# 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.

27 questions
1answer
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### 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 ...
5answers
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3answers
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### 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 ...
2answers
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### 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 ...
1answer
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### 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 ...
1answer
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### 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 ...
3answers
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### 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 ...
1answer
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### 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 ...
2answers
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### 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.
0answers
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### 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
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### 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 ...
1answer
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### 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 ...
1answer
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### 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 ?
2answers
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### 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 ...
2answers
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### 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 ...
0answers
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### 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 ...
2answers
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### 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 ...
3answers
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### 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 ...
1answer
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### 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 ...