Questions tagged [abelian-groups]
For questions about groups whose elements commute.
254 questions
94
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2
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$A$ 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 $A \not\cong A \oplus \mathbb{Z}$?
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votes
9
answers
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Is there a slick proof of the classification of finitely generated abelian groups?
One of 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 $...
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71
votes
28
answers
8k
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Results from abstract algebra which look wrong (but are true)
There are many statements in abstract algebra, often asked by beginners, which are just too good to be true. For example, if $N$ is a normal subgroup of a group $G$, is $G/N$ isomorphic to a subgroup ...
37
votes
5
answers
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When is Aut(G) abelian?
Let $G$ be a group such that $\operatorname{Aut}(G)$ is abelian. Is then $G$ abelian?
This is a sort of generalization of the well-known exercise, that $G$ is abelian when $\operatorname{Aut}(G)$ is ...
- 63.1k
37
votes
1
answer
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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?
$\DeclareMathOperator\Hom{Hom}$The question is in the title. If the isomorphism $\Hom(A, G) \cong \Hom(B, G)$ is natural in $G$ then this is just the Yoneda Lemma. If $A$ and $B$ are finitely ...
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votes
3
answers
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Is there a nice explanation for this curious fact about cyclic subgroups?
Here's something that I noticed that quite surprised me.
Let $G$ be a finite abelian group. Consider the following expression.
$$
\nu(G) = \sum_{\substack{H \leq G \\ H \text{ is cyclic}}} |H|
$$
It ...
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30
votes
1
answer
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Do the algebraic integers form a free abelian group?
It is a well-known fact, proved in every introductory textbook on algebraic number theory, that if $K$ is an algebraic number field, i.e. a finite extension of $\mathbb{Q}$, then its ring $\mathcal{O}...
- 1,298
29
votes
0
answers
877
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The field of fractions of the rational group algebra of a torsion free abelian group
Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions.
...
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28
votes
2
answers
863
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$A^2$ 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 ...
- 18.7k
26
votes
3
answers
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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 ...
uuu
23
votes
2
answers
1k
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Are the p-adics a direct summand of the direct product of the groups $\mathbb{Z}/p^n\mathbb{Z}$?
The p-adic integers $\mathbb{Z}_p$ can be thought of as a subgroup of the direct product group $P = \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z}$. Are they a direct summand of this group? That is, is ...
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22
votes
1
answer
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Alternating forms on abelian groups
Let $G$ and $H$ be abelian groups. By an alternating form, I mean a bilinear function $A\colon G\times G\to H$ such that $A(x,x)=0$ for all $x\in G$.
Question. If $A\colon G\times G\to H$ is an ...
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20
votes
5
answers
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Constructively, is the unit of the “free abelian group” monad on sets injective?
Classically, we can explicitly construct the free Abelian group $\newcommand{\Z}{\mathbb{Z}}\Z[X]$ on a set $X$ as the set of finitely-supported functions $X \to \Z$, and so easily see that the unit ...
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20
votes
4
answers
2k
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Categorical proof subgroups of free groups are free?
This is a crossport of this question from MSE.
Is there a categorical proof that subgroups of free groups are free?
How about the result that subgroups of free abelian groups are free abelian?
What ...
- 935
18
votes
2
answers
876
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Groupoid cardinality of the class of abelian p-groups
$\DeclareMathOperator\Aut{Aut}\newcommand\card[1]{\lvert#1\rvert}$So, after going over the classification of finite abelian groups in a class I was teaching this winter, I got curious about whether it ...
- 42.4k
17
votes
1
answer
2k
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Subgroups of $\mathbb{Z}^n$
I hope that the following problem isn't actually elementary (at least, for the sake of the fact that I'm posting it here), and I apologize if it is. I did try hard to solve it first.
Let $V$ be a $\...
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17
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1
answer
1k
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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 ...
- 63.1k
16
votes
4
answers
1k
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Zero-sum partition of an abelian group
This is a question I have been asking myself some 5 years ago. I later got bored by lack of progress, but maybe some additive combinatorialists here know further. I'm not claiming it is conceptual or ...
- 33.8k
16
votes
1
answer
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A possible mistake in Walter Rudin, "Fourier analysis on groups"
I have the following lemma 4.2.4 on page 80 in the book (we have locally compact abelian topological groups $G_1, G_2$ and their duals $\Gamma_1, \Gamma_2$):
Suppose $E$ is a coset in $\Gamma_2$ ...
- 556
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3
answers
1k
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The existence of non-trivial homomorphisms $\prod_{n=1}^{\infty}\mathbb{Z}/\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$
Let $\prod_{n=1}^{\infty}\mathbb{Z}$ be the Baer-Specker group and $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ be the natural free abelian subgroup. It is known that if $G$ is a countable abelian group with ...
- 7,203
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votes
1
answer
753
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Finite abelian groups with fewer automorphisms than a subgroup
It is not too hard to find examples of finite groups which have fewer automorphisms than one of their subgroups. For example $\mathcal D_4 \times \mathbb Z/2\mathbb Z$ (where $\mathcal D_4$ is the ...
- 1,322
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 \...
- 1,742
15
votes
1
answer
1k
views
Is the class of additive groups of rings axiomatizable?
I know that it is impossible to axiomatize the multiplicative structures of rings, called $R$-semigroups. Is anything known about the first-order axiomatizability of the class of abelian groups which ...
- 1,445
14
votes
1
answer
696
views
$\mathbb{Z}$-module structure of the subring generated by an algebraic number
Let $a$ and $b$ be algebraic numbers which are not necessarily algebraic integers. Is there some invariant that allows us to determine whether $\mathbb Z[a]$ and $\mathbb Z[b]$ are isomorphic as $\...
- 151
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0
answers
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Is $\mathbb{Z}^{\omega}$ ever the union of a chain of proper subgroups each isomorphic to $\mathbb{Z}^{\omega}$?
Recall that the covering number $cov(B)$ is the least cardinal $\kappa$ such that $\kappa$ meagre sets cover the real line. Andreas Blass and John Irwin http://www.math.lsa.umich.edu/~ablass/bb.pdf ...
- 2,111
13
votes
1
answer
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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 ...
- 151k
13
votes
1
answer
1k
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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 ...
- 63.1k
12
votes
2
answers
2k
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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?
- 307
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votes
4
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How many non-isomorphic abelian subgroups of the permutation group $S_n$?
I am interested in how many (pairwise non-isomorphic) subgroups of the permutation group $S_n$ are abelian. ($n \in \mathbb{N}$ arbitrary and possibly big)
Are you aware of any references which treat ...
- 263
12
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1
answer
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Looking for concrete description of a category derived from abelian groups
The category of abelian groups $\mathsf{Ab}$ is the $\mathcal{Ind}$-completion of the full subcategory of finitely presentable abelian groups $\mathsf{Ab}_{fp}$. This is not so special, since the ...
- 1,291
11
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1
answer
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Where can I easily look up / calculate (abelian) group cohomology?
For an example I'm trying to understand, I need to calculate some cohomology group of some $\mathbb Z$-module with coefficients in some other $\mathbb Z$-module (with no interesting actions). (In ...
- 54.6k
11
votes
1
answer
498
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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 ...
- 64k
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1
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Are condensed vector spaces over finite fields always solid?
The Clausen-Scholze theory of condensed mathematics offers an abelian category with enough projective objects that embraces the study of arbitrary locally compact (and Hausdorff) groups. The behaviour ...
- 1,638
11
votes
1
answer
424
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Functorial description of mod-2 homology of an abelian group $A$ in terms of $A/2$ and ${}_2A.$
Let $A$ be an abelian group and $p$ be a prime. If $p\ne 2,$ there is a very nice functorial description of the homology algebra $H_*(A,\mathbb Z/p):$
$$H_*(A,\mathbb Z/p)\cong \Lambda^*(A/p)\otimes \...
- 1,484
11
votes
0
answers
430
views
Is $(\mathbb{R}, +)$ still injective as long as $(\mathbb{Q},+)$ is?
It is known that the existence of nontrivial injective abelian groups is independent of choice in ZF (or, rather, ZFA). In particular, $\mathbb{Q}$ is not provably injective, much less $\mathbb{R}$, ...
- 433
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votes
2
answers
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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 $...
- 1,484
10
votes
1
answer
2k
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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.
- 599
10
votes
1
answer
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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 $p^n$.
...
- 198
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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, ...
10
votes
1
answer
845
views
Reference request: a locally cyclic group is isomorphic to a section of the rational numbers
A group $G$ is locally cyclic if whenever $H \le G$ is a finitely generated subgroup then $H$ is cyclic. If $G$ is a locally cyclic group then $G$ is isomorphic to a quotient of a subgroup of the ...
- 11.2k
10
votes
1
answer
514
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Positivity of power of positive PSD matrices
Background: Let $M$ be an $n\times n$ matrix with nonnegative entries. It is immediate that for any integer $k$, $M^k$ has nonnegative entries.
Suppose now that, on top of having nonnegative entries, ...
- 505
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0
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428
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Sets which are unions of translates of each other but aren't single translates
I'm a hobbyist mathematician so any question I ask here might be at risk of closure. I hope this one is good enough, but I'm not sure. This is a continuation of two questions I asked on math....
- 145
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votes
3
answers
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Why are divisible abelian groups important?
I just quote wikipedia:
"Divisible groups are important in understanding the structure of abelian groups, especially because they are the injective abelian groups."
I am asking for detail ...
9
votes
1
answer
3k
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Automorphism group of a finite group
I would like to ask if there exists an explicit description of $\mathrm{Aut}(G)$, the group of automorphisms of a finite group $G$, in particular, when $G$ is abelian. E.g., if $G = \mathbb{Z}/m\...
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2
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On the existence of a direct summand containing a fixed subgroup
Let $G$ be a finite abelian group, and $g_1, \ldots, g_n \in G$ such that the cyclic groups that they generate are in direct sum $\langle g_1 \rangle \oplus \cdots \oplus \langle g_n \rangle$. Is it ...
- 133
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votes
1
answer
995
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constructive Serre classes
A Serre class (of abelian groups) is a class of abelian groups closed under subgroups, quotients, and extensions. For instance, finitely generated groups and finite groups are both Serre classes.
...
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votes
1
answer
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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}/\...
- 65.4k
9
votes
0
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298
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An abstract zero-sum problem
I would like to know whether the problem described below has appeared in the literature and/or whether similar questions have been studied. I would be very happy to find some references or, if none ...
- 1,169
8
votes
2
answers
501
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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 ...
- 463
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votes
2
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Locally compact abelian groups
First, some preliminaries:
Define an "LCA group" to be a locally compact Hausdorff abelian topological group.
Define "smooth manifold" in a way that requires Hausdorffness, but not connectedness or ...
- 22.3k