All Questions
Tagged with abelian-groups gr.group-theory
141 questions
0
votes
0
answers
42
views
What do you know about Jacobian group of the Cayley graphs? [closed]
Please can you help with some information like:
a) examples
b) general principles
c) references
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 ...
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 ...
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.
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 ...
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 ...
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, ...
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 $S$ ...
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:\...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 $...
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
$$
G\cong ...
5
votes
1
answer
401
views
Retract of a product
Let $G$ be a retract of a product $\prod_I\mathbb{Z}$ of copies of $\mathbb Z$.
This means there are group homomorphisms $\pi:\prod_I\mathbb{Z}\to G$ and $\sigma:G\to\prod_I\mathbb Z$ such that $\pi\...
3
votes
0
answers
70
views
Admissibility of Ulm's invariants
Let $G$ be a reduced abelian $p$-group. We set $G_0=G$. Let $\alpha$ be an ordinal. Inductively, if $\alpha=\beta+1$ is a successor ordinal, we define
$$G_{\alpha}=pG_{\beta}.$$
If $\alpha$ is a limit ...
0
votes
0
answers
90
views
Invariants of primary groups
In Kaplansky's book "Infinite Abelian Groups", an abelian group $G$ is called primary if every element has order power of $p$ for some fixed prime number $p$. It is well-known that every ...
3
votes
1
answer
203
views
normalizer quotient is $\operatorname{GL}_2(p)$
Let $p$ be a prime and let $w$ be a primitive $p$-th root of unity in $\mathbb{C}$. There is an element $e$ of order $p$ in $G=\operatorname{PGL}_n(\mathbb{C})$ where $n=pk$ and
$$e=\left[\left(\begin{...
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
$$
f:H^{n}(G,G^{\vee})\...
3
votes
0
answers
107
views
(Non)complete abelian groups in the “transfinite p-adic topology”
For an abelian group $A,$ a prime $p$ and an ordinal $\alpha,$ we recursively define $p^\alpha A$ as a subgroup of $A$ such that $p^0A=A,$
$$p^{\alpha+1}A=p(p^\alpha A) \hspace{5mm} \text{and} \...
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 ...
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 ...
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 $\sigma$ ...
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 ...
5
votes
1
answer
204
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 "...
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 ...
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 ...
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.) $[...
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}$?
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 ...
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}})$ ...
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 &...
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 ...
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 ...
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 $...
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 ...
4
votes
1
answer
182
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 ...
3
votes
1
answer
474
views
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 ...
5
votes
1
answer
308
views
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 ...
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 ...
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, ...
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 $\...
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 ...
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 $p^n$.
...