What is the identity element in abstract algebra groups?

Click For Summary

Homework Help Overview

The discussion revolves around identifying the identity element in the context of abstract algebra groups, specifically through the lens of symmetric difference operations on sets.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the properties of the identity element, with one suggesting that the identity must be the empty set based on set operations. Others question the implications of certain equations and whether they lead to undefined scenarios.

Discussion Status

The conversation is ongoing, with participants sharing insights and questioning the definitions and properties of the identity element. Some guidance on the nature of symmetric difference is provided, but no consensus has been reached regarding the identity element itself.

Contextual Notes

Participants are navigating the constraints of group theory definitions and the implications of set operations, particularly in relation to the identity element and its properties.

IKonquer
Messages
47
Reaction score
0
The .pdf can be ignored.

Let A + B = (A - B) U (B - A) also known as the symmetric difference.

1. Look for the identity and let e be the identity element

A + e = A
(A - e) U (e - A) = A

Now there are two cases:

1. (A - e) = A
This equation can be interpreted as removing from A all elements that belong to e to yield the set A. In order for this statement to be true, the identity element e must be the empty set.

2. (e - A) = A
This equation can be interpreted as removing from e all elements that belong to A to generate a set A. Is this statement undefined?
 

Attachments

Last edited:
Physics news on Phys.org
If A=A'(inverse) then why does A+A'={}(empty set)?
 
A + A' is the symmetric difference, and not by means of normal addition.
 
Ah. Well I learned something :)
 
(e-A) must equal something else and not A. Moreover it must equal something such that the union of (A-e)=A with (e-A)=X is A U X=A. I am sure you are aware of such a set =).

It can't be undefined or else were breaking the conditions of what it is to be a group. A and e are elements of the group so (A-e)U(e-A) must be too. right?
 
Last edited:

Similar threads

Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 26 ·
Replies
26
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 25 ·
Replies
25
Views
12K