Determine if the following is an equivalence relation

[polarbears]

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.

 

[Mark44]

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.

 

[lanedance]

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...

 

[polarbears]

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

 

[lanedance]

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

 

[lanedance]

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 = ()

 

[polarbears]

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

 

[vela]

Yes, the equivalence class would be the entire group because all the elements commute with all the other elements and are therefore related.