Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

Questions tagged [abelian-groups]

Should be used with the (group-theory) tag. A group $(G,*)$ is said to be abelian if $a*b=b*a$ for all $a,b\in G.$

47
votes
2answers
37k views

If $G/Z(G)$ is cyclic, then $G$ is abelian

Continuing my work through Dummit & Foote's "Abstract Algebra", 3.1.36 asks the following (which is exactly the same as exercise 5 in this related MSE answer): Prove that if $G/Z(G)$ is cyclic, ...
29
votes
13answers
36k views

Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.

Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian. This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem ...
50
votes
5answers
9k views

Prove that if $(ab)^i = a^ib^i \forall a,b\in G$ for three consecutive integers $i$ then G is abelian

I've been working on this problem listed in Herstein's Topics in Algebra (Chapter 2.3, problem 4): If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in ...
3
votes
4answers
12k views

Need to prove that $(S,\cdot)$ defined by the binary operation $a\cdot b = a+b+ab$ is an abelian group on $S = \Bbb R \setminus \{-1\}$.

So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity ...
18
votes
2answers
728 views

How to recognize a finitely generated abelian group as a product of cyclic groups.

Let $G$ be the quotient group $G=\mathbb{Z}^5/N$, where $N$ is generated by $(6,0,-3,0,3)$ and $(0,0,8,4,2)$. Recognize $G$ as a product of cyclic groups. Honestly, I do not know how to solve these ...
105
votes
4answers
21k views

The direct sum $\oplus$ versus the cartesian product $\times$

In the case of abelian groups, I have been treating these two set operations as more or less indistinguishable. In early mathematics courses, one normally defines $A^n := A\times A\times\ldots\times A$...
27
votes
2answers
6k views

Group of positive rationals under multiplication not isomorphic to group of rationals

A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{>0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} ,...
7
votes
1answer
436 views

Finding an explicit isomorphism from $\mathbb{Z}^{4}/H$ to $\mathbb{Z} \oplus \mathbb{Z}/18\mathbb{Z}$

There was a past qualifying exam problem, I was having trouble with, it is stated below as follows: In the group $G= \mathbb{Z} \times \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}=\mathbb{Z}^{4}$, ...
23
votes
3answers
4k views

A nonsplit short exact sequence of abelian groups with $B \cong A \oplus C$

A homework problem asked to find a short exact sequence of abelian groups $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ such that $$B \cong A \oplus C$$ although the sequence does ...
16
votes
3answers
1k views

If $|\lbrace g \in G: \pi (g)=g^{-1} \rbrace|>\frac{3|G|}{4}$, then $G$ is an abelian group.

Assume that $\pi$ is an automorphism of a finite group $G$. Let $S$ denote the set $\lbrace g \in G: \pi (g)=g^{-1} \rbrace$. Show that if $|S|>\frac{3|G|}{4}$, then $G$ is an abelian group. ...
17
votes
3answers
1k views

A group such that $a^m b^m = b^m a^m$ and $a^n b^n = b^n a^n$ ($m$, $n$ coprime) is abelian?

Let $(G,.)$ be a group and $m,n \in\mathbb Z$ such that $\gcd(m,n)=1$. Assume that $$ \forall a,b \in G, \,a^mb^m=b^ma^m,$$ $$\forall a,b \in G, \, a^nb^n=b^na^n.$$ Then how prove $G$ is an abelian ...
8
votes
2answers
2k views

Product of elements of a finite abelian group

Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$. Prove that if there is more than one element of order $2$ then $x=e$. What I've done so far: (#1 is just for ...
21
votes
2answers
7k views

Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$?

I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$. ...
15
votes
4answers
14k views

Proof that all abelian simple groups are cyclic groups of prime order

Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks! Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$,...
30
votes
1answer
4k views

Are $(\mathbb{R},+)$ and $(\mathbb{C},+)$ isomorphic as additive groups?

Are $(\mathbb{R},+)$ and $(\mathbb{C},+)$ isomorphic as additive groups? I know that there is a bijection between $\mathbb{R}$ and $\mathbb{C}$, and this question asks whether they are isomorphic as ...
25
votes
1answer
4k views

Is it true that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as abelian groups?

I think the answer is yes. Sketch of the proof Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Let $\{e_\lambda:\lambda\in\Lambda\}\subset\mathbb{R}$ be its Hamel basis. Then $\{(e_{\...
2
votes
2answers
233 views

proving that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l $ as groups, where $l=lcm(m,n$) and $d=gcd(m,n)$

How would one go about proving that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l $ as groups, where $l=$lcm($m,n$) and $d=$gcd($m,n$)? I am attempting to use the fundamental ...
22
votes
1answer
13k views

Computing the Smith Normal Form

Let $A_R$ be the finitely generated abelian group, determined by the relation-matrix $$R := \begin{bmatrix} -6 & 111 & -36 & 6\\ 5 & -672 & 210 & 74\\ 0 & -255 &...
17
votes
2answers
3k views

Find an abelian infinite group such that every proper subgroup is finite

I found this question in Arhangel'skii and Tkachenko's book Topological Groups and Related Structures. The first chapter of the book is devoted to algebraic preliminaries. The question actually reads:...
35
votes
2answers
4k views

Structure Theorem for abelian torsion groups that are not finitely generated

I know about the structure theorem for finitely generated abelian groups. I'm wondering whether there exists a similar structure theorem for abelian groups that are not finitely generated. In ...
5
votes
1answer
2k views

Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G $ is an abelian group.

Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group. Thank you in advance. Any help is appreciated.
6
votes
1answer
2k views

Sum of elements of a finite field

Let $F$ be a finite field and $i$ an integer. Calculate the sum of all the elements of $F$, each raised to the $i$th power. My approach so far:...
29
votes
1answer
1k views

Recovering a finite group's structure from the order of its elements.

Suppose you know the following two things about a group $G$ with $n$ elements: the order of each of the $n$ elements in $G$; $G$ is uniquely determined by the orders in (1). Question: How difficult ...
12
votes
2answers
3k 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 ...
23
votes
2answers
7k views

Showing that a finite abelian group has a subgroup of order m for each divisor m of n

I have made an attempt to prove that a finite abelian group of order $n$ has a subgroup of order $m$ for every divisor $m$ of $n$. Specifically, I am asked to use a quotient group-induction ...
13
votes
1answer
344 views

Problem on abelian group

Let $G$ be an abelian group, and $\Phi:G\to \mathbb{R}$ is a function with the following property: $$\forall a,b\in G,~~ |\Phi(a+b)-\Phi(a)-\Phi(b)|<c$$ The problem asks to prove the existence of ...
8
votes
3answers
1k views

If a group is $3$-abelian and $5$-abelian, then it is abelian

In a group $(Z,*)$, $(a*b)^{5}=a^{5}*b^{5},\forall a,b\in Z$ and $(a*b)^{3}=a^{3}*b^{3}$ then prove that $Z$ is abelian. I know that for three consecutive integer if $(a*b)^{i}=a^{i}*b^{i},\forall a,b\...
4
votes
1answer
692 views

Prove that any two nontrivial subgroups of $\mathbb{Q}$ have nontrivial intersection

I need to prove that any two nontrivial subgroups of $\mathbb{Q}$ have a nontrivial intersection as part of a larger proof that $\mathbb{Q}$ cannot be represented as a nontrivial direct product. (...
9
votes
3answers
2k views

Finite abelian $p$-group with only one subgroup size $p$ is cyclic

My goal is to prove this: If $G$ is a finite abelian $p$-group with a unique subgroup of size $p$, then $G$ is cyclic. I tried to prove this by induction on $n$, where $|G| = p^n$ but was not ...
7
votes
4answers
1k views

Subgroups containing kernel of group morphism to an abelian group are normal. [closed]

Let $\varphi:G\rightarrow H$ be a group homomorphism from group $G$ to group $H$. Show that, if $H$ is abelian, all subgroups of $G$ that contain $\mathrm{ker} (\varphi)$ are normal in $G$.
10
votes
1answer
2k views

Finite abelian groups - direct sum of cyclic subgroup

Let $G$ be a finite abelian $p$-group. It is quite elementary to see that if $g \in G$ is an element of maximal order (and thus its span is a cyclic subgroup of $G$ of maximal order) then $G$ can be ...
9
votes
1answer
257 views

For which $n$, $G$ is abelian?

My question is: For Which natural numbers $n$, a finite group $G$ of order $n$ is an abelian group? Obviouslyو for $n≤4$ and when $n$ is a prime number, we have $G$ is abelian. Can we consider ...
5
votes
2answers
5k views

Order of products of elements in a finite Abelian group

We want to show that if $a,b\in G$ where $G$ is a finite Abelian group, we have $\operatorname{LCM}(|a|,|b|) = |ab|$ given that $ab \neq e$. How I approached this question was by saying let $\...
14
votes
5answers
1k views

Prove that $(a_1a_2\cdots a_n)^{2} = e$ in a finite Abelian group

Let $G$ be a finite abelian group, $G = \{e, a_{1}, a_{2}, ..., a_{n} \}$. Prove that $(a_{1}a_{2}\cdot \cdot \cdot a_{n})^{2} = e$. I've been stuck on this problem for quite some time. Could someone ...
7
votes
4answers
2k views

Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$ [duplicate]

Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$. My try: Let $\mathbb Z^n\cong \mathbb Z^m $. To show that $m=n$. Case 1:Let $m>n$.Now that $\mathbb Z^m$ has $m$ ...
34
votes
3answers
2k views

In a group we have $abc=cba$. Is it abelian?

Let $G$ be a group such that for any $a,b,c\ne1$: $$abc=cba$$ Is $G$ abelian?
19
votes
2answers
6k views

Proving that a subgroup of a finitely generated abelian group is finitely generated

A question says: Using the isomorphism theorems or otherwise, prove that a subgroup of a finitely generated abelian group is finitely generated. I would say that for a finitely generated abelian ...
12
votes
1answer
6k views

Converse of Lagrange's theorem for abelian groups

I'm trying to prove that the converse of Lagrange's theorem is true for finite abelian groups (i.e. "given an abelian group $G$ of order $m$, for all positive divisors $n$ of $m$, $G$ has a subgroup ...
12
votes
4answers
2k views

Additive group of rationals has no minimal generating set

In a comment to Arturo Magidin's answer to this question, Jack Schmidt says that the additive group of the rationals has no minimal generating set. Why does $(\mathbb{Q},+)$ have no minimal ...
8
votes
3answers
4k views

Computing $\mathrm{Hom}(\mathbb Z_n,\mathbb Z_m)$ as $\mathbb Z$-module

My algebra is weak I need help computing $\mathrm{Hom}(\mathbb Z_n,\mathbb Z)$, $\mathrm{Hom}(\mathbb Z_n,\mathbb Z_m)$ and also $\mathrm{Hom}(\mathbb Z,\mathbb Z)$ as $\mathbb Z$-modules. Also books ...
22
votes
3answers
13k views

Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian

Trying to get my head around the commutator subgroup. This is an excercise from Artin's Algebra: Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian. Here is what I've done: Let $...
11
votes
2answers
371 views

Does there exist an $n$ such that all groups of order $n$ are Abelian?

I know that all groups of order $\leq$ 5 are Abelian and all groups of prime order are Abelian. Are there any other examples? If so is there something special about the orders of these groups?
19
votes
2answers
2k views

Status of the classification of non-finitely generated abelian groups.

From the Wikipedia on abelian groups: By contrast, classification of general infinitely-generated abelian groups is far from complete. How far are we from a classification exactly? It seems like ...
12
votes
1answer
4k views

Showing that a cyclic automorphism group makes a finite group abelian

From a bank of previous masters exams: Let $G$ be a finite group such that its automorphism group $\operatorname{Aut}(G)$ is cyclic. Prove that $G$ is abelian. Here's what I was thinking. Let $\...
6
votes
1answer
2k views

Subgroups of a finitely generated abelian group without torsion

If $G\cong \mathbb{Z}\times \mathbb{Z}\times \dots \times \mathbb{Z}$ is a finitely generated abelian group without torsion of rank $n$, where $n$ is the number of copies of $\mathbb{Z}$. Then any ...
3
votes
2answers
1k views

A finite abelian group $A$ is cyclic iff for each $n \in \Bbb{N}$, $\#\{a \in A : na = 0\}\le n$

Let $A$ be a finite abelian group. Prove that $A$ is cyclic iff for each $n \in \Bbb{N}$ $$\#\{a \in A : na = 0\}\le n.$$ Any help or hint will be appreciated.
6
votes
3answers
764 views

Embedding torsion-free abelian groups into $\mathbb Q^n$?

Glass' Partially Ordered Groups states without proof: Every torsion-free abelian group can be embedded into a rational vector space (as a group). Can someone link me to a proof of this? It seems ...
8
votes
2answers
1k views

Finite Abelian groups with the same number of elements for all orders are isomorphic

Let $A$ and $B$ be finite abelian groups. Suppose that for every natural number $m$, the number of elements of order $m$ in $A$ is equal to the number of elements of order $m$ in $B$. Prove that $A$ ...
6
votes
2answers
1k views

All subgroups normal $\implies$ abelian group

This is , I think an easy problem just that I am not getting the catch of it. How to show whether or not the statement is true? All subgroups of a group are normal$\implies$ the group is an abelian ...
4
votes
4answers
2k views

Give an example of a nonabelian group in which a product of elements of finite order can have infinite order. [duplicate]

So, I let a,b be elements in such a group. So |a|=n and |b|=m, n and m are finite. But |ab| needs to be infinite, but since |ab|=lcm(n,m), how can that be possible?