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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 $a,b\in G.$

2
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1answer
87 views

Non-abelian Groups of Order $p^3$

I am trying to show that a non-abelian group $G$ of order $p^3$ is isomorphic to one of two groups constructed on page 48 of these group theory notes (see examples 3.14 and 3.15 on that page). Here ...
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2answers
32 views

What is the difference between Cauchy's Theorem and Cauchy's Theorem for Abelian Groups?

These theorems seem to be identical but for some reason, the requirement that a group is finite AND abelian is sometimes stated instead of just finite. Could someone let me know if there is a ...
4
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2answers
49 views

Showing there is a unique group table for $\{1, a,b,c\}$ such that there is no element of order $4$. [duplicate]

Assume $G = \{1, a,b,c\}$ is a group of order $4$ with identity $1.$ Assume also that $G$ has no elements of order $4$. Show that there is a unique group table for $G$. Also show that $G$ is abelian. ...
4
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2answers
54 views

Infinite group with the order of abelian subgroup bounded

In Isaac's Finite Group Theory Page 28, it states: There exist infinite groups in which the abelian subgroups have bounded order. I fail to construct such group. In fact, I'm only able to deduce ...
2
votes
1answer
34 views

Abelian groups about rationals

I cannot delete it so i edited it
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1answer
25 views

Direct product decomposition of the group of complex roots of unity

I'm studying $p$-adic numbers (Robert's "A course in $p$-adic analysis) and, at page 41, the author states that, for every prime $p$, the group $\mu$ of all complex roots of unity has a direct product ...
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0answers
28 views

Abelian subgroups of a simple group

Let $G$ be a finite non-abelian simple group, and $A \leqslant G$ be an abelian subgroup. How large $A$ can be? There exists any bound of the type $|A| \leq |G|^r$ for some $r<1$? How can I prove ...
3
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1answer
47 views

Commutative subtraction

It is well known that subtraction is not commutative in general. However, it is commutative in some groups: $\mathbb I$, $\mathbb C_2$, $\mathbb K_4$. I am trying to understand the logic. ...
3
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2answers
74 views

Why does $a^{m_1}=a^{m_2}$ imply $a^{m_1-m_2}=e$?

I was reading this answer. I understand almost all of it. However, there is still one thing that continues to puzzle me. How should I prove for sure that, in this example, if $m_1\neq m_2$ and $a^{...
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0answers
31 views

Condition for exponents, for a group G to be Abelian [closed]

We say that a group $G$ has exponent $e$ if $e$ is the smallest positive integer such that $x^e = 1$ for every $x\in G$. My question is the following: For what integers e is a group having ...
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0answers
14 views

Exercise about a family of homomorphisms between abelian groups

It is considered the category $\mathscr{Ab}$ of abelian groups; a family of homomorphisms: $\{ f_i : M_i \rightarrow N_i | i \in I, M_i,N_i \in \mathscr{Ab} \}$ and constructions $\prod_i f_i$, $\...
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1answer
27 views

Find the isomorphism type

Consider the abelian group $G$ generated by $a$, $b$ and $c$ and determined by the following relations \begin{aligned} 3 a+9 b+9 c &=0 \\-3 b+9 c &=0 \end{aligned} determine the isomorphism ...
5
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1answer
103 views

G is a group and $(ab)^3=a^3b^3$ for all $a,b \in G$. Prove (or disprove with a counterexample) that if $(ab)^3=(ba)^3$, then $ab=ba$.

Proposition. Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a,b \in G$. If $(ab)^3=(ba)^3$, then $ab=ba$. Is it true or false? So far I've only been able to prove that powers of $a$ commute ...
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2answers
40 views

prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelian

I need to prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelian. I use that, from Lagrange's theorem, the order ...
2
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1answer
18 views

Relationship between Archimedean and Divisible ordered groups

Let $(G,+,\leq)$ be a linearly ordered abelian group (i.e. the order is total and compatible with the sum) and $n\cdot x$ denote the classical action of $\mathbb{Z}$ over $G$ (i.e. $0$ for $n=0$, sum ...
2
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0answers
31 views

Modifying long exact sequences

Let $$\dots A_i\stackrel {f_i}\to B_i \stackrel {g_i}\to C_i \stackrel {h_i}\to A_{i+1}\to \dots$$ be a long exact sequence of Abelian groups. Is it true that if there are maps $k_i:D_i\to E_i$ such ...
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0answers
14 views

All the $m$th powers commute with each other and all the $n$th powers commute with each other, $m$ and $n$ relative prime, is abelian. [duplicate]

Show that a group in which all the $m$th powers commute with each other and all the $n$th powers commute with each other ,$m$ and $n$ relatively prime, is abelian. How to do this ? we know $1=mx+yn$ ...
5
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2answers
77 views

Associocommutativity

One thing I've noticed is that addition and multiplication both form commutative groups over the reals, but subtraction, division, and exponentiation are neither associative nor commutative. Ignoring ...
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0answers
33 views

Group Homomorphisms into an Abelian Group

The following comes from Hungerford's Algebra. [Prove that if] $f: G \to H$ is a homomorphism, $H$ is abelian and $N$ is a subgroup of $G$ containing $\ker f$, then $N$ is normal in $G.$ A ...
3
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1answer
28 views

if $G$ is a divisible group then any subgroup of $G$ is also divisible

Is it true that if $G$ is a divisible group then any subgroup of G is also divisible? I know that if $H \leq G$ then for any $h \in H$ and for any $k \in \mathbb{N}$ then there exists $x \in G$ such ...
5
votes
1answer
112 views

finite abelian groups tensor product. [closed]

Is the following question obvious ? Let $G$ be an abelian group, such that for any finite abelian group $A$, we have $G\otimes_{\mathbf{Z}}A=0$, does it mean that $G$ is a $\mathbf{Q}$-vector space ?...
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0answers
12 views

Let be $G$ a group if order $p^n$ where p is a prime and n a natural number. Prove that there exist normal subgroups… [duplicate]

Let be $G$ a group of order $p^n$ where $p$ is a prime and $n$ a natural number. Prove that there exist normal subgroups $N_{1}, N_{2}, ..., N_{n}$ of $G$ such that $N_{1} < N_{2}<...<N_{n}$ ...
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0answers
22 views

Let $G$ be a group of order $p^n$ where $p$ is a prime and $n \in \mathbb{N}$. Prove that exist normal subgroups [duplicate]

Let $G$ be a group of order $p^n$ where $p$ is a prime and $n \in \mathbb{N}$. Prove that exist normal subgroups $N_{1},N_{2},N_{3},...,N_{n}$ with $|N_{i}|=p^i$ for all $i \in ${$1,2,3,...,n$}.
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1answer
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Let $G$ be Abelian. Then any subgroup of $G$ is normal. Does the converse hold? [duplicate]

I need a little help with the following problem of abstract algebra: Let $G$ an Abelian group. Clearly, any subgroup of $G$ is normal. Is the opposite true, that is if every subgroup of $G$ is ...
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0answers
34 views

Showing commutativity in sets [duplicate]

Given $*$ is a binary closed operation on a non-empty set $G.$ $*$ is assosciative. for all $g$ in $G$, there exists $e$ such that $e*g=g$ and for all $g$ in $G$, there exists $h$ such that $h*g=...
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0answers
24 views

Find invariant factors from elementary divisors of an abelian group of order 720

For any abelian group of order $720=2^43^25^1$, by writing out all the partitions of each of the powers $4,2,1$. I found that the abelian group must be isomorphic to only one of the following: $$ \...
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0answers
45 views

Proving Abelian Group Decomposition Uniqueness

Supposing I've already got handy a proof that every finite abelian group can be written in at least one way as the direct product of some cyclic groups of prime-power order, is the following correct ...
1
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1answer
25 views

Quotients of finitely generated abelian groups

Let $G=\mathbb{Z}^n\times F$ be a finitely generated abelian group, where $F$ is finite, and let's assume that the order of $F$ is odd. Let $H_1,H_2\leq G$ be two subgroups of $G$ such that the ...
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vote
1answer
66 views

why splitting lemma fails for nonabelian groups?

Like the title said, I want to ask why does it fail. The claim and the counterexample here https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Splitting_lemma.html are very ...
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0answers
13 views

Coset Progression is Freiman Isomorphic to Bohr Set

For an abelian group $G$, $H$ a finite subgroup of $G$, $x_1, \dots, x_r \in G$ and $L_1, \dots, L_r \in \mathbb N$, let: $P(x ; L) = P(x_1, \dots, x_r ; L_1, \dots, L_r) = \{l_1x_1 + \dots + l_rx_r ...
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1answer
50 views

An exact sequence of abelian groups

Consider an exact sequence of abelian groups $$ 0 \to A \to \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to B \to 0, $$ where we make no assumption on the map $\mathbb{Z} \oplus \mathbb{Z} \to \mathbb{...
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1answer
51 views

Is $|\{z\in\Bbb Z_3\times\Bbb Z_9: |z|=9\}|=18?$

This is Exercise 8.10 of Gallian's "Contemporary Abstract Algebra". Answers that use only the preceding material from the textbook are preferred. The Question: How many elements of order $9$ does $...
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2answers
36 views

Are all abelian subgroups of symmetric groups generated by disjoint cycles?

That is, do all the subgroups of, say, S6, look sort of like the subgroup generated by the cycles (123) and (45)? This seems like an overly simplistic characterization of the Abelian subgroups, but I ...
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2answers
60 views

Homomorphisms Group Theory [closed]

Does anybody know how I would go about proving this question ? Let G and H be groups and let φ : G −→ H be a group homomorphism. Suppose that G is abelian and φ is a surjection. Prove that H is ...
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0answers
29 views

If $G = \mathbb{Z}_{p}^{*} $, show that $G$ is cyclic

On the group of the invertible elements in $\mathbb{Z}_p$, the question asks to show that the group is cyclic. This must have something to do with the representation of $G$ as a product of groups with ...
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2answers
39 views

Cyclic quotient group of permutation group $\mathbb{S}_n$

I have a group-theoretic lemma that I am having difficulty with: $\mathbb{S}_n$ has a cyclic quotient group of prime order $p$ iff $p=2$ and $n\geq2$ Proof: $n=2$ is trivial so suppose $n\geq3$. ...
0
votes
1answer
30 views

Prove that every irreducible real representation of an abelian group is one or two dimensional. [closed]

Knowing the following theorem: " Every irreducible complex linear representation of an abelian group is one-dimensional " how can I prove the question mentioned in the title ..... Could anyone help me ...
3
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0answers
132 views

Torsion-free abelian groups, tensor product and $p$-adic integers

I'm studying torsion-free abelian groups and I know (see Fuchs, "Infinite Abelian Groups", vol. $2$, pp $154$) that, if $\mathbb{Z}_p$ is the set of $p$- adic integers and $\mathbb{Z}_{(p)}$ denotes ...
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1answer
27 views

Product of all elements in finite abelian group times random group element

I'm just starting a course on group theory and the question in struggling with is as follows: Let $G$ be a finite Abelian group and let $x\in G$. Proof: $\prod_{g\in G}xg=\prod_{g\in G}g$ I've ...
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0answers
37 views

Prove that if $G$ is a finite group, then the index of $Z(G)$ cannot be prime. [duplicate]

This appears to be new to MSE. I'm reading "Contemporary Abstract Algebra (Eighth Edition)," by Gallian. This is Exercise 7.38 ibid. Answers that use only the tools available in the textbook so far ...
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1answer
79 views

Torsion-free Abelian Groups of Finite Rank and Free Groups (Fuchs) - Self study

I want to solve the following problem (Fuchs, "Infinite Abelian Groups", Vol.$2$, pp. $153$ Ex. $4$): "Let $A$ be a torsion-free group of finite rank $n$ and $F$, $F'$ free subgroups of $A$ of rank $...
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0answers
15 views

How to show that $g^**g$ is a function of positive type, $g \in L^1(G) \cap L^2(G)$?

Let $G$ be a locally compact abelian Hausdorff group (with Haar measure $d\mu$). Call a function $h \in L^\infty(G)$ a function of positive type if $$ \int_G (f^* *f)h \, d\mu \geq 0 \ \ \ \forall f \...
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votes
1answer
37 views

Proper $\Bbb Z$-submodules of $\Bbb Q$ are finitely generated or not? [closed]

Let $M$ be a proper $\Bbb Z$-submodule of $\Bbb Q.$ Can we say that $M$ is finitely generated? I know that $\Bbb Q$ is not finitely generated as a $\Bbb Z$-module. Please help me in this regard. ...
2
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1answer
37 views

Characters of $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 $

Follow up question to $\mathbb{Z}_2 \oplus \mathbb{Z}_2$, how would I find the characters of $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$? The Cayley table for this group: \begin{align*} \...
4
votes
1answer
51 views

Characters of $\mathbb{Z}_2 \oplus \mathbb{Z}_2$

From the Cayley table: \begin{align*} \begin{array}{c | c c c c } & (0,0) & (0,1) & (1,0) & (1,1)\\ \hline (0,0) & (0,0) & (0,1) & (1,0) & (1,1)\\ (0,1) &...
2
votes
1answer
27 views

Epimorphism between abelian groups

I am new to group theory, and I need some help with surjective group homomorphisms for abelian groups. Let's say I have two finitely generated abelian groups: $G=\mathbb{Z}^n\oplus \mathbb{Z}_{m_{1}}...
2
votes
1answer
38 views

Prove that if $|A+A| \leq K|A|$ then $2A - 2A$ is a $K^{16}$-approximate group.

Let $A$ be a finite subset of an abelian group, $G$ (call the operation addition). We say $A$ is a $K$-approximate group if: 1) $e_G \in A$ 2) $A^{-1} = \{ a^{-1} \mid a \in A \} = A$ 3) $\exists X ...
2
votes
1answer
69 views

What is the order of the subgroup $\langle 5\rangle \times \langle 3\rangle$ in $Z_{30} \times Z_{12} ?$

What is the order of the subgroup $\langle 5\rangle \times \langle 3\rangle$ in $Z_{30} \times Z_{12} ?$ I think it should be $12$, since $O(5) = 6$ in $Z_{30}$ and $O(3) = 4$ in $Z_{12}$. The order ...
3
votes
0answers
45 views

Are $\mathbb{Z}$ and $\mathbb{Z}_n$ the only rings (with identity) whose modules are equivalent to abelian groups?

Let $R$ be a ring with identity. Let $M$ and $N$ be $R$-modules. Let $f$ be an (arbitrary) group homomorphism from $M$ to $N$. Under what conditions on $R$,$M$, and $N$ is $f$ also a $R$-module ...
1
vote
1answer
56 views

dual isomorphism

Let $C$ be the category of finitely generated abelian groups. If $M$ is a finitely generated abelian group, then define its dual as $M^* = \operatorname{Hom}(M,\mathbb{Q}/\mathbb{Z})$. Now I want to ...