Questions tagged [abelian-groups]

For questions about groups whose elements commute.

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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
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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
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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
144 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's user avatar
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3 votes
1 answer
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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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2 votes
1 answer
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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
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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
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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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Dimension inequality for primary groups

Let $p$ be a prime number and $G$ an abelian group. The group $G$ is said to be $\textbf{primary}$ if every element of $G$ has order power of $p$. For every natural number $n$, we define $$\ker(p^n)=\{...
Nini's user avatar
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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
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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
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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
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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
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(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
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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
445 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
4 votes
1 answer
225 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
207 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
384 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
7 votes
1 answer
318 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
191 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
383 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
177 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
170 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
121 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
697 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
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2 votes
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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
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4 votes
1 answer
239 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
113 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
91 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
249 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
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1 vote
0 answers
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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
69 votes
28 answers
7k 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 ...
5 votes
1 answer
179 views

Characteristic subgroups of a finite abelian $2$-group

I have recently stumbled across the problem of describing the characteristic subgroups of a finite abelian group. With some discussions with some mathematicians in my lab, I managed to obtain a "...
Nataniel Marquis's user avatar
2 votes
0 answers
177 views

Existence of fully supported element in a finite-dimensional vector space over $\mathbb{F}_p$ (and in finite abelian groups)

Let $V$ be an $n$-dimensional vector space over $\mathbb{F} = \mathbb{Z} / (p)$, the field of $p$ elements, $p$ a prime, with $\{v_1, \dotsc, v_n \}$ a basis for $V$. An element $x \in V$ is called &...
Nick's user avatar
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1 vote
0 answers
58 views

Finding a particular kind of basis of subgroup of a lattice generated by non-negative part

For $\mathbf v=(v_1,\ldots,v_n)\in \mathbb Z^n$, let $\operatorname{supp}(\mathbf v):=\{j: v_j \ne 0\}$. For a subset $X$ of $\mathbb Z^n$, define $\operatorname{supp}(X):=\bigcup_{\mathbf v \in X} \...
uno's user avatar
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1 vote
0 answers
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elementary abelian subgroups with centralizers not connected

Let $G =$ PGL$_{8}(\textbf{C})$. Let $a, b, c, d$ be four representatives of conjugacy classes of involutions in $G$ where $$a = \begin{pmatrix} -1 & 0\\ 0 & I_{7} \end{pmatrix}, b = \begin{...
user488802's user avatar
0 votes
0 answers
49 views

Complemented subalgebra in a free Lie ring

A Lie ring is a triple $(G,+, [\ ,\ ]),$ where $(G,+)$ is an abelian group and $ [\ ,\ ]$ is a bilinear map satisfying $[x,x]=0$ $[\ ,\ ]$ is bilinear $[[x,y],z]+[[y,z],x]+[[z.x],y]=0,\ \forall\ x,...
MANI's user avatar
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1 vote
0 answers
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Compute 01-vectors in the orbit of a given vector wrt a finitely-generated abelian subgroup of SL(n,ℤ)

Given a vector $v\in\mathbb Z^n$ and pairwise commutative matrices $M_1,\dotsc,M_k\in \operatorname{SL}(n,\mathbb Z)$, how to compute all 01-vectors in the orbit of $v$ with respect to multiplication ...
Max Alekseyev's user avatar
28 votes
2 answers
808 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 ...
Pace Nielsen's user avatar
1 vote
1 answer
258 views

Closed form roots for polynomial $x^9 + ax^6 + bx^5 + cx^3 + d = 0$

I know about Abel–Ruffini theorem, but I have a polynomial of special form. From "Beyond the Quartic Equation" by R.B. King (a very interesting book, btw) I've learned about Tschirnhaus ...
Moonwalker's user avatar
7 votes
1 answer
327 views

$N_{G}(E)/C_{G}(E)$ is the Weyl group of $G$?

In the algebraic group $G = \operatorname{PGL}_4(\mathbb{C})$, let $E$ denote the subgroup of elements of order dividing 2 in the diagonal maximal torus; it is generated by the images of the three ...
user488802's user avatar
0 votes
0 answers
148 views

Finite-exponent abelian groups

Let $G$ be an abelian group and $G=\bigoplus_{i=1}^t{{\Bbb{Z}}_{p_i}^{n_i}}^{(\Lambda_i)}$ where each $\Lambda_i$ is a set (at least one of $\Lambda_i$ is infinite). Since $G_{\Bbb{Z}}$ is a finite-...
Najmeh Dehghani's user avatar
1 vote
0 answers
59 views

Eigenvalue function of the representation variety of free abelian group

$\DeclareMathOperator\SL{SL}$Let $\rho:\mathbb Z^n\rightarrow \SL(n,\mathbb C)$ be a representation of a finitely generated free abelian group, by simultaneously triangularization, we can assume that $...
Fredy's user avatar
  • 492
0 votes
0 answers
67 views

countable direct sum of cyclic abelian $p^{2}$ groups

Let $G={{\Bbb{Z}}_{p^{2}}}^{(\aleph)}$ (countable direct sum of copies of ${\Bbb{Z}}_{p^2}$). It is clear that every subgroup of $G$ is a homomorphic image of $G$. Now this is my question: Is it true ...
Najmeh Dehghani's user avatar
4 votes
1 answer
489 views

Why does the category of abelian groups satisfy the axiom AB6?

In his Tohoku article, Grothendieck says that the category $\mathbf{Ab}$ of abelian groups satisfies the axiom AB6, namely "All small colimits exist in $\mathbf{Ab}$. Moreover for any index ...
rtwo's user avatar
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4 votes
1 answer
160 views

How far is a countably infinite reduced abelian $p$-group from being an infinite direct sum?

Question Let $G$ be a countably infinite reduced abelian $p$-group. Is it always possible to write it has an infinite direct sums of non-trivial groups? If it is not true, how far is $G$ from being an ...
PHL's user avatar
  • 276
1 vote
1 answer
86 views

Element of order $p$ and finite height $\geq1$ in a reduced abelian group $p$-group with an element of order $p^2$

This is a reference request for the following statement: Fact: Let $G$ be a reduced abelian $p$-group with an element of order $p^2$. Then $G$ contains an element of order $p$ and of finite height at ...
PHL's user avatar
  • 276
14 votes
2 answers
776 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 ...
Simon Henry's user avatar
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1 vote
1 answer
154 views

Number of orbits for abelian group actions

Suppose $G$ is an abelian group acting faithfully on two sets, $X$ and $Y$, of the same size. None of $G$, $X$ and $Y$ is finite. Now suppose $G$ is the union of abelian groups $G_i$, where $i$ varies ...
THC's user avatar
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