Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [divisible-groups]

For questions about the structure and properties of a divisible group, which are Abelian groups in which one can "divide" by positive integers.

An abelian group $G$ is called divisible if for every $x \in G$ and $n \in \mathbb{Z}^+$, there exists a $y \in G$ for which $x = ny$.

The group of rational numbers $(\mathbb{Q}, +)$ is divisible, as are the quasicyclic groups $\mathbb{Z}_{p^\infty}$ for prime numbers $p$. In fact, every divisible group can be decomposed as a direct sum of copies of $\mathbb{Q}$ and quasicyclic groups for various primes.

Divisible groups are precisely the injective $\mathbb{Z}$ modules, and so are important in studying the structure of abelian groups.

Reference: Divisible group.

60 questions
19
votes
2 answers

Is $SO_n({\mathbb R})$ a divisible group?

The title says it all ... Formally, if $SO_n(\mathbb R)=\lbrace A\in M_n({\mathbb R}) |AA^{T}=I_n, {\sf det}(A)=1 \rbrace$ and $W\in SO_n(\mathbb R)$, is it true that for every integer $p$, there is a $V\in SO_n(\mathbb R)$ satisfying $V^p=W$ …
Ewan Delanoy
  • 63,436
7
votes
2 answers

Divisible groups, exercise from Rotman's theory of groups

The following exercise is from Rotman, An Introduction to the theory of groups, 4th ed, p324. "The following conditions on a group G are equivalent: (i) G is divisible, (ii) Every nonzero quotient of G is infinite, (iii) G has no maximal…
user129773
  • 113
6
votes
1 answer

Extension of divisible fields

Assume that $F$ is an infinite subfield of a field $K$ such that its multiplicative group, $F^\times$, is divisible. Also, $a\in K$ and $[F(a):F]<\infty$. Can we conclude that the multiplicative group of $F(a)$ is a divisible group? For example…
Reza Fallah Moghaddam
  • 692
5
votes
3 answers

What are the finite order elements of $\mathbb{Q}/\mathbb{Z}$?

I need to find what are the at the group $\mathbb{Q}/\mathbb{Z}$. I think that any element at this group has a finite order, but I don't know how to prove it... I'd like to get help with the proof writing... If I'm wrong, I'd like to to know it…
CS1
  • 2,091
5
votes
1 answer

Socle of abelian divisible periodic group

I'm trying to prove that the socle of a periodic divisible abelian group J is a proper subgroup of J. I know that J is direct sum of quasicyclic groups, say $$ J={\oplus}_{i\in I} P_i $$ and that the socle of J is $$ Soc(J)={\oplus}_p J[p] $$ Now,…
Rob
  • 129
5
votes
0 answers

Splitting of homomorphism onto Prüfer group

Let $G$ be a totally disconnected locally compact abelian group. Let $U \leq G$ be open and such that $G/U \cong \mathbb{Z}(p^\infty)$, the Prüfer $p$-group. In general $U$ is not necessarily a direct summand of $G$: for instance when $G =…
frafour
  • 3,125
5
votes
5 answers

Divisibility 1,2,3,4,5,6,7,8,9,&10

Tried: Seems the ten-digit number ends with $240$ or $640$ or $840$ (Is not true, there are more ways the number could…
user608997
5
votes
3 answers

Isomorphism between $\mathbb{T}$ and $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$.

I'm trying to prove that the circle group $\mathbb{T}$ is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ with a little bit of cardinal arithmetics. First, I know that $\mathbb{T}$ can be decomposed (structure theorem of divisible groups) as…
Gilberto López
  • 400
5
votes
1 answer

A group is divisible if and only if it has no maximal subgroup ?

Is it true that a group is divisible if and only if it has no maximal subgroup ?
user123733
4
votes
1 answer

If $G, H, K$ are divisible abelian groups and $G \oplus H \cong G \oplus K$ then $H \cong K$

This is an exercise in Hungerford. But can somebody explain why is the following not a counter-example? Let $G$ be the direct sum of $|\mathbb{R}|$ copies of $\mathbb{Q}$. Let $K$ be the direct sum of $|\mathbb{N}|$ copies of $\mathbb{Q}$. Then $G…
Pedro
  • 6,936
4
votes
1 answer

The pure subgroups of a divisible abelian group are just the direct summands.

This is Exercise 4.3.3 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE. (NB: I have left out the modules tag for a reason: the tools available here are entirely group theoretic.) The…
Shaun
  • 47,747
4
votes
1 answer

Right exactness of quotienting out the maximal divisible subgroup

For every abelian groups $G$ let $\mathrm{d}G$ be its maximal divisible subgroup. Then $G \mapsto G/\mathrm{d}G$ is a right exact functor $\mathbf{Ab} \to \mathbf{Ab}$. Let $$ 0 \to G \xrightarrow{i} H \xrightarrow{p} K \to 0 $$ be an exact…
scsnm
  • 1,321
4
votes
1 answer

If $M$ and $N$ are divisible, abelian groups, show that their tensor product $M \otimes_\mathbb{Z} N$ is uniquely divisible

Recall that an abelian group $M$ is divisible if for each $m \in M$ and $r \in \mathbb{Z}$, there is an $m' \in M$ such that $rm' = m$. It is uniquely divisible if that $m'$ is unique. If $M$ and $N$ are divisible, abelian groups, show that their…
michiganbiker898
  • 1,573
4
votes
1 answer

Relationship between Archimedean and Divisible ordered groups

Let $(G,+,\leq)$ be a linearly ordered abelian group (i.e. the order is total and compatible with the sum) and $n\cdot x$ denote the classical action of $\mathbb{Z}$ over $G$ (i.e. $0$ for $n=0$, sum of $|n|$ copies of $x$ if $n>0$ and of $-x$ if…
AlienRem
  • 4,130
  • 1
  • 27
  • 38
4
votes
1 answer

Homomorphism from divisible group to finite group is always trivial

Let $A$ be a divisible group, let $B$ be a finite group, and let $f: A \rightarrow B$ be a homomorphism. Show that $f$ is trivial. (A group $A$ is divisible if for each $a \in A$ and $n \ge 1$ there exists some $b \in A$ such that $b^n = a$) I…
ChikChak
  • 2,062
1
2 3 4