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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2
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1answer
82 views

homomorphisms of abelian groups

Describe: 1) Hom(Q/Z->Q) 2) Hom(Q->Q/Z) Q rational numbers Z integers My thoughts 1) Q/Z i can describe as a/b, a,b coprime and smaller than 1. So I thought Hom(Q/Z->Q) I can describe as [f:a+N->b, N ...
0
votes
4answers
35 views

Automorphism iff G is abelian

Let $G$ be a group. Prove the mapping $\alpha(g)=g^{-1}\forall g \in G$ is an automorphism iff $G$ is abelian. Proof (forwards): Assume $G$ is an automorphism. Show $ab=ba$. How would I even go about ...
2
votes
3answers
264 views

How many distinct subgroups of order 10 are there in a non-cyclic abelian group of order 20?

We are currently working with free abelian groups and finitely generated groups. The homework problem asks us to find the number of distinct subgroups of order 10 in a non-cyclic abelian group of ...
1
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0answers
42 views

Abelian Group (Alternative Proof)

Is there an alternative method to prove $(ab)^{2}=a^{2}b^{2}$ for all elements $a,b \in G \implies$ $(ab)^{-1}=a^{-1}b^{-1}$ for all elements $a,b \in G$. then the one I give below? Let $G$ be ...
1
vote
2answers
103 views

Order of the group.

An abelian group $G$ is generated by $x$ and $y$ with $$O(x)=16,O(y)=24,x^2=y^3$$ What is the order of $G$? My attempt:There are $24+16-1=39$ elements generated by $x$ and $y$ separately. Also ...
1
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0answers
74 views

Description of an abelian group

I'm again stuck in an algebra exercise. I'm not sure if I understand the problem right. Could it be that I have to show that $\mathbb{Z}[i]/\gamma$ can be expressed by a product of finite abelian ...
1
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1answer
36 views

Can we calculate the order of $\hom (G,G')$ in terms of $|G| $ and $ |G'|$ , when $G,G'$ are finite abelian groups?

Let $G,G'$ be abelian groups and let $\hom (G,G')$ be the set of all homomorphisms from $G$ to $G'$. We define an operation $\ast$ on $\hom (G,G')$ as: for $f,g \in \hom(G,G') \space , (f\ast ...
0
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2answers
51 views

|ab|=lcm(|a|,|b|) in an abelian group

Assume in an abelian group $G$ that $\langle b\rangle\cap \langle a\rangle=e$, then the order of $(ab)$ is the lcm of the orders of $a$ and $b$. Essentially, $|ab|=\operatorname{lcm}(|a|,|b|)$. So ...
5
votes
2answers
130 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 ...
1
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2answers
93 views

Least common multiple of orders and abelian groups.

I am a little stuck here and would like some minor help. The quesiton I am dealing with is: Assume in an abelian group G that $<b>{\large\cap} <a>=e$, then the order of $(ab)$ is the lcm ...
0
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1answer
133 views

$G$ infinite abelian group with $[G:H]$ finite for every non trivial subgroup $H$ , to prove $G$ is cyclic

Let $G$ be an infinite abelian group such that for any non-trivial subgroup $H$ of $G$ , $[G:H ]$ is finite ; then how to prove that $G$ is cyclic ? Please don't use any structure theorem of abelian ...
6
votes
1answer
406 views

Direct sum and direct product of infinitely many abelian groups are not isomorphic

Let $I$ be an infinite set, and for each $i$ let $A_i$ be an abelian group with order $o(A_i) \ge 2$. Prove that the direct product $\prod A_i$ and the direct sum (coproduct) $\bigoplus A_i$ are ...
1
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1answer
32 views

Show $a\in G$ is contained in $Z(G)$ iff $Z(a)=G$ for center and centralizer? [closed]

The center of a group $G$ is defined as the set $Z(G):= \{a\in G\mid \forall b\in G : ab=ba\}$ and the centralizer of an element $a\in G$ is defined as the set $Z(a) := \{b\in G\mid ab=ba\}$. How can ...
0
votes
1answer
26 views

The Dihedral group $D_1$ is non-abelian?

Same as above. I'm trying to show that for any n being odd, $D_n$ has exactly n elements of order 2 where $D_n$ is non-abelian. I know that for $n\ge3$ this is true, but what about for $n=1$.
-2
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2answers
90 views

Two abelian groups with the same order are isomorphic? [closed]

True of false: if G and H are two groups with the same order and both are abelian, then they are isomorphic.
1
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0answers
36 views

Measure theory mapping sets to groups?

This is a question from a physicist wondering if a certain idea in mathematics has been developed. Intuitively, suppose I have a number of objects distributed in space. I want a function that given a ...
-1
votes
1answer
176 views

On groups with none of their quotient groups divisible [closed]

Does there exist a group $G$ that satisfies the following conditions: Any proper subgroup of $G$ is contained in a maximal subgroup. There is some $N\unlhd G$ such that $\frac{G}{N}$ is divisible. ...
0
votes
1answer
43 views

Rank of abelian groups

I have read that given a $\mathbb{Z}$-module $M$, the maximal number of $\mathbb{Z}$-linear independent elements is given by $\operatorname{rank}M=\dim_\mathbb{Q}(\mathbb{Q}\otimes_\mathbb{Z}M)$. ...
1
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2answers
24 views

Clarification on finding another subgroup given the order of two existing subgroups

If we assume that G is abelian and that it has a subgroup of order 7 and another of order 11. If we were asked to find another subgroup of this group would we take the least common multiple of the ...
0
votes
1answer
63 views

Subgroup of an abelian group isomorphic to a given quotient group

STATEMENT: Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroups that is isomorphic to $G/H$. QUESTION: Could someone offer a proof using dual groups. I have found one ...
3
votes
4answers
157 views

Prove that no finite abelian group is divisible.

A nontrivial abelian group $G$ is called divisible if for each $a \in G$ and each nonzero integer $k$ there exists an element $x \in G$ such that $x^k=a$. Prove that no finite abelian group is ...
0
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0answers
81 views

Number of Abelian Groups of Order 36

GRE Subject Test Question: Up to isomorphism, how many abelian groups are there of order 36? The answer given is 4 and the explanation is as follows: Let G be an abelian group with order n. Then G ...
0
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1answer
48 views

Submodule iff subgroup?

It is late at night and time for another silly question: Is it true that a subset $S$ of an $R$-module $M$ is a submodule if and only if it is a subgroup of $M$ as an abelian group? Of course, by ...
0
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0answers
53 views

Finding an isomorphism from $\mathbb{R}^\times$ to a defined group $G$

Here's the problem I am solving: $G=\{x\in \mathbb{R}:x\not = 0\}$. The operation for $G$ is "$*$", with $x*y=\frac{1}{2}xy.\mathbb{R}^\times$ is the multiplicative group $\mathbb{R}.$ Find an ...
1
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0answers
51 views

Proof of elliptic curves being an abelian group

What are some simple proofs that the points on an elliptic curve form an abelian group under addition? I am mostly looking for proofs of closure and associativity, since the other three requirements ...
0
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1answer
36 views

Proving that some property on a chain complex of groups implies isomorphism between direct sums of these groups.

Let $C_*$ be a chain complex such that every $C_i$ is a torsion-free finitely generated abelian group, with $C_i=0$ for every $i<0$ and every $i>N$ for some sufficiently large integer $N$. If ...
0
votes
4answers
346 views

Show that an abelian group $G$ of order 55 must be cyclic.

I know that in order to be cyclic: A group G is called cyclic if there exists an element g in G such that G = ⟨g⟩ = { $g^n$ | n is an integer } by wikipedia. But I just get lost in how simple it looks ...
0
votes
2answers
50 views

abelian groups?

let p and q be distinct prime numbers. how does the number (up to isomorphism) of abelian groups of order p^r compare with the number (up to isomorphism) of abelian groups of order q^r? I am just not ...
0
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3answers
72 views

Can $\mathbb{Z}/n\mathbb{Z}$ (not $(\mathbb{Z}/n\mathbb{Z})^{\times}$) be a group under multiplication?

I was wondering why we usually say $\mathbb{Z}/n\mathbb{Z}$ is a group under addition and invent notation like $(\mathbb{Z}/n\mathbb{Z})^\times$ specifically for the multiplicative group modulo $n$. ...
0
votes
2answers
50 views

Suppose G is a group which has only one element a such that |a| = 2. Prove that xa = ax, for all x ∈ G.

I know the following are true. 1) There is an inverse of a 2) There is an identity element (e*a) = a In this case, e = 1 and the inverse of a is 1/|2|. However, if a is the only element in G and a ...
1
vote
6answers
465 views

Let G be a group and a; b ∈ G. Suppose |a| = |b| = |ab| = 2. Then show that ab = ba.

I'm having trouble understanding this question and help would be appreciated. If |ab|=2 and |a|=2, |b|=2, wouldn't this imply that |a||b|=|ab|=4? How would I go about proving that this is Abelian? ...
0
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1answer
54 views

Order of Group with Elements of Order 2 [duplicate]

Let G be a finite group such that every element in G which isn't the identity has order of 2. Show that $|G| = 2^{n}$ for some $n \in \mathbb{N}$. I know that G is necessarily going to be abelian. ...
0
votes
1answer
36 views

The number, up to isomorphism, or abelian grips of order 40 is

The number, up to isomorphism, or abelian groups of order 40 is: I got: 2*2*10 2*20 40 So the total number is 3. However, the answer says 7, where 40 10*4 8*5 20*2 10*2*2 5*4*2 I think the ...
2
votes
1answer
35 views

Property of abelian groups without using Lagrange's theorem

I need to prove the following without using Lagrange's Theorem: Show that for an abelian group $G$, $\forall \; a \in G:$ $a^{o(G)}=e$ . This is a generalization of the Euler-Phi Theorem. So I ...
0
votes
1answer
50 views

Groups of Order 2 with subgroups

Let G be an abelian group and $a,b\in G$ be two distinct elements with a and b or order $2$. Show that $H=\{e,a,b,ab\}$ forms a subgroup and write out its multiplication table. Justify why all the ...
1
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1answer
45 views

$(\mathbb C^{\times}, \cdot)$ is a subgroup of $(GL(n,\mathbb C), \cdot)$

I am learning groups and subgroups in my algebra course. Today, we talked about examples of subgroup but I am not sure why the following holds: $(\mathbb C^{\times}, \cdot)$ is a subgroup of ...
1
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0answers
35 views

Direct limits of free abelian groups and diagonalization

So, say I have a matrix $A\in M_d(\mathbb{Z})$ and would like to describe the group $\lim(\mathbb{Z}^d,A)$, i.e. the limit of the stationary system $$ \mathbb{Z}^d\to^A \mathbb{Z}^d \to^A ...
1
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0answers
30 views

$G$ a finite group $n$-abelian goup and g.c.d.$\big(|G|,n(n-1)\big)=1$ , then to show $G$ is abelian [duplicate]

Let $G$ be a finite group and $n$ be a given positive integer such that $(ab)^n=a^nb^n , \forall a,b \in G$ and g.c.d.$\big(|G|,n(n-1)\big)=1$ , then how to prove that $G$ is abelian ? If I can show ...
1
vote
1answer
91 views

Complement but not direct summand; Lam, Lectures on Modules and Rings, Example 6.17(5)

Let $S$ be a submodule of an $R$-module $M$. A submodule $C⊆M$ is said to be a complement to $S$ (in $M$) if $C$ is maximal with respect to the property that $C∩S=0$. (This does exist by Zorn Lemma.) ...
0
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1answer
44 views

$R$ is not a direct Sum of its subgroups

How to prove the set of real numbers under addition i.e $(R,+)$ is not the direct sum of two of its proper subgroups?
1
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1answer
92 views

When an Abelian group is cyclic

Let G be a finite abelian group.It contains a non trivial subgroup which is contained in every non trivial subgroup.Then G must be cyclic. This is a problem of Herstein book(Pg 108,#11 2nd edition).I ...
6
votes
1answer
90 views

$G$ is torsion-free $[G:Z(G)]$ is finite $\implies$ $G$ is abelian ?

If $G$ is a a group having no non-identity element of finite order and $Z(G)$ , the center of the group , has finite index , then is it true that $G$ is abelian ?
1
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1answer
47 views

From Automorphism to abelian ness … in a finite group

Let $G$ be a finite group such that for any two non-identity elements $a,b$ in $G$ , there is an Automorphism of $G$ sending $a$ to $b$ , then is it true that $G$ is abelian ?
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0answers
41 views

divisible subgroup without axiom of choice

the theorem asserting that the divisible subgroup of an Abelian group is a direct summand depends on Zorn's lemma. in ZF without AC is there a construction which yields a model of an Abelian group ...
2
votes
1answer
78 views

Exercise on characterization of free abelian groups

I was wondering if someone can please check my work on a homework problem. This is from the graduate Hungerford text. Chapter 2.1, number 3. Let $X=\{a_i\ |\ i\in I\}$ be a set. Then the free abelian ...
3
votes
3answers
100 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.
1
vote
1answer
71 views

Rational group algebras and maximal orders

Let $G$ be a finite group, and $\mathbb{Q}[G]$ be the rational group algebra. Then the group ring $\mathbb{Z}[G]$ is an order in $\mathbb{Q}[G]$, but is not in general a maximal order. What can we ...
3
votes
2answers
48 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 ...
3
votes
3answers
77 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
36 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 ...