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

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17
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1answer
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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 ...
34
votes
4answers
5k 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 ...
14
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5answers
11k views

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 to crack it. (please note that $e$ in the question is the ...
13
votes
3answers
530 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. ...
21
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1answer
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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 ...
16
votes
2answers
3k 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
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2answers
1k 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 ...
2
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4answers
3k views

Need to prove that (S,*) defined by the binary operation a*b = a+b+ab is an abelian group on S = R \ {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 ...
60
votes
4answers
8k 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 ...
12
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1answer
247 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 ...
21
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1answer
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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 ...
13
votes
3answers
368 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 $\gcd(m,n)=1$ and $$ \forall a,b \in G, \,a^mb^m=b^ma^m,$$ $$\forall a,b \in G, \, a^nb^n=b^na^n.$$ Then how prove $G$ is an abelian group? ...
29
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2answers
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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 ...
23
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1answer
784 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 ...
14
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2answers
2k 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 ...
8
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1answer
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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 ...
13
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5answers
855 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
2answers
237 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 $A$ ...
10
votes
4answers
734 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 ...
11
votes
1answer
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Computing the Smith Normal Form

This question is related to the Smith Normal Form of Matrices: Let $A_R$ be the finitely generated abelian group, determined by the relation-matrix $R :=$ $$ \begin{bmatrix} -6 & 111 & ...
14
votes
2answers
960 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 ...
8
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2answers
707 views

Show that any abelian transitive subgroup of $S_n$ has order $n$

Can anybody tell me what is known about the classification of abelian transitive groups of the symmetric groups? Let $G$ be a an abelian transitive subgroup of the symmetric group $S_n$. Show that ...
8
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1answer
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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 ...
5
votes
1answer
576 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.
3
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3answers
684 views

On Groups of Order 315 with a unique sylow 3-subgroup .

in Dummit and Foote , an exercise asked me to prove that , if $G$ is a group of order $315$ , $G$ has a normal sylow $3$-subgroup then , $G$ is abelian . this is exercise number $27$ , section $5$ , ...
1
vote
1answer
133 views

Let G be an abelian group, and let a∈G. For n≥1,let G[n;a] := {x∈G:x^n =a}. Show that G[n; a] is either empty or equal to αG[n] := {αg : g ∈ G[n]}… [closed]

We were given questions to study for our exam coming up. We have not covered much of this topic, so any help would be greatly appreciated! Let $G$ be an abelian group, and let $a\in G$. For $n≥1$, ...
10
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6answers
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Are cyclic groups always abelian?

If a group $C$ is cyclic, is it also abelian (commutative)? If so, is it possible to give an “easy” explanation of why this is? Thanks in advance!
12
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3answers
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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 ...
7
votes
1answer
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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 ...
9
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2answers
1k views

Equivalences and isomorphisms of short exact sequences

In case it's necessary, I'm working in the category $\mathbf{Ab}$ of abelian groups. My question concerns what I find to be a strange way of viewing the elements of the Ext group $\mbox{Ext}(A,B)$ of ...
9
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1answer
205 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 ...
12
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2answers
3k 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 ...
8
votes
2answers
1k 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 ...
6
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6answers
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Give an example of a noncyclic Abelian group all of whose proper subgroups are cyclic.

I've tried but I could not find a noncyclic Abelian group all of whose proper subgroups are cyclic. please help me.
5
votes
1answer
120 views

To show that group G is abelian if $(ab)^3 = a^3 b^3$ and the order of $G$ is not divisible by 3

Let $G$ be a finite group whose order is not divisible by $3$. suppose $(ab)^3 = a^3 b^3$ for all $a,b \in G$. Prove that $G$ must be abelian. Let$ $G be a finite group of order $n$. As $n$ is ...
4
votes
4answers
253 views

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

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 $m$ ...
6
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3answers
607 views

Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.

Suppose $G$ is an abelian group and $a\in G$ and $$f:\left<a \right>\to\Bbb T$$ is a homomorphism. Can $f$ be extended to a homomorphism on $G$: $$g:G\to \Bbb T$$ ? $\Bbb T$ is the circle ...
1
vote
2answers
127 views

If $G$ is non-abelian, then $Inn(G)$ is not a normal subgroup of the group of all bijective mappings $G \to G$

Let $(G,\cdot)$ be a group and let $\mathfrak{S}(G)$ be the set of all bijective mappings from $G$ to $G$. Show that: If $G$ is non-abelian, then $Inn(G):=\{\kappa_a \vert a\in G\}$ is not a ...
31
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2answers
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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?
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2answers
947 views

When is the automorphism group $\text{Aut }G$ cyclic?

Let $G$ be a finite group. Under which conditions on $G$ is the automorphism group $\text{Aut }G$ cyclic? More precisely, does "$G$ is abelian" or "$G$ is cyclic" imply "$\text{Aut }G$ is cyclic"?
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3answers
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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 ...
6
votes
2answers
336 views

Whence this generalization of linear (in)dependence?

I recently came across a definition of (in)dependence that is supposed to be a generalization of linear (in)dependence among a set of vectors: An element $x$ is dependent on a set of elements ...
6
votes
2answers
649 views

A condition for a subgroup of a finitely generated free abelian group to have finite index

Let A be a free Abelian group of finite rank and B be a subgroup of A such that $A=B+pA$ for some prime number p, then how to prove $B$ is a subgroup of finite index in A? And if $A=B+pA$ holds for ...
5
votes
1answer
346 views

Finite Abelian groups: $G \times H \cong G\times K$ then $H\cong K$

Let $G,H,$ and $K$ be finite abelian groups. If $G \times H \cong G\times K$ then $H\cong K$. I am trying to use the fundamental theorem for abelian groups to solve this, it is clear intuitively ...
5
votes
2answers
893 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 ...
4
votes
2answers
406 views

Direct limit of $\mathbb{Z}$-homomorphisms

What is the direct limit of the following sequence of $\mathbb{Z}$-homomorphisms (as groups)? $$ \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{3} \mathbb{Z}\xrightarrow{5} ...
3
votes
0answers
110 views

$\gcd(|G|, |\text{Aut}(G)|)=1$ means G is abelian?

Prove the following assuming that $G$ is finite group with $\gcd(|G|, |\text{Aut}(G)|)=1$. a) G is abelian (done). b) Every Sylow subgroup of $G$ is cyclic of prime order. Since G is ...
10
votes
2answers
195 views

The existence of a group automorphism with some properties implies commutativity.

Let $G $ be a finite group, $T$ be an automorphisom of $ G $ st $ Tx = x \iff x=e $. Suppose further that $ T^2 =I $. Prove that $ G $ is abelian. I was thinking if I show $ T aba^{-1} b^ ...
9
votes
1answer
170 views

Prove $G$ is abelian if $f(f(x)) = x$?

Let $G$ be a finite group and $f$ an automorphism such that $f(f(x)) = x$, and $f(x) = x$ if and only if $x=e$. Prove that $G$ is abelian and $f(x) = x^{-1}$. My attempt: ...
4
votes
2answers
176 views

Prove that the group $G$ is abelian if $a^2 b^2 = b^2 a^2$ and $a^3 b^3 = b^3 a^3$

In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$. Prove that the group $G$ is abelian. My approach was the following: Let $a,b\in G$ Then, $a^2b^2=b^2a^2$ and ...