This is for questions about abelian groups, each with a basis.
Questions tagged [free-abelian-group]
207 questions
31
votes
0 answers
A conjecture about Catalan sequence
For $n=2$, consider the free abelian group generated by polynomials corresponding to $\frac{(2 n)!}{2^n n!}=3$ partitions of $2n=4$ vertices into pairs (chords):
$$
(x_1-x_3)(x_2-x_4),(x_1-x_2)(x_3-x_4),(x_1-x_4)(x_2-x_3)
$$
Euler's Identity:…
hbghlyj
- 5,361
20
votes
1 answer
Is this group free abelian?
Let $K$ be the subgroup of $\mathbb{Z}^\mathbb{Z}$ consisting of those functions $f : \mathbb{Z} \to \mathbb{Z}$ with finite image. Is $K$ free abelian?
My guess is no, because $K$ feels too much like the Baer-Specker group $\mathbb{Z}^\mathbb{Z}$,…
diracdeltafunk
- 16,589
18
votes
6 answers
What is the definition of a free abelian group generated by the set $X$?
What is the definition of a free abelian group generated by the set $X$?
If $X=\{A,B,C,D\}$ what are the elements of the free abelian group on this set? What is the identity of this group?
Oliver G
- 5,132
12
votes
2 answers
Reference request for characteristic subgroups of free abelian groups
The characteristic subgroups of a free abelian group $F$ are the $k F$ with $k \in \mathbb{N}$. (In particular, they are verbal and hence fully invariant.) I know a proof when $F$ has finite rank, but I would like to know a reference of this fact in…
Martin Brandenburg
- 181,922
10
votes
1 answer
Is $\text{Hom}(A,\mathbb{Z})$ a product of free abelian groups for all abelian groups $A$?
Let $A$ be an abelian group, and consider the abelian group $\text{Hom}(A,\mathbb{Z})$ of homomorphisms from $A$ to $\mathbb{Z}$. What can be said about this group?
Since $\mathbb{Z}$ is torsion-free, so is $\text{Hom}(A,\mathbb{Z})$.
If $A$ is…
Lukas Lewark
- 373
10
votes
6 answers
Prove that the group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$ implies that $m = n$
Let $m$ and $n$ be two nonnegative integers. Assume that there is a group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$. Prove that $m = n$.
I tried using a contrapositive, ($m \neq n$ implies $\mathbb{Z}^m \ncong \mathbb{Z}^n$), and I think the…
chubbycantorset
- 2,091
8
votes
1 answer
Does every finitely presented group have a finite index subgroup with free abelianisation?
Let $G$ be a finitely presented group. Does there exist a finite index subgroup $H$ such that its abelianisation $H^{\text{ab}} = H/[H, H]$ is free abelian?
Note, if $G^{\text{ab}}$ is not already free abelian, then it is non-zero and surjects…
Michael Albanese
- 104,029
7
votes
3 answers
Order of the set of group homomorphisms from $\mathbb{Z}^n$ into an arbitrary finite group $G$.
Question: Let a finite $G$ act on itself by conjugation, and let $N$ be the number of conjugacy classes. Find a formula for $|\mathrm{Hom}\left(\mathbb{Z}^n, G\right)|$, denoting the set of group homomorphisms from $\mathbb{Z}^n$ into…
Charles
- 356
6
votes
1 answer
Question about accessibility of category of free abelian groups.
I've read, that the accessibility of the category of all free abelian groups is independent on basic set theory (say ZFC). What is the reason for that? And how can I interpret this result? Does it mean that the property of been a free abelian group…
bbxlmnistvii
- 177
- 5
6
votes
2 answers
Proof that any element of a free abelian group which is not divisible by any $k>1$ can be extended to a basis
Let $A$ be a free abelian group of rank $n$, and let $\alpha_1 \in A\setminus \{0\}$ such that $\alpha_1 \not \in kA$ for all $k > 1$. Do there always exist $\alpha_2,\ldots, \alpha_n \in A$ such that $\alpha_1,\ldots, \alpha_n$ is a basis for…
Sebastian Monnet
- 3,979
6
votes
1 answer
Let $G$ be a torsion-free abelian group, prove that for $g\neq h\in G$ there exists a homomorphism $\phi:G\to\Bbb{R}$ such that $\phi(g)\neq\phi(h)$
I have the following question:
Let $G$ be a torsion-free abelian group, prove that for every distinct elements $g, h \in G$ there exists a homomorphism $\phi: G \rightarrow \mathbb{R}$ such that $\phi(g)\neq \phi (h)$.
I have some observations, if…
Odi
- 129
6
votes
1 answer
On free abelian groups
I'm learning about the concept of a free abelian group. First question: nowhere it is stated that these groups cannot be finite, but the definition seems to imply it. Is this true?
Second question: an isomorphism to a free abelian group $A = \left<…
Jos van Nieuwman
- 2,142
5
votes
1 answer
There are only finitely many integer lattices with bounded covolume
I have an uninsightful proof for the following lemma (I discuss motivation below). Let $C>0$ and $n\geq 1$ be fixed, set $\def\Z{\mathbb{Z}}M=\Z^n$.
Lemma. There are only finitely many subgroups $H\subset M$ with $\#(M/H) \leq C$.
Below is a…
Olivier Bégassat
- 21,101
5
votes
2 answers
The existence of group isomorphism between Euclidean space.
Is there any group isomorphism for addition $\mathbb{R}^n$ to $\mathbb{R}^m$ where $n\neq m$? I could prove that there exists any vector space isomorphism or smooth map, but I still could not know that if we consider only abelian group structure for…
afdsfasdf
- 357
4
votes
0 answers
Does the category of free abelian groups have sequential colimits?
Does the category $\mathbf{FreeAb}$ of free abelian groups have sequential colimits?
A sequential colimit is a colimit of the countable shape
$\bullet \to \bullet \to \bullet \to \cdots,$
i.e. indexed by $(\mathbb{N},\leq)$. This is a special case…
Martin Brandenburg
- 181,922