3
votes
3answers
66 views

An epimorphism from $\mathbb Z⊕\mathbb Z⊕\cdots$ to $\mathbb Q$

I want an explicit example of an epimorphism from $\mathbb Z⊕\mathbb Z⊕\cdots$ to $\mathbb Q$. Thanks.
3
votes
2answers
41 views

Abelian group and morphism equivalent statement

Exercise Show that the following statements are equivalent: $(i) \space G \space \text{is abelian.}$ $(ii) \space \text{the map f: G} \to \text{G defined as} \space f(x)=x^{-1} \space \text{is a ...
2
votes
3answers
49 views

When is a non-trivial homomorphism injective?

I noticed that over the natural numers $(\mathbb{Z},+)$ any group homomorphism $f : \mathbb{Z} \rightarrow \mathbb{Z}$ that is not the trivial one, is automatically injective. Where exactly does ...
1
vote
2answers
23 views

Does the following binary operation form a group on a set with 3 elements? (multiple identities?)

Let S = {a, b, c}. *| a b c ----------- a| a b c b| b a a c| c a a This seems to have all the desired characteristics of a group, however, both b and c ...
1
vote
0answers
58 views

A sequence of subsets of $\Bbb Z$ not containing nontrivial subgroups

Is there a sequence $(A_n)$ of subsets of $\Bbb Z$ such that always $\{a-b\mid a,b\in A_{n+1}\}$ is a proper subset of $A_n$ and no $A_n$ contains an infinite subgroup of $(\Bbb Z,+)$?
2
votes
1answer
49 views

About a nested sequence of subsets of integers

Let $(H_n)$ be a sequence of nonempty subsets of $\Bbb Z$ such that always $\{a-b\mid a,b\in H_{n+1}\}\subsetneqq H_n$. Can we deduce that there is some $n$ such that $\{a-b\mid a,b\in H_{n}\} = ...
0
votes
0answers
30 views

If the automorphism group of a group is cyclic, then the group is commutative [duplicate]

Let $G$ be a group and the $Aut(G)$ group is cyclic $\Rightarrow$ the group $G$ is commutative. I looked at the homomorphism $\varphi : G \rightarrow Aut(G) \ g \mapsto (x \mapsto gxg^{-1})$. Let ...
4
votes
2answers
49 views

Is $<\mathbb Q^+, \times>$ the free abelian group on countably infinitely many generators?

It seems to make sense to me that it should be, with the generators being the set of primes. However, I'm not sure that my intuition is right. Additionally, would this not be contradicted by the fact ...
0
votes
1answer
40 views

easy short exact sequence question

Suppose I have have a short exact sequence of finitely generated Abelian groups $0 \longrightarrow G \overset{f}\longrightarrow H \overset{g}\longrightarrow K \longrightarrow 0$. Suppose I have a ...
1
vote
2answers
88 views

On finite exponent abelian $p$-groups

Let $G$ be an abelian $p$-group non-isomorphic to any group of the form $H\times K$ where $H$ and $K$ are nontrivial groups. And let $\{|a|\mid a\in G\}$ have an upper bound in $\Bbb N$ . Is $G$ ...
3
votes
3answers
124 views
+50

Whether two quotients of $\mathbb{Z}^2$ are isomorphic.

Let $H_1$ be the subgroup of $\mathbb{Z}^2$ generated by $\{(1,2),(4,1)\}$, let $H_2$ be the subgroup of $\mathbb{Z}^2$ generated by $\{(3,2),(1,3)\}$. Is it true that $\mathbb{Z}^2/H_1\cong ...
-1
votes
3answers
57 views

$G$ is an abelian group of order a product of distinct primes $\implies G$ is cyclic?

If $G$ is an abelian group of order $p_1p_2...p_k$ , where $p_1,p_2,...,p_k$ are distinct primes , then is it true that $G$ is cyclic ?
6
votes
3answers
100 views

Automorphisms of $\mathbb{Z}/p\oplus\cdots\oplus \mathbb{Z}/p$

Consider the abelian group $$G = \underbrace{\mathbb{Z}/p\oplus\cdots\oplus \mathbb{Z}/p}_{n},$$ where $p$ is prime and $1\le n \le p$. I want to show that $G$ has no automorphism of order $p^2$. I ...
6
votes
1answer
83 views

About translating subsets of $\Bbb Z.$

This is a continuation of About translating subsets of R2. Is it possible to find a pair of sets $A,B\subseteq\Bbb Z$ such that A is a union of translated (only translations are allowed) copies of ...
1
vote
1answer
42 views

Are these exactly the abelian groups (2)?

This is a continuation of Are these exactly the abelian groups? I would like to consider another condition on a group and see if it implies commutativity. The condition is $$(\forall A,B\subseteq ...
0
votes
1answer
43 views

Under what conditions can the symmetric group be isomorphic to the abelian group?

The symmetric group is the set of all permutations. My question addresses the representability of the symmetric group using only additions. I am guessing that on the finite field $\mathbb{Z}/n ...
2
votes
2answers
49 views

How to count the number of elements of given order?

I am trying to prove the following result. Let $G$ and $G'$ be two finite abelian groups. Besides, they have the same number of elements of any given order. Prove that $G\cong G'$. My attempt is ...
3
votes
3answers
155 views

Basic Group Theory question

This is not so much a plea of ignorance, but rather me trying to see whether intuitively I actually understand what is going on in group theory. The question asks What group is ...
0
votes
1answer
48 views

generators of groups from exact sequence

Suppose I have a middle term exact sequence of finitely generated abelian groups $G \longrightarrow H \longrightarrow K$. How do I get the generators of $H$ if I know the same for other two groups?
0
votes
0answers
48 views

Finite generated abelian group $G$ and $H<G$. What is the rank of $(G/H)/(G/H)_t$?

I saw another question about this problem here. However there are quite different answers from my expectation. Anyway, here are my trials. Trial 1 : By structure theorem, $G\cong G_t\oplus F_1$ ...
2
votes
1answer
49 views

If direct limits of matrices are isomorphic, is the direct limit of the transpose matrices also isomorphic?

On the one hand, the following conjecture seems reasonable, but on the other hand it doesn't seem natural because some objects are being dualised while others are not. I would appreciate if anyone ...
5
votes
1answer
98 views

Additive non-abelian group?

Sometimes I see in books the term "additive abelian groups". In my opinion, when we use addition to represent the group operation, we already have in mind that the operation is commutative. So ...
0
votes
3answers
73 views

A question on Abelian Groups

Prove that every subgroup of an Abelian group is Abelian but the converse is not true. I recently stumbled onto this question , but not able to solve it . Please help me out!
3
votes
5answers
88 views

Proof that a group is abelian.

If $(G,*)$ is a group and $(a * b)^2 = a^2 * b^2$ then $(G, *)$ is abelian for all $a,b \in G$. I know that I have to show $G$ is commutative, ie $a * b = b * a$ I have done this by first using ...
1
vote
1answer
49 views

Commutators Calculus

I was trying to understand the above Corollary but I have a problem, namely why in the second to last line $A_0 \leq \zeta_p(G)$? Any ideas? Definitions By recurrence we define $[x,_0\, y]=x$; ...
1
vote
4answers
65 views

Proving that a group $(G, \ast)$ is abelian if $x^3=x$ for all $x\in G$

If $(G, \ast)$ is a group so that $x^3=x$ for all $x\in G$ then $G$ is abelian
1
vote
1answer
38 views

A group with bounded element orders and its minimal and maximal subgroups.

Let $n>1$ be an integer. Is there an abelian group $G$ with all elements of order less than $n$ for which exactly one of these conditions is correct: 1) every non-trivial subgroup of $G$ contains ...
0
votes
0answers
18 views

maximal and minimal subgroups of torsion abelian groups

Is there a torsion abelian group $G$ for which exactly one of these conditions is correct: 1) every non-trivial subgroup of $G$ contains a minimal (non-trivial) subgroup of $G$. 2) every proper ...
0
votes
2answers
51 views

What can I say about the quotient group?

Let $G$ be a group of order $24$, and let $H$ be a normal subgroup of order $6$. So the quotient group $ {G\over H} $ is Abelian group?. What can I say about the quotient group beside her order?
1
vote
0answers
23 views

Automorphism group of an abelian p-group

I'd like to know if it's known the structure of the automorphism group of an abelian $p$-group with the minimal condition on subgroups, for some prime number p. I know that if $A$ is an abelian ...
4
votes
4answers
72 views

$n$th power map is an automorphism implies abelian group?

If $G$ is a finite group and $\phi(x) = x^n$ is an automorphism of $G$ does this imply $G$ is abelian? I've been reading this page. Def: A group $G$ is said to be $n$-abelian if $(ab)^n=a^nb^n$ ...
2
votes
1answer
37 views

Show that, given an element $\bar{x}\in\mathbb{Q}/\mathbb{Z}$, there is an integer $n \ge 1$ such that $n\bar{x} = 0$. [duplicate]

Consider $\mathbb{Z}$ as a subgroup of the additive group $\mathbb{Q}$ of rational numbers. >Show that, given an element $\bar{x}\in\mathbb{Q}/\mathbb{Z}$, there is an integer $n \ge 1$ such that ...
2
votes
1answer
53 views

How to prove that $\mathbb Z/p$ is not a direct summand of any direct sum of copies of $\mathbb Z/n$?

How can I prove that $\mathbb Z/p$ ($p$ is a prime) cannot be a direct summand of any arbitrary direct sum of copies of $\mathbb Z/n$, where $p^2$ divides $n$?
5
votes
1answer
44 views

Let $G$ be a group and $a,b,c \in G$. Given that $abc$ and $cba$ are conjugated, prove that $G$ is abelian.

Let $G$ be a group and $a,b,c \in G$. Given that $abc$ and $cba$ are conjugated, prove that $G$ is abelian. In other words, if for any $a,b,c \in G$ there is a $g \in G$ so that $a b c = g c b a ...
2
votes
1answer
43 views

$\operatorname{Hom}(G,\Bbb C\setminus \{0\})$ non- isomorphic to $\operatorname{Hom}(G, \Bbb T)$?

Do you have an example of an abelian group $G$ for which $\operatorname{Hom}(G,\Bbb C\setminus \{0\})$ is not isomorphic to $\operatorname{Hom}(G, \Bbb T)$? $\Bbb C$ is the complex plane and $\Bbb ...
2
votes
2answers
62 views

The factor group $\mathbb{R}^{*}/\{-1,1\}$ is isomorphic to $\mathbb{Z}_{2}$. True or False.

I have been told that the answer for this question is False. And I'm trying to understand why. For what I understand so far, $\mathbb{Z} _2$ is an abelian group. I also understand that for a group ...
2
votes
1answer
76 views

What we know about $\mathbb{Q}/\mathbb{Z}$ as a group? [closed]

What we know about $\mathbb{Q}/\mathbb{Z}$ as a group? Are there any interesting properties? Best regards.
0
votes
0answers
59 views

Free groups vs. free abelian groups

I'm trying to solve this question in page 74 of Hungerford's book: A free abelian group is a free group (Section I.9) if and only if it is cyclic. I have no idea how to proceed, a solution or a ...
0
votes
3answers
567 views

What does it mean for a group to be Abelian?

I'm revising questions on groups for exams, and I still can't quite understand what an Abelian group is. Please help me understand, if anyone could give me a more simple explanation.
3
votes
1answer
65 views

Normal Abelian Subgroup does not imply Abelian Quotient Group

I'm a bit confused and just need some clarification about what I am missing in this: I have $S_4$ with normal subgroup $N=\lbrace(),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\rbrace$. I know that $N$ is ...
1
vote
1answer
31 views

homomorphisms between amenable (discrete) groups

Let $\theta \,: G \to H$ be a group homomorphism between amenable groups (s.t. $\theta(G)$ is a normal subgroup of $H$, if needed). Is it possible to define amenable means $m_G$ on $L^\infty(G, ...
0
votes
0answers
53 views

Is my solution correct? Finite abelian groups are CLT groups.

I want to solve the following exercise from Dummit & Foote's Abstract Algebra text: Use Cauchy's Theorem and induction to show that a finite abelian group has a subgroup of order $n$ for each ...
1
vote
1answer
40 views

Nilpotent group with torsion divisible abelian quotient

Just want to make sure this is true: If $G$ is a nilpotent group such that $G/[G,G]$ is a torsion divisible abelian group (like $\mathbb{Q}/\mathbb{Z}_{(p)}$), then $G$ is abelian. I get that ...
0
votes
0answers
58 views

A doubt in Atiyah-Macdonald's “Introduction to Commutative Algebra”

"Introduction to Commutative Algebra" by Atiyah-Macdonald says the following: Let $G_n$ be the subgroup containing elements of order $p^n$ in the group $\Bbb{Q/Z}$ for all $n\in\Bbb{N}$. Here $p$ ...
0
votes
2answers
31 views

Show that $ℤ^{m}$ is a subgroup (and a free abelian group) of $ℤ^{n}$ for all $m≤n$

My question is: Show that $ℤ^{m}$ is a subgroup (and a free abelian group) of $ℤ^{n}$ for all $0≤m≤n$.
2
votes
2answers
62 views

Determine the isomorphism class of M/T(M)

Let $M=\Bbb{Z}\oplus\Bbb{Z}\oplus\Bbb{Z}$ and $T: M\rightarrow M$ given by $T(x,y,z)=(4x+2z,2y,2x+10z)$. Show the cokernel $M/T(M)$ is an abelian group of order $72$, and determine its isomorphism ...
3
votes
2answers
60 views

Is this a proof by contradiction?

Below is a proof that any group of order $p^2$ is abelian $(p$ prime of course). Let $Z \left({G}\right)$ be the center of $G$. We know $|Z(G)|>1$. $\color{blue}{\text{Suppose}} \left\vert{Z ...
4
votes
1answer
53 views

$\biggl ( \prod_p G_p \biggr) /\biggl( \bigoplus_p G_p\biggr)$ is divisible

Let $G$ be an abelian group, $p$ a prime, then $G_p$ is the $p$-primary component of $G$, i.e. $$G_p = \lbrace g \in G \ | \exists \ n \in \mathbb{N} \ , p^ng = 0\rbrace$$ I have to prove that ...
2
votes
4answers
97 views

Homomorphism from $\mathbb{Z}\oplus \mathbb{Z}_2$ to $\mathbb{Z}$.

What sort of homomorphisms can I have from $\mathbb{Z}\oplus \mathbb{Z}_2$ to $\mathbb{Z}$? What about if I know the homomorphism sends the $\mathbb{Z}$ part to zero. In other words, if I know my ...
4
votes
2answers
115 views

Simple proof of the structure theorems for finite abelian groups

Many proofs of the structure theorems for finite abelian groups first reduce to the problem to $p$-groups, which is fine and is an important technique. However, it seems to me that a simple proof can ...