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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 $A \not\cong A \oplus \mathbb{Z}$?
Martin Brandenburg's user avatar
12 votes
2 answers
2k views

Do all exact sequences $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split for finitely generated abelian groups?

Suppose $A$ and $B$ are finitely generated Abelian groups. Are all exact sequences of the form $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split? If not, is there an example?
Felix Y.'s user avatar
  • 307
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 ...
Martin Brandenburg's user avatar
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. ...
Jeremy Rickard'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
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 $...
Ben Webster's user avatar
  • 44.7k
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 ...
13 votes
1 answer
791 views

How nearly abelian are nilpotent groups?

It is not uncommon to read that "nilpotent groups are 'close to abelian'."1,2 Can this sentiment be made precise in the sense of the Turán and Erdős definition of "the probability that two elements of ...
Joseph O'Rourke's user avatar
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
10 votes
1 answer
2k views

A group whose automorphism group is cyclic

Is there an Abelian group $A$ which is not locally cyclic whose automorphism group is cyclic ? This question was first posted here.
W4cc0's user avatar
  • 599
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
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
5 votes
2 answers
907 views

What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$?

What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$? I have a reference that says the group in question is $\mathbb{Q}_p,$ the additive group of the quotient field of the $p$-adic integers. Can ...
Chris Leary's user avatar
4 votes
2 answers
449 views

Number of torsion-free abelian groups

Let $\mathfrak{c}$ be the cardinality of the continuum. How much Choice, if any, is needed to prove that there are $2^{\mathfrak{c}}$ distinct (mutually nonisomorphic) torsion-free abelian groups of ...
Bruce Blackadar's user avatar
4 votes
1 answer
383 views

Finite dimensional compact abelian group that is not a product of connected and a totally disconnected

Let $G$ be a compact abelian group. A compact abelian group is said to have dimension $n$ if $\dim_\mathbb{Q} \mathbb{Q}\otimes \hat G = n$. Equivalently one can show that this holds if $G$ is ...
TopGroups's user avatar
2 votes
1 answer
682 views

One-dimension Algebraic groups

I am searching for a possible analogue of a result in algebraic groups in a non-commutative setting, so I am looking for different proofs of the following : Let $K$ be an algebraically closed field. ...
Drike's user avatar
  • 1,555
2 votes
2 answers
474 views

Non-archimedean group over the reals

I have a totally ordered group $(\mathbb{R};\leq,\oplus,0,-)$ with the reals as base set satisfying monotonicity, i.e. for all $x,y,z$ we have that if $y\leq z$ then $x\oplus y \leq x\oplus z$, and I ...
chros's user avatar
  • 43
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