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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.

17
votes
2answers
484 views

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 ...
7
votes
2answers
172 views

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 ...
6
votes
1answer
141 views

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 ...
5
votes
5answers
230 views

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 end) $8325971640,$ $8365971240,$ $8317956240,$ $8291357640,$ $8325971640,$ $...
5
votes
3answers
175 views

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 ...
4
votes
1answer
66 views

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 ...
4
votes
1answer
68 views

Torsion-free, divisible multiplicative groups of fields

For which cardinalities $\kappa$ is $\def\Q{\mathbb Q}\Q^{(\kappa)}$—by which I mean the direct sum of $\kappa$-many copies of $\Q$—isomorphic (as an abelian group) to the multiplicative ...
4
votes
1answer
304 views

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 ...
3
votes
3answers
1k views

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 ...
3
votes
2answers
248 views

Irreducible subgroups of the additive rationals

Let $G$ be a group. A proper subgroup $H$ is called irreducible if $H$ can't be written as an intersection of two subgroups which contain it properly. I'd like to know if $(\mathbb Q,+)$ (and more ...
2
votes
1answer
261 views

group law of complex torus is divisible?

I need help with this exercise: Show that the group law of a complex torus (the definition I have is that of Rick Miranda's book Algebraic curves and Riemann surfaces, the one that he constructs from ...
2
votes
1answer
22 views

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 ...
2
votes
2answers
321 views

$(G,+)$ abelian group is divisible $\Longleftrightarrow$ it's an homomorphic image of $\Bbb Q^{(X)}$

Let $(G,+)$ be an additive abelian group. Let us suppose $G$ divisible (i.e. $G=nG\;\;\;\forall n\ge1$). Let then $x,y\in G$. Then there exists $z\in G$ and $n,m\ge1$ such that $x=nz$ and $y=...
2
votes
1answer
627 views

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 ?
2
votes
1answer
51 views

Does every ordered divisible abelian group admit an expansion (and how many) to an ordered field?

Let $(G,+,<)$ be an ordered divisible abelian group. $1)$ Is it always the case that there exists a binary function $*:G\times G \rightarrow G$ such that $(G,+,*,<)$ is an ordered field? ...
2
votes
1answer
209 views

If an abelian group $A$ is injective as $\mathbb{Z}$ module then $A$ is a divisible group

I was reading a proof to this proposition from this link: http://planetmath.org/abeliangroupisdivisibleifandonlyifitisaninjectiveobject they proceed by contradiction, so $A$ is not divisible and ...
2
votes
1answer
106 views

Relation between open divisible subgroup and the quotient of the group with subgroup

I wanted to prove the following proposition: Let H be an open divisible subgroup of an abelian topological group G. Then G is topologically isomorphic to H x G/H. As for the proof, using extension of ...
2
votes
1answer
89 views

Unit group of an algebraically closed field is divisible

In the lecture notes on Valuation theory, in Ex. $1.16$ on page $11$ we are asked to show that: If $k$ is an algebraically closed field, then $k^{\times}$ is a divisible abelian group. Isn't $k = \...
2
votes
0answers
63 views

universal nonabelian divisible group

For this post, a group $G$ shall be referred to as generally divisible, in case $\forall{x\in G:}~\forall{n\in\mathbb{N}^{\times}:}~\exists{y\in G:}~y^{n}=x$. Note. Here is no commutativity ...
1
vote
2answers
33 views

Let G be a finite group. Let $a,b\in G$ be two distinct elements or order 2. Prove that if $ab=ba$, then the order of G is divisible by 4.

Let $G$ be a finite group. Let $a,b\in G$ be two distinct elements or order 2. Prove that if $ab=ba$, then the order of $G$ is divisible by 4.
1
vote
2answers
158 views

Prove that$ k^3n-kn^3$ is divisible by $6 $ for all n∈N. [closed]

Hello I have problem with solution of task. Prove that $k^{3}n-kn^3$ is divisible by $6$ for all $n∈N$, $k∈N$ . Help me, please. I know, when $n^3-n$ is divisible by 6. $n^3-n= (n-1)(n)(n+1)$ and ...
1
vote
1answer
58 views

$\mathbb{Z} [1/p] / \mathbb{Z}$ is divisible

I can't see why this group $\mathbb{Z} [1/p] / \mathbb{Z}$ is divisible. It is $p$-divisible, but how is it divisible by any integer $n$ when the denominator is only powers of $p$?
1
vote
1answer
72 views

If every nonzero quotient of $G$ is infinite, then $G$ is divisible

I tried proving this by contradiction (assuming $G$ is abelian, else $G$ could not possibly be divisible): If $G$ is not divisible then there exist $g\in G$ and $n\in \mathbb {N}$ such that for all $...
1
vote
0answers
103 views

Structure theorem for divisible modules

I would like to know if there is some analogue theorem of structure for divisible modules as there is for divisible abelian groups. More exactly, given a divisible $R$-module $M$, where $R$ is a PID, ...
0
votes
2answers
113 views

Largest divisible subgroup of an abelian group

How do I prove that any abelian group $G$ contains divisible subgroup $H$, such that $G / H$ has no divisible subgroups other than $\{0\}$? Attempts: 1) Using Zorn's lemma was suggested to me in ...
-1
votes
3answers
192 views

Remainder when $2^{108}$ is divided by $11$? [closed]

What is the remainder obtained when $2^{108}$ is divided by $11$? I tried bringing in $11$ in the given no. such as in $(11-3)^{36}$ and then using binomial expansion...but its not helping. Any ...
-2
votes
1answer
29 views

Show that $Q_p / \mathbb Z$ is divisible as $\mathbb Z$-module. [closed]

Let $p \in \mathbb N$ be a prime. Let $$Q_p : = \left \{ x \in \mathbb Q : (\exists k \in \mathbb Z)\ \mathrm {and}\ (\exists n \in \mathbb N)\ \mathrm {such}\ \mathrm {that}\ x= \frac {k} {p^n} ...