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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 $a,b\in G.$

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2answers
38 views

Proof for this identity in an abelian group [duplicate]

I need to show the following identity in an abelian group. $$\Bigl(\prod_{g \in G}g\Bigr)^2 = e$$ I think I have the basic idea for the proof, in that you can reorder the factors of the product such ...
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1answer
28 views

How do we show that the set of module homomorphisms is a commutative group?

Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 \in ...
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1answer
68 views

Find all non-isomorphic abelian groups s.t. $|G| \leq 30$ and $g^{12}=1, \, \forall g\in G$

Assume the prime factorization of the order of $G$: $$ |G|=p_1^{a_1}p_2^{a_2}\dots p_r^{a_r} $$ The condition $$ g^{12}=1,\, \forall g\in G\tag{1} $$ in other words means that we must find those ...
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2answers
77 views

An abelian group $G$ of order $35$ with $g^{35}=e$ for all $g\in G$ is cyclic.

I'm reading "Contemporary Abstract Algebra," by Gallian. This is Exercise 4.20. Suppose that $G$ is an Abelian group of order $35$ and every element of $G$ satisfies the equation $x^{35}=e$. Prove ...
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0answers
26 views

What is $H_G^{(n)}:=\{h\in G: \operatorname{ord}(h)\mid n\}$ called for any fixed abelian group $G$ and $n\in\Bbb N$?

I'm reading "Contemporary Abstract Algebra," by Gallian. This is based on exercises 3.45 and 4.15 ibid. What is $H_G^{(n)}:=\{h\in G: \operatorname{ord}(h)\mid n\}$ called for any fixed abelian ...
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1answer
55 views

Are the groups $(\Bbb Q^2,+)$ and $(\Bbb Q,+)$ isomorphic? [duplicate]

Are the groups $(\Bbb Q^2,+)$ and $(\Bbb Q,+)$ isomorphic ? I don't know how to proceed because I am getting all the structural properties same. Can I get some help?
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1answer
33 views

Is there an algorithmic way to compute the quotient of two subgroups of $\mathbb Z^n$?

In algebraic topology one sometimes has to compute the homology groups via CW complexes, reducing the problem to the calculation of $Z/B$ where $B \subseteq Z \subseteq \mathbb{Z}^n $. It is difficult ...
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0answers
15 views

How do you call groups formed as direct sums of cyclic groups?

What do you call groups of the form $\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/\mathbb{Z}d_1\oplus\mathbb{Z}/\mathbb{Z}d_2\oplus\cdots$ These kind of groups usualy appear as homology ...
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1answer
37 views

Why does not generator of multiplicative group generate all the members of group? [duplicate]

Generator(g) of multiplicative group (Zp where p is primary) is an element the power of ...
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1answer
38 views

Is this true for about abelian generating sets?

Given a finite abelian group $G$ which is genertaed by a set $S$. Question : Is there always exists a $S' \subseteq S$ which is a basis of $G$? Seems true to me for example $\mathbb{Z}_6$, take $S ...
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29 views

Classification of abelian groups of order 72, and subgroups of order 4

Problem : Find all abelian groups of order 72, up to isomorphism. Which one of those groups are cyclic? Amont these subgroups, find all groups with a unique subgroup of order 4. My ...
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1answer
29 views

Let $N$ be an Abelian normal subgroup of $G$, if $G/N$ is perfect, then also $G'$ is perfect.

I'm reading Kurzweil & Stellmacher's "The Theory of Finite Groups", its 1.5.3 says: Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect, then also $G'$ is perfect. Proof. ...
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0answers
74 views

On a characterization of abelian groups $G$ based on special commutator relations ($\exists n\in\Bbb N$ s.t. $[x^n,y]=[x,y^{n+1}],\forall x,y \in G$).

Let $G$ be a group. If $\exists n\in \mathbb N$ such that $x^n=yx^n(y^{n+1}x)^{-1}xy^n,\forall x,y\in G$, then how to prove that $G$ is abelian? Thoughts: the condition is same as saying $[x^n,y]=[x,...
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56 views

Representations and subgroups of the translation group [closed]

Let $H_\lambda=\lambda\mathbb{Z}$ for $\lambda\in \mathbb{R}$ be a subgroup of the 1-dimensional translation group $T$. Consider then the factor group $K_{\lambda}=T/H_{\lambda}$ with representation $...
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1answer
46 views

Abelian normal subgroups of A-groups

Let $G$ be a finite solvable group, where every Sylow subgroup is abelian. I want to show that if $A\lhd G$ is an abelian normal subgroup, then $$ A=(A\cap Z(G))(A\cap G')$$ This is easy if $A$ is a ...
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1answer
45 views

For which abelian groups $G$ is there a short exact sequence $0 \rightarrow \mathbb{Z}/p^2 \rightarrow G \rightarrow \mathbb{Z}/p^2 \rightarrow 0$?

I am trying to find for which abelian groups $G$ is there a short exact sequence. $0 \rightarrow \mathbb{Z}/p^2 \rightarrow G \rightarrow \mathbb{Z}/p^2 \rightarrow 0$? I have reasoned as follows: ...
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0answers
71 views

Existence of abelian group which has no “square-root” but whose “cube” has a “square-root”

Does there exist an abelian group $G$ such that $G \ncong H \times H$ for every abelian group $H$ but $G \times G \times G \cong K \times K$ for some abelian group $K$ ? Also see Existence of ...
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1answer
91 views

Puzzle group of $4\times 4$ “flip” game

I have been goofing around with the game "flip," which can be played at the following link. The puzzle consists of an $n\times n$ grid of squares that are either black or white, and when one clicks on ...
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0answers
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Are groups constructed using semidirect product always non-abelian? [duplicate]

When using semidirect product to construct new groups based on smaller groups, we have to define a group homomorphism from the non-normal subgroup to the group of automorphism of the normal one, i.e. ...
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2answers
94 views

Existence of subgroup of order power of prime in a finite abelian group?

Say I have a finite abelian group $G$ such that $\left | G \right |=p_1^{n_1}...p_m^{n_m}$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^{n_1}$? I'm thinking ...
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0answers
57 views

2-generated abelian group

Given a non-cyclic 2-generated abelian group $G=\langle a,b\rangle$, it seems straightforward from fundamental theorem of abelian groups that $$G\cong\langle x\rangle\times\langle y\rangle$$ for some ...
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1answer
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Finite abelian group $G$, with $N \triangleleft G$ cyclic and $G/N$ cyclic [closed]

Let $G$ be a finite abelian group and $N \triangleleft G$. If both $N$ and $G/N$ are cyclic and $GCD(\vert N \vert,\vert G/N \vert)=1$ then $G$ is cyclic. I don't know how to prove that.
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1answer
27 views

Commutator subgroup of Pauli's group of order 16 [closed]

The Pauli matrices are $$e = \begin{bmatrix}1&&0\\0&&1\end{bmatrix},\ X = \begin{bmatrix}0&&1\\1&&0\end{bmatrix},\ Y = i\begin{bmatrix}0&&-1\\1&&0\end{...
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1answer
40 views

Constructing an isomorphism between groups

I am trying to come up with an isomorphism of $X$ and ker$(f \oplus g)$, where $f \oplus g: X \oplus Y \to Z$, and $f: X \to Z$ is a homomorphism and $g:Y \to Z$ is an isomorphism of commutative ...
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1answer
39 views

Construct a non-trivial homomorphism: $\mathbb{Z}_2 \times \mathbb{Z}_4 \longrightarrow \mathbb{Z}_8$

We shall construct a map $$ \varphi : \left\{0,1,2,3\right\} \times \left\{0,1\right\} \longrightarrow S \subseteq \mathbb{Z}_8 $$ which satisfies $$ \varphi (a+b \bmod4,c+d \bmod 2)=\varphi(a,c) +\...
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1answer
59 views

Classify $\mathbb{Z}_{108}$ up to an isomorphism [closed]

Note: My question is a lot more general. Can you provide me with an answer about what "classify" implies when it comes to groups and what are the steps that one should follow to carry through with ...
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5answers
351 views

A curiosity: how do we prove $\mathbb{R}$ is closed under addition and multiplication?

So I tried looking around for this question, but I didn't find much of anything - mostly unrelated-but-similarly-worded stuff. So either I suck at Googling or whatever but I'll get to the point. So ...
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1answer
41 views

Epimorphism between free Abelian groups

I have found a statement (without proof) that there is no epimorphism in Group category $\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z} $ existing. Could you please help,...
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0answers
19 views

Constructing injection into injective group

$\newcommand{\ZZ}{\mathbb{Z}}$ $\newcommand{\QQ}{\mathbb{Q}}$ $\newcommand{\Hom}{\mathrm{Hom}}$ Some time ago I tried to construct for a given abelian group $M$ functorially a group $I(M)$ which ...
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2answers
33 views

Let $A$ be an abelian finitely generated free group and $A/B$ be a torsion group. Show that $rank(A)=rank(B)$.

Let $A$ be an abelian free group that is finitely generated, and let $B\subset A$ be a subgroup of $A$ such that $A/B$ is a torsion group. Show that $rank(A)=rank(B)$. From the hypothesis, I know ...
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2answers
29 views

Show that $R^n/Im(\rho)=R^{n-1}\bigoplus R/2R$, where $R$ is an abelian group and $\rho$ is the following function.

Consider the following group homomorphism $\rho$, where $R$ is an abelian group, \begin{align*} \rho:&R\rightarrow R^n\\ \rho(r)=&(2r,2r,\cdots,2r). \end{align*} Show that $R^n/Im(\rho)=R^{n-...
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1answer
45 views

Cayley Table of Elementary Abelian Group $E_8$

I read about elementary abelian group $E_8$ at https://groupprops.subwiki.org/wiki/Elementary_abelian_group:E8#Definition. I've performed some searches on other sites and have yet to come across a ...
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1answer
75 views

Question about construction of The Grothendieck group.

In the Algebra by Serge Lang, he constructed a Grothendieck group of commutative monoid $M$, namely $K(M)$:(page 39-40) $M$ is a commutative monoid. Let $F_{ab} (M)$ be the free abelian group ...
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0answers
21 views

Classification of indecomposable abelian groups and direct product

I am having many questions about abelian groups, indecomposable groups and the direct product. Here goes : 1) Are all the indecomposable abelian groups (which can't be written as non trivial direct ...
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1answer
48 views

Hatcher 2.1.14 last part

This exercise asks to find all abelian groups that fit in the short exact sequence $0\to \mathbb{Z}\to A\to \mathbb{Z}_n\to 0$ I've proved that $A$ must be isomorphic to $\mathbb{Z}\oplus \mathbb{Z}...
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0answers
21 views

Homomorphism and relative order questions

1) I have to prove that if $\exists$ a nontrivial homomorphism $\phi:A\rightarrow B$, where A and B are finite and Abelian, then $|A|$ and $|B|$ are not relatively prime. I know that $\phi(A)$ is a ...
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1answer
39 views

Is there an abelian linearly ordered group of the multiplicative convention for complex number excluding $\{0\}$?

Evidently, $\{0\}$ has to be excluded since it has no inverse. My question is reduced to: Is there any total order on complex numbers w/o $0$? From what I sense (but not 100% sure), the ...
2
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1answer
29 views

How to prove that semidirect product of $Z_{13}$ and $Z_3$ is non Abelian for a non-trivial homomorphism

The semidirect product of $Z_{13}$ and $Z_3$ is given here Finding presentation of group of order 39 as $\{x,y | x^{13} = y^3 = 1, yxy^{-1} = x^3\}$. I understand how this is arrived at but to show ...
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1answer
47 views

Prove that a group $G$ is abelian [duplicate]

Suppose we have a group $(G, *).$ Prove that the group is abelian if $b * a^2 = b$ where $(a, b)$ are part of the group.
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1answer
36 views

Proof of Cauchy's theorem for finite groups in Dummit and Foote

The proof in Dummit & Foote is the familiar induction proof. $\require{amssymb}$ $G$ is an abelian group. Let $N = \langle x \rangle$ for some $x \in G$. Since $G$ is abelian, $N \...
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1answer
29 views

Finite group of size n for each n > 1 example

So I am trying to think of an example of a finite group of size $n$ for each $ n \gt 1 $, but nothing is coming to mind. If it is a finite group denoted as $G$, then the order of G is is $|G|$, but I ...
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1answer
79 views

Understanding rules on an Abelian group decomposition

In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $\langle g\rangle \oplus \ G/\langle g\rangle$ it concludes that by induction $G \cong \mathbb{Z}/d_1\mathbb{...
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0answers
34 views

How prove a group is not free abelian group [duplicate]

Anyone can give me a hint please? "Show $\mathbb{Z}^{\mathbb{N}}$ is not a free-abelian group"
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2answers
114 views

How to show that $\langle a,b \mid aba^{-1}ba = bab^{-1}ab\rangle$ is not Abelian?

I'd like to show that $$ G = \langle a,b \mid aba^{-1}ba = bab^{-1}ab\rangle $$ is non-Abelian. I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one....
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0answers
29 views

What are the almost periodic functions on the complex plane?

I was wondering if there is a nice characterization of almost periodic functions on the complex plane that is similar to that on the real line - the uniform closure of trignometric polynomials. ...
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0answers
46 views

Isomorphism $\{1\}\times \mathbb C^*\to \{1\}\times \mathbb C^*$

I am reading a text where I have trouble understanding an argument: Let $f: \mathbb C^*\times \mathbb C^*\to \mathbb C^*\times \mathbb C^*$ an isomorphism, such that $f(\{1\}\times \mathbb C^*)= \{1\}...
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2answers
34 views

Evaluate $a_1\cdot \dots \cdot a_n$ where $G=\{a_1,…,a_n\}$ is an abelian group with no element $a\ne e$ such that $a^2=e$

Let $G=\{a_1,...,a_n\}$ be a finite abelian group such that $\nexists a\ne e$ with $a^2=e$ Evaluate $a_1\cdot\dots\cdot a_n$ I thaught that a finite abelian group with the property of $a^2=e\implies ...
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0answers
18 views

Commutators and abelianisations of congruence subgroups in function fields

Let $A = F_q[T]$ be the ring of polynomials in one variable with coefficients in a finite field, and let $r>1$ be an integer. I'm currently looking for the abelianisation of the congruence subgroup ...
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1answer
22 views

Abelian group of prime exponent $p$ is $\mathbb{F}_p$-vector space?

In reading about group cohomological, I came across the following in a statement of a lemma. "Let $p$ be a prime, and let $A$ be an abelian group of exponent dividing $p$..." Is this just a ...
0
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1answer
22 views

Abelian Lie group representation

I have the following doubt: If I have an abelian Lie algebra with 3 generators (let's call them $t_1, t_2, t_3$), by definition of adjoint representation all of them will be equal to the $3\times 3$ ...