# Order of group elements ab and ba

1. Aug 26, 2007

### antiemptyv

1. The problem statement, all variables and given/known data

Prove that in any group the orders of $$ab$$ and $$ba$$ equal.

2. Relevant equations

n/a

3. The attempt at a solution

Let $$(ab)^{x} = 1.$$

Using associativity, we get

$$(ab)^{x} = a(ba)^{x-1}b = 1.$$

Because of the existence of inverses--namely $$a^{-1}$$ and $$b^{-1}$$--this implies

$$(ba)^{x-1} = a^{-1}b^{-1} = (ba)^{-1}.$$

Multiplying both sides by $$(ba) = ((ba)^{-1})^{-1}$$ yields

$$(ba)^{x} = 1.$$

So,

$$(ab)^{x} = (ba)^{x} = 1$$,

and the orders $$ab$$ and $$ba$$ are the same.

---

How is that?

Last edited: Aug 26, 2007
2. Aug 26, 2007

Looks correct.

3. Aug 26, 2007

### antiemptyv

Thanks!

4. Aug 26, 2007

### Hurkyl

Staff Emeritus
Well, technically you only proved that
$$(ba)^{\mathop{\mathrm{ord}}(ab)} = 1$$
which leads to the conclusion that the order of ba is a divisor of the order of ab. You have to do a little bit more work to prove they are equal.

5. Aug 26, 2007

### antiemptyv

Ahhh, I see... I think. So, if $$(ab)^{x} = (ba)^{x} = 1$$, then ord(ba) divides ord(ab), AND ord(ab) divides ord(ba). Thus, ord(ab) = ord(ba)?

6. Aug 26, 2007

### Hurkyl

Staff Emeritus
Right. It's important to pay attention to the difference between proving the order is equal to something, and the order simply divides something. I know I've made mistakes before by messing that up.

7. Aug 26, 2007

8. Nov 1, 2008

### sairalouise

Hi,
I dont understand the step that goes:
Using associativity, we get
(ab)^{x} = a(ba)^{x-1}b = 1.
Could someone elaborate, thanks!

9. Nov 1, 2008

### Hurkyl

Staff Emeritus
So I know where you're coming from... what have you done to try and understand it? Have you worked through any special cases? Made attempts at proving it?

10. Nov 1, 2008

### sairalouise

I can show that the orders of an element and its inverse are equal, and have tried supposing that ab and ba have different orders to reach a contradiciton but i cant work the problem though.

11. Nov 1, 2008

### Hurkyl

Staff Emeritus
(ab)^{x} = a(ba)^{x-1}b​

12. Nov 1, 2008

### sairalouise

i just dont see how to get from one side of the equation to the other.

13. Nov 1, 2008

### sairalouise

Hey, now i can!

14. May 12, 2009

### Firepanda

is saying $$(ab)^{x} = 1.$$ the same as saying $$(ab)^{x} = e$$?

15. May 12, 2009

### Firepanda

Can someone confirm this for me please? it would greatly help my understanding

16. May 12, 2009

### Hurkyl

Staff Emeritus
Yes and no. You can name the identity element of a group whatever you want, just like you can name the group operation whatever you want, as well as the inverse operation. The identity in antiemptyv's group was named '1'. That group doesn't have any elements named 'e', so $(ab)^{x} = e$ can't even make sense.

17. Nov 16, 2011

### Bachelier

I hope I don't get a warning for necro-posting but I was doing research on order of elements of a group and came across this.

Hurkyl's comment that we only proved that |ba|= x means |ba| | x and not equal x.

But the proof that it is equal appeas to be weak.

So we're saying (ba)x=(ab)x = e implies |ba| | |ab| and |ab| | |ba| which makes them equal. seems weak to me.

Let |ba| = d s.t. d < x and x = k.d for some pos. integer k, then surely d ≠ x.

any ideas?