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Questions tagged [abelian-groups]

For questions about groups whose elements commute.

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Cohomology of "symplectically self-dual" chain complex

Let $G,H$ be abelian groups (denoted additively) with their Pontryagin duals denoted as $G^*$ and $H^*$. (The cases I'm interested in are products of $\mathbb Z_n$, $\mathbb Z$, $\mathbb R$, and $\...
Andi Bauer's user avatar
  • 3,001
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
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 $\mathbb{R}$, ...
Garrett Figueroa'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
0 votes
1 answer
187 views

Quotient of a ring by a left ideal

This is a simple algebra question I'm struggling with. Let $A$ be a ring (with unity) and $I\subset A$ a left ideal and $B\subset A$ a two sided Ideal. $A/I=B$ and $A/B=I$ (in the category of left $A$...
lun's user avatar
  • 71
1 vote
1 answer
209 views

A question about automorphism group of abelian group

Does anyone know any references that describe automorphism group $\operatorname{Aut}(\mathbb R^n\times \mathbb T^m)$? I searched for a long time but couldn't find it.
free's user avatar
  • 71
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
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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22 votes
1 answer
958 views

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 ...
Charles Rezk's user avatar
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2 votes
0 answers
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Adjoint to "strict twocategory of strict twofunctors"

Let C be the category of strict twofunctors, featuring the addition of a Grothendieck universe. Strict twocategories are categories enriched over the category of categories. C has an internal hom ...
user avatar
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 ...
Nathan Dunfield's user avatar
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 ...
Nathan Dunfield's user avatar
5 votes
1 answer
791 views

Can we test if an abelian group is finitely generated by taking tensor product?

If $A$ is a finitely generated abelian group, then we know that for all fields $k$, the tensor product $A\otimes_{\mathbb{Z}}k$ is finite dimensional as a $k$-vector space. The converse is not true, ...
Bingyu Zhang's user avatar
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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-1 votes
1 answer
209 views

Every abelian group can be embedded into a ring [closed]

Let $(G,0,+)$ be an abelian group. Does there always exist a ring with unity $(R,0,1,+,\cdot)$ and an injective homomorphism of groups $ \psi:(G,0,+)\rightarrow (R,0,+)$? Is this hard to prove, or are ...
Ândson josé's 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
1 vote
1 answer
129 views

Small total variation distance between sums of random variables in finite Abelian group implies close to uniform?

Let $\mathbb{G} = \mathbb{Z}/p\mathbb{Z}$ (where $p$ is a prime). Let $X,Y,Z$ be independent random variables in $\mathbb G$. For a small $\epsilon$ we have $\operatorname{dist}_{TV}(X+Y,Z+Y)<\...
alon's user avatar
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5 votes
2 answers
331 views

Computing the Abelian invariants of a subgroup of a f.g. Abelian group

We have a f.g. Abelian group $A$ given as a direct sum of $N$ cyclic subgroups $C_{k_j}=\langle x_j\rangle$, with $k_j\in \{2,\dots,\infty\}$, $1\leq j \leq N$, and the associated homomorphism $\phi:\...
Dima Pasechnik's user avatar
1 vote
0 answers
134 views

Isomorphic quotients of a countably infinitely-generated free abelian group

Let $F$ denote the free abelian group on countably infinite generators. I am trying to understand the relationship between normal subgroups $A$ and $B$ of $F$ with isomorphic quotients. So is there a ...
medvjed's user avatar
  • 11
1 vote
0 answers
64 views

Groups with prescribed Ulm invariants

In Kaplansky's book infinite abelian groups he provides (through some exercises) a complete classification of $p^{\infty}$-torsion countable abelian groups in terms of Ulm invariants. In other words ...
Richard's user avatar
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3 votes
2 answers
205 views

Descending chain in $\mathbb{Z}$ with certain confining property, but not strongly

Call a sequence $A_1 \supseteq A_2 \supseteq \cdots$ of subsets of $\mathbb{Z}$ confining if for all $i$ we have $A_i \supseteq A_{i+1}+A_{i+1}$. (Let us insist that the $A_i$ are symmetric and ...
Matt Zaremsky'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
243 views

Can a non-free Whitehead group embed as a discrete subgroup of a normed space?

Every countable discrete subgroup of a normed space is isomorphic to the direct sum of the group of integers. I wonder whether it is possible to push this beyond such direct-sum (free abelian) groups ...
Tomasz Kania's user avatar
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3 votes
1 answer
182 views

Almost free group without the Specker group as a subgroup

An Abelian group is almost free whenever every countable subgroup is free Abelian. Famously, the Specker group $\mathbf Z^{\mathbf N}$ is almost free. What are examples of almost free groups that are ...
user754245's user avatar
3 votes
0 answers
70 views

Admissibility of Ulm's invariants

Let $G$ be a reduced abelian $p$-group. We set $G_0=G$. Let $\alpha$ be an ordinal. Inductively, if $\alpha=\beta+1$ is a successor ordinal, we define $$G_{\alpha}=pG_{\beta}.$$ If $\alpha$ is a limit ...
Nini's user avatar
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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 ...
Terry Tao's user avatar
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3 votes
0 answers
173 views

Abelian characters and odd perfect numbers?

This question is about applications of abelian characters to odd perfect numbers: Context and Definitions: Let $n$ be a natural number and $D_n$ be the set of divisors. We can make this set to a ring ...
mathoverflowUser's user avatar
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
  • 10.6k
0 votes
0 answers
90 views

Invariants of primary groups

In Kaplansky's book "Infinite Abelian Groups", an abelian group $G$ is called primary if every element has order power of $p$ for some fixed prime number $p$. It is well-known that every ...
Nini's user avatar
  • 31
1 vote
0 answers
79 views

A general theory of pairings

Bilinear forms and bilinear maps for vector spaces over a field are standard material for an introductory course in linear algebra. There are also text books for bilinear forms and related quadratic ...
Thomas Preu's user avatar
3 votes
0 answers
107 views

(Non)complete abelian groups in the “transfinite p-adic topology”

For an abelian group $A,$ a prime $p$ and an ordinal $\alpha,$ we recursively define $p^\alpha A$ as a subgroup of $A$ such that $p^0A=A,$ $$p^{\alpha+1}A=p(p^\alpha A) \hspace{5mm} \text{and} \...
Sergei Ivanov's user avatar
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
8 votes
1 answer
523 views

Trivial group cohomology induces trivial cohomology of subgroups

From the answer to another question I asked (Projective representations of a finite abelian group) and from the structure theorem of finite abelian groups it follows that if $A$ is a finite abelian ...
Andrea Antinucci's user avatar
5 votes
1 answer
313 views

Projective representations of a finite abelian group

Projective representations of a group $G$ are classified by the second group cohomology $H^2(G,U(1))$. If $G$ is finite and abelian, it is isomorphic to the direct product of cyclic groups $$ G\cong ...
Andrea Antinucci'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
6 votes
1 answer
408 views

Is there a clear pattern for the degree $2n$ cohomology group of the $n$'th Eilenberg-MacLane space?

Let $G$ be a finite abelian group, and its higher classifying space is $B^nG=K(G,n)$. For $n=1$ it is well known that $H^2(B G, \mathbb{R}/\mathbb{Z}) \cong H^2(G,\mathbb{R}/\mathbb{Z})$ is isomorphic ...
Andrea Antinucci's user avatar
8 votes
1 answer
393 views

Pontryagin dual of a group-cohomology class

Let $A, B, C$ be finite Abelian groups fitting in a short exact sequence $$ 1 \rightarrow A\overset{\iota}{\rightarrow} B\overset{\pi}{\rightarrow} C\rightarrow 1 $$ This determines a class $[\...
Andrea Antinucci's user avatar
3 votes
1 answer
203 views

normalizer quotient is $\operatorname{GL}_2(p)$

Let $p$ be a prime and let $w$ be a primitive $p$-th root of unity in $\mathbb{C}$. There is an element $e$ of order $p$ in $G=\operatorname{PGL}_n(\mathbb{C})$ where $n=pk$ and $$e=\left[\left(\begin{...
user488802's user avatar
5 votes
1 answer
401 views

Retract of a product

Let $G$ be a retract of a product $\prod_I\mathbb{Z}$ of copies of $\mathbb Z$. This means there are group homomorphisms $\pi:\prod_I\mathbb{Z}\to G$ and $\sigma:G\to\prod_I\mathbb Z$ such that $\pi\...
user avatar
1 vote
1 answer
198 views

A map in group cohomology from $H^n(G,G^{\vee})$ to $H^{n+1}(G,U(1))$

Let $G$ be a finite abelian group and denote by $G^{\vee}=\mathrm{Hom}(G,U(1))$ its Pontryagin dual. For any positive integer $n$ one can define a homomorphism of abelian groups $$ f:H^{n}(G,G^{\vee})\...
Andrea Antinucci's user avatar
2 votes
0 answers
193 views

Decomposition of finite abelian groups of even order if there is an involution

Let $G$ be a finite abelian group and $\sigma :G\rightarrow G$ an automorphism of order two ($\sigma\circ \sigma =id_G$). Denote by $F$ and $A$ the subgroups of fixed and anti-fixed points of $\sigma$ ...
Andrea Antinucci's user avatar
3 votes
1 answer
160 views

Linking form for homology with general coefficients

For integral homology groups there is the notion of linking form (http://www.map.mpim-bonn.mpg.de/Linking_form) $$ Tor(H_{l}(X,\mathbb{Z}))\times Tor(H_{n-l-1}(X,\mathbb{Z}))\rightarrow \mathbb{Q}/\...
Andrea Antinucci's user avatar
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 ...
A. Bailleul's user avatar
  • 1,322
2 votes
0 answers
408 views

Conceptual proof of fundamental theorem of finite abelian groups

I'm looking for a conceptual proof of the following statement: Lemma: Let $G$ be a finite abelian $p$-group. Let $a$ be an element of maximal order. Then $G=\langle a \rangle \times H$ for some ...
Dr. Evil's user avatar
  • 2,751
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
1 vote
0 answers
125 views

Is the commutator of the holomorph of generalized quaternion group abelian?

Let $Q_{2^{n}} = \langle x, y \mathrel\vert x^{2^{n-1}}=y^4 = 1, x^{2^{n-2}}=y^2, y^{-1}xy = x^{-1} \rangle$ be the generalized quaternion group of order $2^{n}$. Let $\operatorname{Hol}(Q_{2^{n+1}})$ ...
bidermeyer'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
3 votes
0 answers
251 views

Commuting real elements in finite groups

Let $p$, $q$, $r$ be three distinct odd primes, and $G$ a finite group with $|G|$ divisible by $p$, $q$, $r$ to the first power only. Let $x,y,z \in G$ be of order $p,q,r$ respectively. Assume (a.) $[...
Nick's user avatar
  • 191
1 vote
0 answers
97 views

A duality of finite groups coming from a surjective homomorphism with finite kernel of algebraic tori

$\newcommand{\Hom}{{\rm Hom}} \newcommand{\Gm}{{{\mathbb G}_{m,{\Bbb C}}}} \newcommand{\X}{{\sf X}} $ I am looking for a reference for the following lemma (for which I know a proof): Lemma. Let $\...
Mikhail Borovoi's user avatar
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 ...

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