All Questions
17 questions
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 $S$ ...
- 44.4k
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 ...
- 63.9k
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 ...
- 55.9k
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 ...
- 63.9k
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 ...
- 35.2k
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 ...
- 12.9k
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 ...
- 2,928
6
votes
4
answers
2k
views
Is a torsion-free abelian group finitely generated, if all of its localizations at primes $p$ are finitely generated over $\mathbb{Z}_p$?
Background: When proving that the group of $k$-isogenies $\mathrm{Hom}_k(A,B)$ between two abelian varieties is finitely generated, one first shows that the Tate map $$\mathbb{Z}_\ell\otimes_{\mathbb{...
- 63.9k
3
votes
1
answer
474
views
Structures of subgroups of a finite abelian p-group
$\newcommand\la{\langle}\newcommand\ra{\rangle}$Let $G=\mathbb{Z}/p^{i_1}\times\cdots\times\mathbb{Z}/p^{i_r}$ with $i_1\leq\ldots\leq i_r$ be a finite abelian $p$-group. Then there can be many ...
- 37.4k
3
votes
0
answers
98
views
Hales' generalization of the stacked bases theorem (seeking a proof)
In his paper Analogues of the stacked bases theorem, published in the proceedings of a 1976 conference, A.W. Hales claimed some interesting generalizations of the stacked bases theorem for abelian ...
- 2,992
4
votes
2
answers
507
views
Co-finite type abelian groups
Suppose $B$ is an abelian group such that for every integer $n\ge 1$, the $n$-torsion subgroup $B[n]$ is finite.
Let $B_{\rm tor} = \varinjlim_{n\ge 1} B[n]$ be the torsion subgroup of $B$.
Is it ...
- 63.9k
2
votes
0
answers
176
views
Trivial Tate modules
Let $A$ be an abelian group, and $p$ a prime.
I'll call $$T_p(A) := \text{Hom}_{\mathbf{Z}}(\mathbf{Q}_{p}/\mathbf{Z}_{p}, A).$$
If $A$ is finite, then $T_p(A)$ is trivial, but the converse is not ...
CommunityBot
- 1
14
votes
1
answer
696
views
$\mathbb{Z}$-module structure of the subring generated by an algebraic number
Let $a$ and $b$ be algebraic numbers which are not necessarily algebraic integers. Is there some invariant that allows us to determine whether $\mathbb Z[a]$ and $\mathbb Z[b]$ are isomorphic as $\...
- 149k
8
votes
2
answers
501
views
Exact sequence of $n$th powers of abelian groups
Let $A,B,C$ be finitely generated abelian groups. Assume that there is an exact sequence $$0 \to C \to A^n \to B^n \to 0,$$where $A^n = A \oplus \dotsc \oplus A$ as usual. It is not assumed that $A^n \...
- 149k
7
votes
1
answer
373
views
On describing a sort of "well-behaved" subgroups of a free abelian group
I found this question when I tried to figure out what kind of subgroups of a free abelian group behave just as well as in the finitely generated case.
Let $M$ be a free abelian group and $N$ a ...
- 35.2k
29
votes
0
answers
877
views
The field of fractions of the rational group algebra of a torsion free abelian group
Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions.
...
CommunityBot
- 1
9
votes
3
answers
3k
views
Why are divisible abelian groups important?
I just quote wikipedia:
"Divisible groups are important in understanding the structure of abelian groups, especially because they are the injective abelian groups."
I am asking for detail ...
- 57.6k