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.

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Every torsion free divisible abelian group D is direct sum of the copies of the $\mathbb{Q}$

Prove that every torsion free divisible abelian group $D$ is direct sum of the copies of the $\mathbb{Q}$. If $a\in D$ then there exists unique $b \in D$ and $n\neq 0 \in \mathbb{Z} $ such that a=nb....
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41 views

Properties of divisible groups

An abelian group $D$ is said to be divisible if given any $y \in D $ and $n\neq0 \in Z$, there exists $x \in D$ such that $nx = y$. (a) No nonzero finite abelian group is divisible. (b) No nonzero ...
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1answer
93 views

An abelian torsion group has a unique basic subgroup iff it is divisible or bounded.

This is Exercise 4.3.14 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE. The Details: Let $p$ be prime. A $p$-group is a group ...
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1answer
48 views

Is Prüfer group divisible over the ring of $p$-adic integers?

Let $p$ be a prime number. Let $$\mathbb{Z}_{p}=\left\{\sum_{i=0}^\infty a_ip^i\mid a_i\in \{0,1,2,\dots,p-1\}\right\}~~and~~\mathbb{Z}_{p^\infty}=\left\{ \frac{a}{p^n}+\mathbb{Z}\mid a\in \mathbb{Z}\...
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1answer
79 views

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 ...
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0answers
24 views

Abelian divisible p-groups and Prüfer group

I've been asked to show tha if D is an abelian divisible p-group then is isomorphic to the sum of copies of $\mathbb{Z}(p^{\infty})$, which is to say that there exists a set $X$ such that $D\cong\sum_{...
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1answer
51 views

Largest divisible subgroup not an intersection

Is there an abelian group $G$ for which the largest divisible subgroup of $G$ (given by the sum of all divisible subgroups) is not the intersection of all subgroups of the form $nG$ over the positive ...
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43 views

Proving a Certain Subgroup of $\mathbb{Q}/\mathbb{Z}$ is Divisible

In an effort to compute the injective envelope of $\mathbb{Z}_p$ for $p$ prime, I need to show that the group $A \subset \mathbb{Q}/\mathbb{Z}$ generated by $\{1/p^r : r \in \mathbb{Z}^+\}$ is ...
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2answers
53 views

Prove that there is an $R$-module isomorphism between $Q \otimes_R N \cong N$, Q is a quotient field of $R$.

Let $R$ be an integral domain with quotient field Q and $N$ be a unitary, divisible, torsion-free left $R$-module. Show that there is an $R$-module isomorphism so that $Q \otimes_R N \cong N$. Here ...
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81 views

A divisible abelian group is the direct sum of torsion subgroup and a torsion-free divisible subgroup.

Let $D$ be a divisible abelian group. It is known that $D \cong D_t \oplus D/D_t$, where $D_t$ is the torsion subgroup and $D/D_t$ is torsion-free and divisible. I want to find a torsion-free ...
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Divisible totally ordered additive abelian groups [closed]

Let $(G,+,\leq)$ be a divisible totally ordered additive abelian group and $g_{1},g_{2}\in G$. If for every integer $n>1$, $g_{1}\geq (1-\frac{1}{n})g_{2}$, Can we have $g_{1}\geq g_{2}$? Thanks ...
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1answer
95 views

An abelian, characteristically simple group is divisible (supposedly).

This is Exercise 4.13 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE. The Details: On page 69 of Roman's book, we ...
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3answers
79 views

Find common number divisible by six different numbers

If there is recipe to find this - I would like to find the first common number divisible by the following six numbers- 260, 380, 460,560,760 and 960. How does one calculate the numbers I need? Any ...
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0answers
46 views

T.Y. Lam's divisible module definition and factor modules.

In surveying LMR of T.Y.Lam and get the divisible module (for any ring with unity not necessary an integral domain) definition as follows: ``A right $R$-module $I_R$ is called divisible if and only if ...
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1answer
80 views

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 \...
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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 ...
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47 views

prove a field is a divisible group

I saw a statement: every field of characteristic 0, with its underlying additive group structure, is divisible. I met with troubles in checking the above statement. Suppose $\operatorname{char}(F)=...
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177 views

Showing groups are not divisible

Let $G$ be an abelian group and use additive notation. Call $G$ a divisible group if given x in G and a positive integer m we can always find an element y of $G$ such that $my = x$. For example, $\...
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1answer
28 views

Explanation on examples of non-divisible module

So the multiplicative group of complex numbers, $\mathbb{C}^*$, is divisible as an abelian group. Why is the multiplicative group of real numbers, $\mathbb{R}^*$, not divisible as an abelian group? ...
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1answer
213 views

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 $N$ ...
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60 views

$\mathbb{Z}_p$ has cohomological dimension 1

I would like to prove that $\mathbb{Z}_p$ has cohomological dimension 1, without using the fact that it is a free pro-p-group. If someone can suggest me any reference (or a hint for a proof) for ...
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1answer
162 views

Dieudonné module associated to the dual of a $p$-divisible group

Let $k$ be a perfect field of characteristic $p>0$, and consider $X=(X_m,i_m)$ a $p$-divisible group of height $h$ over $\operatorname{Spec}(k)$: it is an inductive system where $X_m$ is a finite ...
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171 views

Is $\mathbb{Z}_{p^\infty}$ a divisible group? [closed]

Let $$\mathbb{Z}_{p^\infty}=\{x\in \mathbb{Q}/\mathbb{Z}: \ \exists n\in \mathbb{N} ,\ \ p^nx=0\}. $$ Is $\mathbb{Z}_{p^\infty}$ a divisible group ?
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1answer
23 views

On set of all $Z$-module homomorphisms as injective module

Let $\mathbb{Z}$ be the integers. The set $M$ of all $\mathbb{Z}$-module homomorphisms from a ring $R$ with unity to a divisible abelian group $A$ is known to be an injective left $R$-module. I am ...
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1answer
53 views

Example of divisible direct product but not the direct sum

Here, divisible abelian group $G$ is defined such that for every $y\in G$ and nonzero integer $n$, we have $x\in G$ with $nx=y$. I am looking for an example where the direct product of groups is ...
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1answer
107 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 ...
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5answers
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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) $8325971640,$ $8365971240,$ $8317956240,$ $8291357640,$ $8325971640,$ $...
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3answers
379 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 ...
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1answer
136 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 ...
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1answer
48 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} \...
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1answer
98 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? $2)$ ...
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1answer
172 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 $...
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3answers
346 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 ...
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2answers
217 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 ...
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1answer
153 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 ...
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1answer
222 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 ...
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3answers
510 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 ...
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1answer
340 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 ...
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187 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.
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567 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=...
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0answers
159 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, ...
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1answer
149 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 $p$?
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1answer
147 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 = \...
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1answer
140 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 ...
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1answer
1k 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 ?
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289 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 ...
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286 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 ...
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0answers
68 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 ...
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3answers
4k 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 ...
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2answers
665 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 ...