All Questions
Tagged with abelian-groups gr.group-theory
141 questions
0
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22
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What do you know about Jacobian group of the Cayley graphs?
Please can you help with some information like:
a) examples
b) general principles
c) references
1
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1
answer
209
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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.
4
votes
1
answer
426
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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 ...
8
votes
1
answer
217
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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 ...
8
votes
1
answer
353
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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 ...
4
votes
1
answer
216
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"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, ...
4
votes
2
answers
227
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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 $S$ ...
5
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2
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331
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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:\...
1
vote
0
answers
134
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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 ...
3
votes
2
answers
205
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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 ...
3
votes
1
answer
243
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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 ...
3
votes
1
answer
182
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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 ...
3
votes
0
answers
70
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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 ...
8
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2
answers
1k
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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 ...
4
votes
1
answer
211
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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 ...
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0
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90
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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 ...
3
votes
0
answers
107
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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 $p^0A=A,$
$$p^{\alpha+1}A=p(p^\alpha A) \hspace{5mm} \text{and} \...
4
votes
0
answers
63
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Possible questions about the Tate-Shafarevich subgroup of a Galois hypercohomology group?
$\newcommand{\wt}{\widetilde}$
Let $n=1,2$. There are infinite torsion abelian groups $H^1$, $H^2$ killed by some natural number $m$.
There are finite subgroups
$$ {\rm Sha}^1 \subset H^1,\quad ...
5
votes
1
answer
313
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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
$$
G\cong ...
3
votes
1
answer
203
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normalizer quotient is $\operatorname{GL}_2(p)$
Let $p$ be a prime and let $w$ be a primitive $p$-th root of unity in $\mathbb{C}$. There is an element $e$ of order $p$ in $G=\operatorname{PGL}_n(\mathbb{C})$ where $n=pk$ and
$$e=\left[\left(\begin{...
5
votes
1
answer
401
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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\...
1
vote
1
answer
198
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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})\...
2
votes
0
answers
193
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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$ ...
15
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 ...
2
votes
0
answers
408
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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 ...
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 ...
1
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0
answers
125
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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}})$ ...
4
votes
0
answers
113
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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 ...
3
votes
0
answers
251
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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.) $[...
1
vote
0
answers
97
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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 $\...
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 ...
5
votes
1
answer
204
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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 "...
2
votes
0
answers
177
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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 &...
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 ...
4
votes
1
answer
182
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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 ...
1
vote
1
answer
89
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 ...
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 ...
12
votes
4
answers
1k
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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 ...
7
votes
1
answer
464
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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 ...
0
votes
1
answer
207
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Fourier transform on lattice strip
I am working with a triangular lattice $L=\{n_1 a_2 + n_2 a_2 : n\in\mathbb{Z}^2 \}$ and $a_1 = \pmatrix{1 \\ 0}$ and $a_2 = \frac{1}{2} \pmatrix{-1 \\ \sqrt{3}}$, and I want to compute the Pontryagin ...
3
votes
1
answer
474
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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 ...
-2
votes
1
answer
158
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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)\...
5
votes
1
answer
308
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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 ...
2
votes
1
answer
105
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Cotorsion-freeness in uncountable products of abelian groups
An abelian group $A$ is cotorsion provided that whenever $A \leq G$ with $G$ abelian and $G/A$ is
torsion-free, we have $G \cong A \oplus B$ for some $B \leq G$. An abelian group $A$ is
cotorsion-...
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 ...
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}$ ...
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$.
...
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 ...
1
vote
1
answer
276
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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 ...
4
votes
1
answer
405
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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,...