views

A group whose automorphism group is cyclic

Is there an Abelian group $A$ which is not locally cyclic whose automorphism group is cyclic ? This question was first posted here.
15
votes
1answer
477 views

Torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z^2$

Is there a torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z \oplus \Bbb Z$ (as groups)? Notice that $\Bbb Z$ is not cancellable, so $A \oplus \Bbb Z \...
8
votes
0answers
285 views

A $\mathsf{ZF}$ example of a nonreflexive group which is isomorphic to its double dual?

Given a group $G$ denote by $G^\ast=\mathrm{Hom}(G,\Bbb Z)$ its dual and by $j\colon G\to G^{\ast\ast}$ the canonical homomorphism $g\mapsto (f\mapsto f(g))$. A group is reflexive iff $j$ is an ...
26
votes
3answers
6k views

Subgroups of a finite abelian group

Let $$G=\mathbb{Z}/p_1^{e_1}\times\cdots\times\mathbb{Z}/p_n^{e_n}$$ be any finite abelian group. What are $G$'s subgroups? I can get many subgroups by grouping the factors and multiplying them by ...
89
votes
2answers
6k views

$A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$

Are there abelian groups $A$ with $A \cong A \oplus \mathbb{Z}^2$ and $A \not\cong A \oplus \mathbb{Z}$?
3
votes
1answer
146 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 ...
2
votes
2answers
434 views

Non-archimedean group over the reals

I have a totally ordered group $(\mathbb{R};\leq,\oplus,0,-)$ with the reals as base set satisfying monotonicity, i.e. for all $x,y,z$ we have that if $y\leq z$ then $x\oplus y \leq x\oplus z$, and I ...
2
votes
1answer
206 views

Example involving partially ordered Abelian groups

Definition 1: Let $(G,\leq)$ be a nonzero partially ordered Abelian group with order unit $u$. (Recall that $u\in G$ is a order unit if, for every $g\in G$, there exists $N\in\mathbb N$ such that $-Nu\...
1
vote
0answers
61 views

Pure (ordered) subgroups

Let $H,G$ be abelian groups with $H \leq G$. We say that $H$ is a pure subgroup of $G$ if for every $n \in \mathbb N$ and $h \in H$ the following holds: If $h$ is $n$-divisible in $G$, then $h$ is $n$-...
2
votes
1answer
312 views

Is $\mathbb{Z}^2$ endowed with the square of the strict order, a lattice-ordered group?

I was looking some lattice-ordered group structure. I have kind of difficulty to figure out about the group $\mathbb{Z}^{2}$ with positive cone is $\mathbb{N}_{>0} \times \mathbb{N}_{>0} \cup \{(...
2
votes
1answer
295 views

When do infinitesimals split in dimension groups?

Let $G$ be a dimension group (i.e. a directed, unperforated abelian group satisfying the Riesz interpolation property) with order unit $u\in G^{+}$. There is a canonical positive group homomorphism $\...