All Questions
Tagged with abelian-groups gr.group-theory
141 questions
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 ...
- 37.4k
-2
votes
1
answer
158
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)\...
- 63.9k
5
votes
1
answer
308
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 ...
- 12.9k
1
vote
1
answer
344
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 ...
- 181
8
votes
2
answers
501
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 ...
- 37.4k
5
votes
1
answer
163
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 `...
- 4,738
3
votes
0
answers
138
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 $G^{\infty}$ ...
- 517
10
votes
1
answer
322
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 $p^n$.
...
- 35.2k
1
vote
1
answer
276
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 ...
- 1,832
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 $$b(x+y,z)=b(x,z)+b(y,z),b(x,z+y)=b(x,z)+b(x,...
- 35.2k
10
votes
2
answers
865
views
Is there a non-degenerate quadratic form on every finite abelian group?
Let $G$ be a finite abelian group. A quadratic form on $G$ is a map $q: G \to \mathbb{C}^*$ such that $q(g) = q(g^{-1})$ and the symmetric function $b(g,h):= \frac{q(gh)}{q(g)q(h)}$ is a bicharacter, ...
- 1,453
4
votes
1
answer
498
views
Short exact sequence $0\to \mathbb Z\to A \to \mathbb R \to 0$
Does every short exact sequence $0\to \mathbb Z\to A \to \mathbb R \to 0$ split in the category of Abelian groups?
- 118k
2
votes
0
answers
70
views
Alternating $n$-homomorphism on abelian group is skew of $n$-cocycle
Let $A$ be a finitely generated abelian group. Let $c$ be a 2-cocycle on $A$, where $A$ acts trivially on $\mathbb{C}^\times$. It is well-known that the skew-map
$$ c(a_1,a_2) \longmapsto \frac{c(a_1,...
- 12.9k
10
votes
2
answers
1k
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 $...
- 31
6
votes
1
answer
337
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 $M \in \text{GL}_n(\mathbb{Z})$ ...
- 4,991
0
votes
1
answer
557
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 ...
- 47.8k
3
votes
0
answers
98
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 ...
- 2,992
1
vote
1
answer
133
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 ...
- 63.9k
10
votes
1
answer
2k
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.
- 63.9k
15
votes
1
answer
567
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 \...
- 63.9k
8
votes
0
answers
309
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 ...
- 63.9k
26
votes
3
answers
7k
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 ...
- 18.7k
94
votes
2
answers
7k
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 $A \not\cong A \oplus \mathbb{Z}$?
- 63.9k
3
votes
1
answer
172
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 ...
- 63.9k
2
votes
2
answers
474
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 ...
- 63.9k
2
votes
1
answer
236
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\...
- 63.9k
1
vote
0
answers
74
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 $n$-...
- 63.9k
2
votes
1
answer
327
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 \{(...
- 63.9k
7
votes
1
answer
266
views
Positive cone of a subgroup of $\mathbb{Z}^n$
This question sounds like it should be very well known, but for some reason I failed to find a decent answer anywhere. Let $G\subset\mathbb{Z}^n$ be a subgroup, and $G_+=G\cap\mathbb{Z}_{\ge0}^n$ be a ...
- 63.9k
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 ...
- 63.9k
2
votes
1
answer
190
views
Logic article on first-order invariants of abelian groups
I remember reading an article published in the 1970s by a Polish mathematician describing the first-order invariants of a torsion-free abelian group. I do not recollect the author's name, the title of ...
- 4,713
1
vote
1
answer
329
views
Is there a free profinite abelian group on a profinite set?
Let $\mathit{Profinite}_{\mathrm{Ab}}$ be the category of profinite abelian groups, and let $\mathit{Profinite}_{\mathrm{Set}}$ be the category of profinite sets. Does the forgetful functor
$$\mathit{...
- 63.9k
12
votes
2
answers
2k
views
Do all exact sequences $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split for finitely generated abelian groups?
Suppose $A$ and $B$ are finitely generated Abelian groups. Are all exact sequences of the form $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split?
If not, is there an example?
- 4,713
3
votes
1
answer
293
views
Abelian torsion-free group with $\mathbb{Z}_2\times\mathbb{Z}$ as automorphism group
Let $A$ be an abelian torsion-free group. If $A$ is isomorphic with the group of rational numbers whose denominators are powers of, say, $2$, then its automorphism group is isomorphic with $\mathbb{Z}...
- 63.9k
6
votes
1
answer
417
views
Finite-by-torsion-free abelian groups (or compact abelian groups with finitely many components)
Here's a question I should know the answer to but don't:
Suppose $1\to F \to G \to G/F \to 1$ is a short exact sequence of abelian groups with $F$ finite and $G/F$ torsion-free. Must the sequence ...
- 9,730
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 ...
- 63.9k
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 $...
- 4,713
13
votes
1
answer
791
views
How nearly abelian are nilpotent groups?
It is not uncommon to read that "nilpotent groups are 'close to abelian'."1,2
Can this sentiment be made precise
in the sense of the
Turán and Erdős definition of "the probability that two elements of ...
- 44.4k
8
votes
1
answer
3k
views
On order of subgroups in abelian groups
I wonder whether any of you guys has already read the homonymous note by R. Beals in the December 2009 issue of the Monthly.
If so, would you be so kind as to let me know about the main ideas in Beal'...
- 4,713
9
votes
1
answer
1k
views
Direct product decomposition for infinite abelian groups with constrained torsion
Let $g$ be a positive integer, and let $G$ be a commutative group with the following constraint on its torsion subgroup: there is an injection $G[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\...
- 63.9k
5
votes
2
answers
907
views
What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$?
What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$?
I have a reference that says the group in question is $\mathbb{Q}_p,$ the additive group of the quotient field of the $p$-adic integers. Can ...
- 63.9k
4
votes
2
answers
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 ...
- 63.9k
2
votes
0
answers
176
views
Trivial Tate modules
Let $A$ be an abelian group, and $p$ a prime.
I'll call $$T_p(A) := \text{Hom}_{\mathbf{Z}}(\mathbf{Q}_{p}/\mathbf{Z}_{p}, A).$$
If $A$ is finite, then $T_p(A)$ is trivial, but the converse is not ...
CommunityBot
- 1
3
votes
1
answer
181
views
Characterisation of a class of group homomorphisms related to a central extension
Let $S$ and $R$ be groups and say $\sigma: S \twoheadrightarrow R$ is a group homomorphism that is a central extension; that is, it is surjective (extension) and its kernel is contained in the centre ...
- 555
2
votes
0
answers
96
views
A kind of cancellation ; exchange problem for groups
For which $(m,n,k,l) \in (\mathbb N\cup \{0\})^4$ , with $m\le n ; k\le l$ , does there exist a group $G$ with a finite subnormal series with torsion-free Abelian quotients such that $G \times \mathbb ...
CommunityBot
- 1
17
votes
1
answer
1k
views
Epimorphisms $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ are split
Consider the additive group of integer sequences $\mathbb{Z}^{\mathbb{N}}$. Why does every epimorphism of groups $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ split? $(\star)$
Actually this ...
- 35.2k
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 ...
- 63.9k
1
vote
0
answers
109
views
Symmetric analogue of "alternating bihomomorphism is skew of 2-cocycle" theorem
Let $G$ be a finite abelian group. It is well-known that every alternating bihomomorphism $\Omega:G\times G \to \mathbb{C}^\times$ arises as the skew $\kappa/\kappa^T$ of a 2-cocycle $\kappa \in Z^2(G,...
- 1,813
5
votes
0
answers
444
views
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 ...
- 93
5
votes
1
answer
472
views
Countable reduced abelian group containing all countable reduced abelian groups
Is there a countable abelian group for which its subgroups are exactly all of the countable "reduced" abelian groups? (Reduced means that its divisible subgroup is zero)
Is the following group a ...
- 63.9k