Questions tagged [abelian-groups]

For questions about abelian groups, including the basic theory of abelian groups as a topic in elementary group theory as well as more advanced topics (classification, structure theory, theory of $\mathbb{Z}$-modules as related to modules over other rings, homological algebra of abelian groups, etc.). Consider also using the tag (group-theory) or (modules) depending on the perspective of your question.

An abelian or commutative group is a group $(G,*)$ in which all elements commute: $$\forall a,b\in G\,\,, a*b=b*a\,.$$ Usually the product is denoted by $+$ in an abelian group, and the identity of the group by $0$. Abelian groups are also known as modules over the ring $\mathbb{Z}$ of integers.

Examples include the integers $\mathbb{Z}$ under addition, as well as the rationals $\mathbb{Q}$ under addition. In fact, every cyclic group is an abelian group. Non-examples include $S_3$, the symmetry group on three elements, as well as $\mathrm{SO}(3)$, the rotations in three dimensions.

The fundamental theorem of abelian groups says that all finite abelian groups are direct products of cyclic groups, themselves abelian.

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Abelian group and commutativity problem

Let us assume that we have a group G with the following property : for every 5 discrete elements of G , at least 2 of them can be commutated. Is G an abelian group?
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how to prove this for abelian groups?

I have an Abelian Group G, with gamma an arbitrary element of G. I am asked to prove the following: $$\prod_{g\in G}g=\prod_{g\in G}\gamma g.$$ I have absolutely no idea how to prove this or why it is true. Could someone give me a hint (not the…
user56834
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Showing that $\mathbb{C}^\times$ is an abelian group

QUESTION Multiplication of complex numbers defines a binary operation on $\mathbb{C}^\times := \mathbb{C} \setminus \{0\}$. Show that $\mathbb{C}^\times$ together with this operation is an abelian group. ATTEMPT I know that for an abelian it has to…
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finite abelian groups and exp(G)

exercise:$$$$ Let $G$ be an abelian finite group and suppose $n$ is the smallest integer such that $g^n =e$ for all $g\in G$. Given a decomposition of G as a product of cyclic groups from prime power orders, find $n$. $$$$ I used these two…
letisya
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Exsitence of element of a certain order in an infinite abelian group

I came up with the following question reading this(Finite Abelian Groups question). Let $G$ be an abelian group. Suppose there is an integer $n \ge 1$ such that $nG = 0$. Let $m$ be the smallest integer $\ge 1$ such that $m G = 0$. Is there an…
user254385
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How can I prove an abelian group is not free?

How can I prove a given abelian group; such as $\mathbb{Z}_4$ with addition mod 4, is not a free group? Should I consider all the subsets of the given group and prove any of them cannot be a basis? But this approach will give me a lot of sub groups…
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Relation between finite Abelian Groups and traces?

I have recently read Kronecker's 1870 paper on finite Abelian groups, on the definition of abstract group and so on. It turns out that such definition is literally taken over (being probably unaware of it, for I am almost certain they did not…
Javier Arias
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Inverse element of the abelian group (P(M),symmetric difference)

What would be the inverse element in this abelian group: $(P(M),\triangle)$? I know the neutral element is the empty set and I thought the inverse element would be $A^{c}$ for every $A$. Turns out $A\triangle A^{c}=P(M)$ but it has to return the…
Arthur
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Non abelian subgroup of a abelian group.

What is the relationship between abelian subgroup of a non-abelian group(when exist, example, theorem)?? any thing such link regarding the question would help. I apologize for previous post-error.
Michael
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Prove that is group is abelian

Let G be a set of all non-zero real numbers. And a * b = (ab/2). We need to show (G,*) is an abelian group. I know I need to show ab = ba, but that seems so trivial, Assuming the group is abelian, a * b = b* a ab/2 = ba/2 Crossing out 2, we get…
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is this operating procedure an Abelian Group?

I have to show if the following procedure gives a (Abelian) Group (G, *). $G = \{ \textrm{true}, \textrm{false} \}$ $a*b := ( a \leftrightarrow b)$ (which means that $a$ is $\textrm{true}$ if and only if $b$ is $\textrm{true}$) 1.) Closure For…
Vazrael
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