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Reference for surjectivity of the canonical map $R^{G_1} \otimes R^{G_2} \to R^{G_1 \cap G_2}$

Let $R$ be a commutative ring, $G$ a finite group with an action over $R$. Let $G_1, G_2 \subset G$ be two subgroups. Then the canonical map $R^{G_1} \otimes R^{G_2} \to R^{G_1 \cap G_2}$ is ...
Zheming Xu'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
5 votes
1 answer
349 views

Commutator subgroup of $\mathrm{GL}_n(R)$ when $R$ is a PID with unity

Hua and Reiner in their paper titled "Automorphisms of the unimodular group" have established what will be the commutator subgroup of $\mathrm{GL}_n(\mathbb{Z})$, $\forall n\geq 0$. I have ...
Guest's user avatar
  • 51
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
1 vote
0 answers
305 views

Presentation of Chevalley groups over Bezout domains

Let $\Phi$ be a root system of type $A_1$, $A_2$, $B_2$ or $G_2$. For a (commutative, unital) ring $R$, consider the group $G_{\Phi}(R)$ defined by Steinberg's presentation as in [1, Theorem 12.1.1 ...
Timothée Marquis's user avatar
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
4 votes
0 answers
219 views

Map $\operatorname{Sym}^{mp}(V^*) \longrightarrow K^{q}$ defined by $q$ points in $\operatorname{Sym}^p(V)$

EDIT : I have edited the question and made it more specific with respect to the kind of answer I expect. Let $V$ be a finite dimensional $K$-vector space and let $x_1, \dotsc, x_q \in V$ be $q$ points,...
Libli's user avatar
  • 7,300
1 vote
0 answers
116 views

List of automorphism groups of low-dimensional complex commutative algebras?

Let $\mathcal{A}$ be a finite-dimensional commutative associative unital $\mathbb{C}$-algebra. I am looking for a list (of further examples of) $\operatorname{Aut}_\mathbb{C}(\mathcal{A})$, the group ...
M.G.'s user avatar
  • 7,127
3 votes
0 answers
71 views

Automorphisms of matrix algebras and Picard group

This is a repost of https://math.stackexchange.com/questions/4692364/automorphisms-of-matrix-algebras-and-picard-group (asked on MSE). Notation. In what follows, $R$ is a commutative ring with $1$, $n\...
GreginGre's user avatar
  • 1,766
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
2 votes
1 answer
232 views

Invariants of the group algebra of a finite group

Consider a finite group $G$ and its complex group algebra $V_G$, on which $G$ acts. I would like to know: what are the polynomial $G$-invariants of $V_G$ i.e., the polynomial functions $p\in \mathbb{C}...
user493645's user avatar
7 votes
2 answers
917 views

Is this exact sequence known?

$\newcommand{\Tors}{{\rm Tors}} \newcommand{\tf}{{\rm\, t.f.}} \newcommand{\Gt}{{\Gamma\!,\,\Tors}} \newcommand{\Gtf}{{\Gamma\!,\tf}} \newcommand{\Q}{{\mathbb Q}} \newcommand{\Z}{{\mathbb Z}} \...
Mikhail Borovoi's user avatar
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
0 votes
1 answer
216 views

A Newton identity and the primes--the Faber partition polynomials and modular arithmetic

[Edit, July 6, 2022: Removed erroneous characterization of Faber polynomials as an Appell sequence.] Dress and Siebeneicher in their tale of the Burnside family express an opinion (1.2) that, if I ...
Tom Copeland's user avatar
  • 10.5k
14 votes
0 answers
432 views

Surprisingly only real points on intersection of certains quadrics

Let $G$ be a finite group and let $X_g$ be variables indexed by $G$. Consider the complex algebraic set defined by \begin{align} X_e &= 0\\ X_g &= X_{g^{-1}}\;\;\text{ for all }g\in G,\\ X_g &...
Peter Mueller's user avatar
5 votes
1 answer
523 views

Is there a non-split algebraic torus (over a finite field) satisfying the following properties?

Is there a non-split algebraic torus $T$ (over a finite field $\mathbb{F}_{\!q}$) satisfying the following properties? $T$ is not $\mathbb{F}_{\!q}$-isomorphic to the direct product of algebraic tori ...
Dimitri Koshelev's user avatar
0 votes
0 answers
151 views

zero divisors of group ring when the group is abelian

Let G be an abelian group with torsion and C[G] be the group ring over complex numbers C. Is there a clear description or classification of zero divisors of C[G]?
HuiRong's user avatar
  • 31
5 votes
0 answers
260 views

When is the profinite completion of a Noetherian group ring also Noetherian?

Let $G$ be a group, and let $\mathbb{Z}[G]$ denote its group ring. Its profinite completion is the inverse limit over all ideals of finite index. By Benjamin Steinberg's answer here, this profinite ...
stupid_question_bot's user avatar
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
3 votes
1 answer
226 views

Absolute Galois group with unique closed non-open subgroup

Is there an absolute Galois group that is not a subgroup of $\hat{\mathbb{Z}}$ and that has one and only one closed non-open subgroup?
klmntn's user avatar
  • 53
0 votes
1 answer
426 views

Generators of $SL(n,\mathbb F_2)$? [closed]

Consider the invertible matrices in $\mathbb F_2^{n\times n}$ which are a multiplicative group structure. Is there a finite set of $2k$ (at a $k\in\mathbb Z_{\geq1}$ independent of $n$) generators for ...
Turbo's user avatar
  • 13.9k
1 vote
0 answers
174 views

What are the irreps in this canonical action of $\operatorname{PGL}_2(F_q)$?

Consider the permutation action of $\operatorname{PGL}_2(\mathbb F_q)$ on $\mathbb P^1(\mathbb F_q)$ by fractional linear transformations. We can consider the associated (complex) representation of ...
Asvin's user avatar
  • 7,746
3 votes
0 answers
180 views

Units in group rings

Let $F$ be any field with $p$ elements and $G$ be any finite $p$-group, combining together they form a group ring $FG$. And $V(FG)$ denotes group of units of coefficient-sum equal to 1 in $FG$. We ...
HIMANSHU's user avatar
  • 381
2 votes
2 answers
265 views

Commuting nilpotent matrices and conjugation isomorphisms

Trying to study isomorphism classes of certain commutative Artinian $\mathbb{C}$-algebras I was lead to the following problem about matrices. Suppose you have a (non-zero) nilpotent matrix $A\in M_n(\...
amateur's user avatar
  • 375
0 votes
0 answers
82 views

Fast double exponentiation in finite fields

Let $p$ be a prime, and let $\mathbb{F}_p$ be the finite field with $p$ elements. Let $a$ be a non-zero element of $\mathbb{F}_p$. Can we quickly evaluate $a^{2^r} \mod{p}$? Using repeated squaring, ...
Gautam's user avatar
  • 1,703
3 votes
1 answer
161 views

Behavior of invariants under reduction mod p

Let $R$ be a finitely generated $\mathbb{Z}$-algebra with an [edit: linear algebraic] action of $G(\mathbb{Z})$ where $G$ is a split simply-connected semisimple group. Then for any prime $p$ we have a ...
user125639's user avatar
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 ...
Carlos Esparza's user avatar
3 votes
0 answers
98 views

Hales' generalization of the stacked bases theorem (seeking a proof)

In his paper Analogues of the stacked bases theorem, published in the proceedings of a 1976 conference, A.W. Hales claimed some interesting generalizations of the stacked bases theorem for abelian ...
Jose Brox's user avatar
  • 2,992
3 votes
0 answers
110 views

Is this a lattice?

Let $R$ be a locally compact ring (commutative with unit) and let $D\subset R$ be a discrete cocompact subring (cocompact means the additive group $R/D$ is compact). Let $G$ be a semisimple linear ...
user avatar
0 votes
1 answer
452 views

Conjugacy in the quaternion group

Let $G$ be a non-commutative group, and suppose we are given two elements $x, y \in G$ which are conjugate, i.e. we know there exists some $z \in G$ such that $zxz^{-1} = y$. Can we find $z$ given $x$ ...
Gautam's user avatar
  • 1,703
3 votes
3 answers
714 views

Cohomology of elementary abelian $p$-groups, i.e. $H(G,{\mathbb F}_p)$ with $G\cong{\mathbb F}_p^r$

I have two questions. $\bf 1.$ First, a reference request. Let $G\cong{\mathbb F}_p^r$ for some integer $r\geq 0$ and let $V=G^*={\rm Hom}(G,{\mathbb F}_p)$. Then $(H(G,{\mathbb F}_p),+,\cup )$ is a ...
Constantin-Nicolae Beli's user avatar
1 vote
0 answers
126 views

Algebraic structures on graphs

There are many algebraic structures linked to graphs. For example one can find zero divisor graphs $[1]$, $[2]$ and many other graphs. Does there exist any survey paper which characterizes all the ...
Charlotte's user avatar
  • 444
3 votes
0 answers
325 views

Intuitive, elementary intros to Hopf algebras/monoids

Motivation: I'm interested in understanding the role that noncrossing partitions play in Hopf algebras/monoids (HAs) as the components of the power series of the compositional inverse of formal ...
Tom Copeland's user avatar
  • 10.5k
1 vote
0 answers
179 views

Matrix factorizations over $GL_2$ of a real quadratic ring of integers

tl;dr: The groups $GL_2(K)$, or $SL_2(K)$, where $K = \mathbb{C,R}$ admits several factorizations (the polar decomposition, the KAN decomposition, the Schur triangular form, etc). Those ...
Adrián González Pérez's user avatar
6 votes
1 answer
299 views

Can a compact object be a nontrivial self-retract?

Let $\mathcal C$ be a locally finitely-presentable category, and let $X$ be a finitely-presentable object of $\mathcal C$. Question: Can there exist a nontrivial idempotent on $X$ whose fixed points ...
Tim Campion's user avatar
  • 63.9k
8 votes
2 answers
2k views

What generalizes symmetric polynomials to other finite groups?

Multivariate polynomial indexed by ${1, \ldots, n}$ are acted on by $S_n$: for $\sigma \in S_n$, define $\sigma(x_i) = x_{\sigma(x_i)}$, etc. Symmetric polynomials are those polynomials which are ...
yberman's user avatar
  • 781
-3 votes
1 answer
234 views

A common name for a functorial construction of Commutative Algebra?

I am interested whether the following construction naturally appearing in Commutative Algebra has some know and acceped name. Given a commutative monoid $(M,+)$ and a set $X$, consider the family $F(...
Taras Banakh's user avatar
  • 41.8k
2 votes
1 answer
413 views

$2$-adic valuation on $\mathbb Q (\sqrt{-15})$

I was reading the book "The structure of groups of prime power order". In the book (page 226, Example 10.1.18) it is proved that $-15$ has a square root is $\mathbb Z_2$ where $\mathbb Z_2$ ...
usermath's user avatar
  • 243
2 votes
0 answers
290 views

tangent space to a (not necessarily algebraic/Lie/..) group

Are there some standard ways to define the tangent space to a group $G$ at its unit element, $e$, when the group is not (pro)algebraic/(pro)Lie, not necessarily over a field, does not have the ``...
Dmitry Kerner's user avatar
7 votes
0 answers
140 views

Is there a homological interpretation for the cokernel of the kernel of a map between complexes induced by tensor product?

Let $A$ be a free abelian group of rank 2, and let $S = \mathbb{Z}[A]\cong\mathbb{Z}[a_1^{\pm1},a_2^{\pm1}]$ the group algebra for $A$. Let $t : S\times S\rightarrow S$ be the $S$-module map given by ...
stupid_question_bot's user avatar
36 votes
3 answers
2k views

The roots of unity in a tensor product of commutative rings

For $i\in\{1,2\}$ let $A_i$ be a commutative ring with unity whose additive group is free and finitely-generated. Assume that $A_i$ is connected in the sense that $0$ and $1$ are unique solutions of ...
Lviv Scottish Book's user avatar
19 votes
1 answer
4k views

How should I think about the module of coinvariants of a $G$-module?

Let $G$ be a group, $M$ a $G$-module, then the group of coinvariants is the module $M_G := M/I_GM$, where $I_G$ is the kernel of the augmentation map $\epsilon : \mathbb{Z}G\rightarrow \mathbb{Z}$. ...
stupid_question_bot's user avatar
1 vote
0 answers
397 views

A functor on the category of commutative rings, algebras or Banach algebras

Edit: According to the comments of abx and Yemon Choi I revise the question as follows: Let $G$ be a group and $\mathcal{A_G}$ be the category of $G$-module commutative algebras, that is the ...
Ali Taghavi's user avatar
7 votes
1 answer
716 views

How to compute the group cohomology of $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ with coefficient in a trivial module?

The group cohomology of cyclic groups can be computed easily due to the periodity. Now how can one compute the group cohomology $H^r(\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z},M)$? As least ...
Heer's user avatar
  • 997
-1 votes
1 answer
224 views

Group isomorphism $R^{\times} \simeq C_n\times C_2$? [closed]

Let $n$ be an positive integer and $R:=\mathbb{Z}[X]/(X^2,nX)$. Do we have $R^{\times} \simeq C_n\times C_2$ (the group of units of $R$) as it seems to be the case for small values of $n$. If so, do ...
Portland's user avatar
  • 2,829
4 votes
2 answers
507 views

Co-finite type abelian groups

Suppose $B$ is an abelian group such that for every integer $n\ge 1$, the $n$-torsion subgroup $B[n]$ is finite. Let $B_{\rm tor} = \varinjlim_{n\ge 1} B[n]$ be the torsion subgroup of $B$. Is it ...
user avatar
2 votes
0 answers
176 views

Trivial Tate modules

Let $A$ be an abelian group, and $p$ a prime. I'll call $$T_p(A) := \text{Hom}_{\mathbf{Z}}(\mathbf{Q}_{p}/\mathbf{Z}_{p}, A).$$ If $A$ is finite, then $T_p(A)$ is trivial, but the converse is not ...
user avatar
2 votes
1 answer
688 views

The Unit Group of $\mathbb{Z}_p$

Let $\mathbb{Z}_p$ the ring of $p$-adic numbers. It's known that the multiplicative unit group $\mathbb{Z}_p ^\times$ can be set theoretically described as $\bigcup _{1 \le a \le p-1} a+ p\mathbb{Z}_p$...
user267839's user avatar
  • 6,018
11 votes
1 answer
558 views

Given a filtration of a finitely generated module over a noetherian ring that "looks" split, is it split?

The following question came up while trying to determine whether the extension problems in a spectral sequence are trivial. Given a noetherian ring $R$ and a finitely generated $R$-module $M$ with a ...
Leon Hendrian's user avatar