Order of group elements ab and ba

[antiemptyv]

Homework Statement

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

Homework Equations

n/a

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.

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How is that?

 

[radou]

Looks correct.

 

[antiemptyv]

Thanks!

 

[Hurkyl]

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.

 

[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)?

 

[Hurkyl]

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.

 

[antiemptyv]

Thanks for your help and advice.

 

[sairalouise]

Hi,
I don't understand the step that goes:
Using associativity, we get
(ab)^{x} = a(ba)^{x-1}b = 1.
Could someone elaborate, thanks!

 

[Hurkyl]

Hi,
I don't understand the step that goes:
Using associativity, we get
(ab)^{x} = a(ba)^{x-1}b = 1.
Could someone elaborate, thanks!

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?

 

[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 can't work the problem though.

 

[Hurkyl]

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 can't work the problem though.

I meant the specific line you were asking about:

(ab)^{x} = a(ba)^{x-1}b​

 

[sairalouise]

i just don't see how to get from one side of the equation to the other.

 

[sairalouise]

Hey, now i can!

 

[Firepanda]

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

 

[Firepanda]

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

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

 

[Hurkyl]

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.

 

[Bachelier]

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)?

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?