Determine if the following is an equivalence relation

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Homework Help Overview

The discussion revolves around determining whether a given relationship C on a group G, defined by aCb if and only if ab=ba, is an equivalence relation. Participants explore the properties of reflexivity, symmetry, and transitivity in the context of group theory.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the three conditions necessary for an equivalence relation and express uncertainty about the transitive condition. Some suggest using counterexamples, particularly from non-Abelian groups, to explore the implications of the relationship.

Discussion Status

There is ongoing exploration of the relationship's validity in both non-Abelian and Abelian groups. Some participants have provided insights into the nature of the identity element and its role in commutativity. The discussion reflects a mix of interpretations regarding the equivalence class in different group types.

Contextual Notes

Participants are considering the implications of the relationship in both non-Abelian and Abelian groups, noting that the relationship may hold in one context but not the other. There is also a focus on the distinction between equality and the equivalence relation.

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Homework Statement


Determined if the following is an equivalence relation, if so describe the equivalence class.
The relationship C on a group G, where aCb iff ab=ba


Homework Equations




The Attempt at a Solution


So i know there's 3 things to check: reflexive condition, symmetric condition, and the transitive condition.
The 1st two passed just by inspection, but I'm really stuck on the last one.
So if ab=ba and bd=db on a group Gdoes that imply ad=da? That's how i set it up but I can't find how to (dis)prove it.
 
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I can't come up with any justification of why this statement wouldn't be true, but I'm not having any success, either. I'll keep gnawing on it for a while.
 
with no general way to shift the order of a & d in a combined multiplication, I'm thinking a counter example may be the way to go, with the two element interchange operations of permutation groups a good place to start...
 
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Would this work?
Let a,b,d exist in a non-Abelian group and then let b be the identity element.
then ab=ba and bd=db just reduce to a=a and d=d but then ad does not have to equal da since the group is not Abelian
 
yeah i think that is reasonable, i think the identity element always commutes, so is equivalent to everything, and in a non-abelian group there exist elements that don't commute, i'd just be a little clear with the difference between = vs ~ as well
 
the counter example i was thinking of was the permutation group of 5 elements with
b = (12)
a = (34)
d = (45)

which works just as well with b = ()
 
Last edited:
Old thread but I think I'm still not done with this problem

So I've establish that this is not an equivalence relationship in a non-abelian group, but it still works in an abelian group. So would the equivalence class be sets of elements in an Abelian group?

I'm still trying to grasp this idea of equivalence class
 
Yes, the equivalence class would be the entire group because all the elements commute with all the other elements and are therefore related.
 

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