Questions tagged [free-abelian-group]

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Set of homomorphisms on a free abelian group is a free abelian group.

If $G$ is a free abelian group with rank $n$, I need to show that ${\rm Hom}(G,\mathbb{Z})$, set of all homomorphisms is also free abelian group of rank $n$, My work: Since $G$ is free abelian group ...
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Abelian group with surjective pushfoward maps and decompsotion of free abelian group

Let $A,B,C$ be abelian groups. $A$ has the property that for any homomorphism $\alpha:A\to B$ and surjective homomorphism $\phi:C\to B$ there exists a homomorphism $\beta:A\to C$ such that $\alpha = \...
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1 vote
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Show that the free abelian group is a group.

Let S be a set and let $F\langle S\rangle = \{\phi : S \to \mathbb{Z}\mid \phi(x) = 0\ \text{ for all but finitely many } x \in S\}$. Show that $F\langle S\rangle$ is an abelian group w.r.t. the ...
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2 answers
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Uniquness and existence of free abelian group and construction

Let $I$ be a set. A pair $(G,\epsilon)$ consisting of a abelian group $G$ and a map $\epsilon:I \to G$ is called a free abelian group over $I$ if and only if for all abelian groups $H$ and maps $\phi :...
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Free abelian group generated by a set

Let $S$ be a set. Then one can construct the free abelian group with basis $S$ as the set of all functions from $S \to \mathbb{Z}$ with finite support: $$\mathbb{Z}^{(S)}:=\{f:S \to \mathbb{Z} \ | \ \...
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  • 241
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show the additive group of rationals has no elements of finite order , but it is not free abelian [duplicate]

show the additive group of rationals has no elements of finite order , but it is not free abelian My attempt : let $( \mathbb{Q}, +)$ bethe additive group of rationals. Since there doesnot ...
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Finding basis elements of submodules of localizations of finitely generated free abelian groups.

Let $M$ be a free finitely generated abelian group of rank $n$ and let $M_0$ be the $\mathbb{Q}$-vector space obtained by localizing at the prime ideal $(0)$. Let $v_1,\ldots,v_n$ be a set of $\mathbb{...
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2 answers
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Proof that any element of a free abelian group can be extended to a basis

Let $A$ be a free abelian group of rank $n$, and let $\alpha_1 \in A\setminus \{0\}$ such that $\alpha_1 \not \in kA$ for all $k > 1$. Do there always exist $\alpha_2,\ldots, \alpha_n \in A$ such ...
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Order of a quotient of a free abelian group

Let $G\subseteq \mathbb{C}$ be a free abelian group of rank $n$ and let $p$ be a prime. Then, we know that $$|G/pG|=p^n$$ and in fact for this we don't even need $p$ to be a prime. Suppose now that ...
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Semidirect product of free abelian groups

Consider the semidirect product $$ G=\mathbb{Z}^n \rtimes \mathbb{Z}^m $$ Is $ G $ always virtually abelian? Is it the case that the abelianization of $ G $ is $ \mathbb{Z}^{n+m} $ if and only if $ G $...
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What is the rank of a subgroup $H$ of finite index $e$ of a free abelian group $G$ of rank $n$?

I've recently been thinking, as it was an implicit corollary to a result in our elementary Abstract Algebra course, about the fact that every subgroup of a free abelian group is also free abelian, ...
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Prove that $G$ is a free abelian group.

Let $G$ denote the set of all 4x4 matrices of the form $$\begin{equation*} M(a,b) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ a & 1 & 0 & 0 \\ 0 & 0 & 1 & b \\ 0 & 0 &...
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Is every surjective group homomorphism $f:\mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ also injective? [duplicate]

I would like to know if every surjective group homomorphism $f:\mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ is also injective. I suspect it is true, but I'm not sure how to ...
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Is $\mathbb{Z_5}$a free abelian group ? Yes/No [closed]

Is $\mathbb{Z_5}$ a free abelian group ? My attempt: I think $\mathbb{Z_5}$ is free abelian group By the definition of free abelian group $X$ generates $G$, and $n_1x_1 +n_2x_2 +\dots+n_rx_r=0$ ...
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Show the following presentation of the free abelian group of rank n

The free group 𝐹𝑆 consists of all reduced words that can be built from members of 𝑆 and formal inverses of members of 𝑆. Show the following presentation of the free abelian group of rank $n$: $$\...
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1 answer
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Prove $(\mathbb Q,+)$ is not free abelian group. [duplicate]

Prove $(\mathbb Q,+)$ is not free abelian group. My solution: First of all I prove $(\mathbb Q,+)$ is not finitely generated. Suppose $\mathbb Q$ is finitely generated then exist $\langle\frac{1}{a_1},...
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In $F=F^{ab}(A)$, define $f\sim f'$ if and only if $f-f'=2g$ for some $g\in F$. Show $F/\sim$ is finite if and only if $A$ is finite.

this is a question from Aluffi's Algebra: Chapter 0, Exercise II.5.10. The full question is the following: Given a free abelian group over a set $A$, denoted $F=F^{ab}(A)$, we define an equivalence ...
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6 votes
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Let $G$ be a torsion-free abelian group, prove that for $g\neq h\in G$ there exists a homomorphism $\phi:G\to\Bbb{R}$ such that $\phi(g)\neq\phi(h)$

I have the following question: Let $G$ be a torsion-free abelian group, prove that for every distinct elements $g, h \in G$ there exists a homomorphism $\phi: G \rightarrow \mathbb{R}$ such that $\...
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Proof that two bases of free abelian groups have same size

So I am reading a proof that implies that if a finitely generated group has two bases then they have the same number of elements. The theorem says Let $P \in \mathbb{Z}^{m\times n} $. The. The ...
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Let $B_0=\{(2,2),(-4,4)\}$. Let $A={\rm span}_\Bbb Z(B_0)$. Prove that $A$ is a free abelian group of rank $2$.

Let $B_0=\{(2,2),(-4,4)\}$. Let $A={\rm span}_\Bbb Z(B_0)$. Prove that $A$ is a free abelian group of rank $2$. I know that A will cover all of the even numbers. However, I am not sure how to prove it ...
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1 answer
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Defining a map on a subgroup of a free group

Given a set $S$, we write $G(S)$ for the free abelian group on the basis $S$. Given a subset $T\subseteq S$, let $H$ be the subgroup of $G(S)$ generated by $T$. I wonder if the following is true: Can ...
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3 votes
1 answer
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Follow up question about homology with coefficients

In my previous question, I asked how one could say that the elements of the $p$-th singular chain group with coefficients in an abelian group, $\Delta_p(X;G) := \Delta_p \otimes_{\mathbb Z} G$, $G \in ...
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2 votes
1 answer
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Defining rank of a finitely generated free abelian group

Consider the two abelian groups $G_r=\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}$ (with $r$ copies of $\mathbb{Z}$, $r\ge 1$) and similarly $G_s=\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}$, $s\ge 1$. If ...
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If $F(X_1)\cong F(X_2)$ then $FA(X_1)\cong FA(X_2)$

I want to prove a bigger statement: Let $F(X_1),F(X_2)$ be free groups of finite rank, then $F(X_1)\cong F(X_2)$ iff $|X_1|=|X_2|$. The only thing missing in my argument is the statement in the title. ...
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  • 447
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2 answers
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Basis for a free abelian group $F(X)/F(X)'$

Let $|X|=n,$ prove that $F(X)/F(x)'$ is a free abelian group of rank $|X|$. For the rank prove and use that $\{xF(X)' : x \in X \}$ is a basis for $F(X)/F(X)'$. The abelian part is because in a group $...
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  • 447
2 votes
2 answers
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an infinite monoid that is not free monoid and does not contain any free monoid [duplicate]

Let $H$ generated by some generators $H=\langle h_1, \ldots, h_n\rangle$ $(n\gt 1)$. My question is whether there exists any monoid $H$ such that $H$, is infinite and $H$ is not a free monoid and $H$ ...
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3 votes
0 answers
105 views

Infinite product not free [closed]

I read that the group $\prod_{n\in{\mathbb N}}\mathbb Z$ is not a free abelian group. Can someone point me to a proof of this fact?
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2 votes
1 answer
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Rank of a free group times a free abelian group.

I know that the rank (i.e. minimal number of generators) of the product $\mathbb{Z}\times F_2$, of the infinite cyclic and free group on two generators, is three, but the only argument I could quickly ...
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On formal sums, free modules and free abelian groups.

I understand (i.e. correct me if I'm wrong) that when we talk about formal sums of a set $X$ with coefficients on a unital ring $R$ we refer to expressions of the form $$ \sum_{x\in X} r_x x $$ where ...
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1 answer
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Extending an element in generating set for free abelian group

Let $G$ be a finitely generated free abelian group and $S$ be a minimal generating set i.e. ${\rm rank}(G)=|S|$. If $w$ is a word in $G$ then when it can be extended in a new minimal generating set ...
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How to lift homotheties of a direct summand $P$ to become homotheties over the direct-sum $P\bigoplus Q$?

Let $R$ be a commutative ring, $M$ be a left $R$-module and $a\in R$. Then the scalar multiplication $f(m)=am$ is an $R$-module endomorphism called homothety. Suppose that $M=M_1\bigoplus M_2$ and ...
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1 answer
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What is the free abelian group on $\mathbb{N}$?

I learnt that the free abelian group on a set $X$ is the group $(\operatorname{Hom}(X, \mathbb{Z}), +)$. Okay, this sounds all right, but I also know the famous result that $\mathbb{Z}^{\mathbb{N}}$ ...
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-1 votes
1 answer
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page 74-75 Problem 2 Hungerford algebra

Consider the following problem from Hungerford algebra: A subset $X$ of an abelian group $F$ is said to be linearly independent if $n_1x_1 + \ldots + n_kx_k = 0 $ always implies $n_i = 0$ for all $i$ (...
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  • 1,586
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1 answer
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Every set has a free abelian group

I know the fact that every set has a free abelian group with basis $S$, (i.e. for any set $S$, free abelian group on $S$ is a group whose elements can be "uniquely" written by $\mathbb Z$-...
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2 votes
4 answers
163 views

How's the 'integer lattice' or the 'direct sum' of $\mathbb{Z} \oplus \mathbb{Z}$ not a Free group?

My question is about the direct sum $\mathbb{Z} \oplus \mathbb{Z}$ which is a Free Abelian group and not a free group. The the integer lattice, or what I think is the direct sum $\mathbb{Z} \oplus \...
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1 answer
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Mapping property of free abelian group, from Artin's Algebra

The question (exercise 7.10.4) is as follows: The group $G = \langle x, y; xyx^{-1}y^{-1}\rangle$ is called a free abelian group. Prove that if $u, v$ are elements of an abelian group $A$, then there ...
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2 votes
1 answer
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What is the canonical isogeny?

Given a (split connected reductive) root datum $\Psi$, there is another associated root datum $\Psi'$ and a morphism of root data $\Psi \to \Psi'$ called the canonical isogeny. My question is, is the ...
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Factor group of a free group

Let $F[A]$ be the free group on the generating set $A$. Let $C$ be the commutator subgroup of $F[A]$, then show that $F[A]/C$ is a free abelian group with basis $\{aC \mid a \in A\}$. It is trivial ...
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Free abelian with basis definition

Let $F$ be an abelian group and $B$ a subset of $F$. $F$ is free abelian with basis $B$ if and only if the cyclic subgroup $\langle b\rangle $ is infinite cyclic for each $b\in B$ and $F$ is the ...
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1 vote
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About order of Base of free abelian group

I am not sure about the connection between order of a base of free abelian group, to the order of the minimal generator set which generates the group. I would like to inherit the conclusion from ...
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2 votes
1 answer
92 views

Free groups as free product of infinite cyclic groups

Let $S$ be an arbitrary set (countable or uncountable). It is clear that the free abelian group generated by $S$ is isomorphic to the direct sum $$\bigoplus_{s\in S}\mathbb{Z}.$$ Is the free group ...
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  • 1,542
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Munkres' Section 67 Exercise 6 ( Rank of the subgroups of free abelian group with finite rank)

Hi. I am studying this topic and I cannot see the statement "If this subgroups is nontrivial, choose $x_m\in B_m$ so that $\pi(x_m)$ is a generator of this subgroup. I want proves that $\langle \...
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1 vote
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Tensor Product with a Free Abelian Group

The following claim is made in Massey's Singular Homology Theory: If $G$ is any abelian group, and $F$ is a free abelian group with basis $\{T_\alpha\}$, then every element of $G \otimes F$ can be ...
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When are free abelian normal subgroups "virtually" central?

See my previous question for the motivation for this question: Are infinite cyclic normal subgroups "virtually" central? Let $G$ be a (finitely generated) group and let $K \lhd G$ be a ...
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1 answer
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Doubt in dummit foote - direct subgroup.

In Section $5.2$ of dummit and Foote, the fundamental theorem of finitely generated ableian group is stated as A finitely generated abelian group $G$ is isomorphic to $\mathbb{Z}^r \times \mathbb{Z}_{...
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2 votes
2 answers
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Lee's definition of free abelian group

In Topological manifolds (Book), Lee defined the free product then free group by construction. i.e. defining a word and making set of all words a group. After that he defined Free Abelian Group in ...
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$G/ H$ is the Direct Product of Finite and Infinite Cyclic Groups.

Let $G$ be a free abelian group of rank $r$, and $H$ a subgroup of $G$. Then $G/ H$ is finite if and only if the ranks of $G$ and $H$ are equal. If this is the case, and if $G$ and $H$ have $\Bbb Z$-...
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1 vote
1 answer
119 views

Elements of quotient group with $\mathbb Z$-basis

Abelian free group $G, H$ have rank $2$ and $G$ has $\Bbb Z$-basis $x, y$, if $H$ has $\Bbb Z$-basis $$2x+y, 2x-3y$$ then what are the elements of $G/ H$ ? I am new to the topics, so don't know how to ...
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0 votes
2 answers
269 views

Example/Demonstration of Quotient group $G/ H$ of Free Abelian Group

Let $G$ be a free abelian group of rank $r$, and $H$ a subgroup of $G$. Then $G/ H$ is finite if and only if the ranks of $G$ and $H$ are equal. If this is the case, and if $G$ and $H$ have $\Bbb Z$-...
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0 votes
1 answer
158 views

Showing elements form a Matrix that is Unimodular

Every subgroup $H$ of a free abelian group $G$ of rank $n$ is free of rank $s \leq n$. Moreover there exists a basis $u_1, ... , u_n$ for $G$ and positive integers $\alpha_1, ... ,\alpha_s$ such that $...
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