Closure in Groups: Definition & Examples

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around the concept of closure in group theory, specifically how it is defined and understood in the context of group operations. Participants explore the implications of closure, the definition of well-defined operations, and provide examples to illustrate their points.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether demonstrating that a*b is a well-defined element of G is necessary, or if it suffices to show that the operation is closed under *.
  • Another participant expresses skepticism about the necessity of the closure axiom, suggesting it is implied by the definition of the operation as a function from G x G to G.
  • A participant notes that different abstract algebra texts may define closure differently, indicating a lack of uniformity in definitions across resources.
  • One participant provides an example involving equivalence classes of integers modulo 3 to illustrate the concept of well-defined operations and closure, emphasizing the need to verify that the operation yields consistent results regardless of representation.
  • Another participant agrees that showing a function is well-defined is analogous to demonstrating closure in the context of groups.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the necessity of demonstrating closure as a separate requirement from showing that an operation is well-defined. There are competing views on the interpretation of closure and its implications in group theory.

Contextual Notes

Some participants highlight that the definitions and implications of closure may depend on the context and the specific mathematical framework being used, suggesting that assumptions about closure may vary between different texts.

bonfire09
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Let G be a group and my book defines closure as: For all a,bε G the element a*b is a well defined element of G. Then G is called a group. When they say well defined element does that mean I have to show a*b is well defined and it is a element of the group? Or do I just show a*b is closed under *(the operation)?
 
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I've always found the closure axiom a bit silly. It's implied if you just write *: G\times G\to G. All it means is that, given a,b\in G, there's a thing named a*b, and that whatever this thing is, it belongs to G.
 
Thanks. I saw that other abstract algebra books have it defined as how you said it. My book apparently has it defined a little differently.
 
This will often depend on the context. * is by definition a binary function *:GxG→G, and a function is well-defined by definition.. I think the problem is best illustrated by examples:

Let G = {[x]: x is an integer not a multiple of 3}, where [x] = {integers y s.t. x~y}, where we write x~y if x and y leave the same remainder upon division by 3 (alternatively, x-y is divisible by 3). The elements of G are sets called equivalence classes of Z modulo 3, and we can easily verify that G={[1],[2]}. Now define a binary operation on G by [a]*=[ab]. At first glance it might not be obvious that * is well-defined since [a] and have many different representations, and [ab] might depend on which of these representations we choose. For example [2]=[5] and [1]=[31], so we better make sure that we get [2]*[1]=[5]*[31] with how we defined *! We can verify that [2]*[1]=[2]=[155]=[5]*[31], since 155 leaves remainder 2 upon division by 3. Indeed, we can prove that [a]*=[ab] gives the same element of G no matter how we choose to write [a] and , i.e. * is well defined!

Edit: G={[0],[1],[2]} is not a group ([0] is not invertible), edited so that G={[1],[2]}
 
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Yes its just like showing a function is well defined. I wasn't sure if it just suffices to show that (G,*) is closed under the operation *.
 

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