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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.
free's user avatar
  • 71
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
  • 1,644
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
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
  • 1,294
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
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
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
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
  • 11.3k
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
  • 31
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
  • 114k
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
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
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
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
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 ...
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 "...
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
  • 191
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 ...
Pace Nielsen's user avatar
  • 18.7k
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 ...
PHL's user avatar
  • 343
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 ...
PHL's user avatar
  • 343
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 ...
Simon Henry's user avatar
  • 42.4k
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 ...
user7427029's user avatar
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 ...
Zhang Yuhan's user avatar
0 votes
1 answer
207 views

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 ...
spaceman's user avatar
  • 595
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 ...
YJ Kim's user avatar
  • 321
-2 votes
1 answer
158 views

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)\...
Jack's user avatar
  • 13
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 ...
Nate River's user avatar
  • 6,313
2 votes
1 answer
105 views

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-...
J.K.T.'s user avatar
  • 517
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 ...
Nourddine Snanou's user avatar
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 $G^{\infty}$ ...
J.K.T.'s user avatar
  • 517
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 $p^n$. ...
Ethan Ackelsberg's user avatar
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 ...
waveman's user avatar
  • 181
1 vote
1 answer
276 views

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 ...
Keshav Srinivasan's user avatar
4 votes
1 answer
405 views

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,...
enjuikuo's user avatar
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, ...
Sebastien Palcoux's user avatar