All Questions
Tagged with abelian-groups gr.group-theory
140 questions
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 $A \not\cong A \oplus \mathbb{Z}$?
- 63.1k
72
votes
9
answers
18k
views
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 $...
- 44.7k
71
votes
28
answers
8k
views
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
11k
views
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
1k
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?
$\DeclareMathOperator\Hom{Hom}$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 ...
- 1,141
32
votes
3
answers
3k
views
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 ...
- 6,290
29
votes
0
answers
877
views
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.
...
- 35.2k
28
votes
2
answers
863
views
$A^2$ 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
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 ...
uuu
20
votes
4
answers
2k
views
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
views
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
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 ...
- 63.1k
15
votes
1
answer
753
views
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
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
14
votes
0
answers
518
views
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
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 ...
- 151k
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?
- 307
12
votes
4
answers
1k
views
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
11
votes
1
answer
3k
views
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
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 $...
- 1,484
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.
- 599
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 $p^n$.
...
- 198
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, ...
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
0
answers
428
views
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
9
votes
3
answers
3k
views
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
views
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\...
- 675
9
votes
2
answers
1k
views
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
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}/\...
- 65.4k
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 ...
- 463
8
votes
2
answers
2k
views
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
8
votes
2
answers
501
views
Exact sequence of $n$th powers of abelian groups
Let $A,B,C$ be finitely generated abelian groups. Assume that there is an exact sequence $$0 \to C \to A^n \to B^n \to 0,$$where $A^n = A \oplus \dotsc \oplus A$ as usual. It is not assumed that $A^n \...
- 63.1k
8
votes
2
answers
1k
views
A conceptual proof that bounded index subgroups of a bounded torsion abelian group contain bounded index complemented subgroups
Call an abelian group $G = (G,+)$ $m$-torsion for some natural number $m$ if one has $m \cdot x = 0$ for all $x \in G$. A subgroup $H$ of $G$ is said to be complemented if one can write $G = H \oplus ...
- 114k
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'...
- 8,842
8
votes
1
answer
353
views
Structure of a single automorphism of a finite abelian p-group
A finite abelian $p$-group $H$ is homogenous when it is the direct sum of cyclic groups of the same order $p^r$, i.e. $H \cong \big(\mathbb{Z}/p^{r}\mathbb{Z}\big)^{e}$. Every finite abelian $p$-group ...
- 365
8
votes
1
answer
217
views
Bilinear forms on abelian $p$-groups: Images of weak metabolizers in the Frattini quotient
Suppose $G$ is a finite abelian $p$-group for $p$ an odd prime and $b : G \times G \to \mathbf{Q}/\mathbf{Z}$ is a non-degenerate symmetric bilinear form. A subgroup $H \leq G$ is a weak metabolizer ...
- 365
8
votes
1
answer
216
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 ...
- 2,197
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 ...
- 3,011
8
votes
0
answers
1k
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 ...
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 ...
- 16.9k
7
votes
1
answer
464
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 ...
- 807
7
votes
1
answer
617
views
Looking for a modern source about Ulm Invariants
I'm looking for a modern, approachable text (preferably a website, textbook, or expository article, and preferably one easily available online or at a library) which can explain the concept of Ulm ...
- 1,979
7
votes
1
answer
373
views
On describing a sort of "well-behaved" subgroups of a free abelian group
I found this question when I tried to figure out what kind of subgroups of a free abelian group behave just as well as in the finitely generated case.
Let $M$ be a free abelian group and $N$ a ...
- 403
6
votes
4
answers
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{...
- 1,527
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 ...
- 9,720
6
votes
2
answers
704
views
Hall polynomial when the subgroup is cyclic?
Does anyone know the formula for a Hall polynomial $g_{u,v}^{\lambda}(p)$ when $v$ is the type of cyclic subgroup (ie. $v=(v_{1})$ ) .
http://en.wikipedia.org/wiki/Hall_algebra
I was hoping this ...
- 152
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 $M \in \text{GL}_n(\mathbb{Z})$ ...
- 27k
6
votes
1
answer
278
views
Nearly slender abelian groups
Let $\prod_{n=1}^{\infty}\mathbb{Z}$ be the Baer-Specker group (infinite direct product of the additive group of integers) and $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ be the natural subgroup which is the ...
- 7,203
6
votes
2
answers
232
views
Finding an "optimal" quotient in a free group
Consider the abelian free group $G = \mathbf{Z}^n$ of rank $n$ and a finite subset $A \subset G \setminus \{0\}$. Since $G$ is residually finite, there is a subgroup $H \subset G$ such that $A \cap H =...
- 373