Proof of commutativity

1. Jan 11, 2010

let G be a group following that whenever a, b and c belong to G and ab = ca, then b = c. prove that G is Abelian.

here is what i have for the proof:

(ab)c = c(ab)
let c = aba^-1 (trying to find a c which which allows for commutativity)

so (ab)aba^-1 = aba^-1(ab)
(ab)aba^-1 = ab(a^-1 a)b
then we see that aba^-1 = b (b = c)

a on both sides on the right: aba^-1 a = ba

then we can see that ab = ba which proves commutativity.

I am not capable of using Latex at the moment, so a^-1 means the inverse of a.

How is this? Where does b = c fit into all of this? I sort of came upon it through trying to prove commutativity and figured that that is logical. So I am not really all the comfortable with it, so I am looking for maybe some insight on how to better approach a problem like this. Thanks for any help.

2. Jan 11, 2010

iomtt6076

Are you assuming that
is true? If so, how do you know this?

I would have started my proof as follows:
Let a,b be elements of G. So a has an inverse a-1. Clearly, b=(a-1a)b and b=b(aa-1).

3. Jan 11, 2010

I'm sorry but i do not follow your proof there. I can see that it shows commutativity, for something, but I just don't see how it connects to the problem at hand. My professor told be to start off with something like (ab)c = c(ab) because that is the definition of commutativity (or something or other). So I am a little lost here. Perhaps it could be showing that substituting in the expression for c into the definition of commutativity and showing that it complies with the hypothesis is the right way to go here? I don't know. Any more suggestions?

4. Jan 11, 2010

Dick

Your suggestion to put c=aba^(-1) is a good idea. You've probably figured out that in that case ab=ca, right? So you can conclude b=c. What does that tell you? As iomtt6067 points out, you can't really start with something like (ab)c=c(ab).

5. Jan 11, 2010

iomtt6076

Well, we are being asked to show that G is Abelian, i.e. any two elements commute. So what I was trying to say is that if a and b are any two elements, we have

$$(a^{-1}a)b=b(aa^{-1}) \Rightarrow a^{-1}(ab)=(ba)a^{-1}$$

which implies ab=ba.

6. Jan 11, 2010

If I can't use (ab)c = c(ab), then what do I plug the equation for c into? And I think I see that if b = c, then the given ab = ca is the same as ab = ba which show commutativity. Right? But still I don't know where to begin now...

7. Jan 11, 2010

Dick

Plug c=aba^(-1) into ab=ca and show it's true. This really isn't as complicated as you think. Just stop thinking about (ab)c=c(ab).

8. Jan 11, 2010

Hopefully this is the last one... c = a b a^-1 ==> b = a b a^-1 ==> ba = ab

9. Jan 11, 2010

Dick

Probably almost the last one. But can you state that in the form of a full proof?

10. Jan 11, 2010

You mean write the whole thing from start to finish?

11. Jan 11, 2010

Dick

Um, yes? That's not so hard is it?

12. Jan 11, 2010

ab = ca
c = aba^-1

==> ab = aba^-1 a ==> ab = ab e = ab (this part I'm not sure what you were really saying, but anyhow, it works, obviously)

so c = aba^-1 and b = c

==> b = aba^-1 ==> ba = aba^-1 a ==> ba = ab e = ab (commutativity so the group is Abelian)

(e is the identity element)

13. Jan 11, 2010

Dick

Yes, that's it exactly. Just checking. Thanks. iomtt6076's version is also pretty nice.

14. Jan 11, 2010

thanks for all the help.