
#1
Aug2607, 01:55 PM

P: 34

1. The problem statement, all variables and given/known data
Prove that in any group the orders of [tex]ab[/tex] and [tex]ba[/tex] equal. 2. Relevant equations n/a 3. The attempt at a solution Let [tex](ab)^{x} = 1.[/tex] Using associativity, we get [tex](ab)^{x} = a(ba)^{x1}b = 1.[/tex] Because of the existence of inversesnamely [tex]a^{1}[/tex] and [tex]b^{1}[/tex]this implies [tex](ba)^{x1} = a^{1}b^{1} = (ba)^{1}.[/tex] Multiplying both sides by [tex](ba) = ((ba)^{1})^{1}[/tex] yields [tex](ba)^{x} = 1.[/tex] So, [tex](ab)^{x} = (ba)^{x} = 1[/tex], and the orders [tex]ab[/tex] and [tex]ba[/tex] are the same.  How is that? 



#3
Aug2607, 02:01 PM

P: 34

Thanks!




#4
Aug2607, 02:28 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

order of group elements ab and ba
Well, technically you only proved that
[tex](ba)^{\mathop{\mathrm{ord}}(ab)} = 1[/tex] 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
Aug2607, 02:37 PM

P: 34

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




#6
Aug2607, 02:41 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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
Aug2607, 02:42 PM

P: 34

Thanks for your help and advice.




#8
Nov108, 01:42 PM

P: 10

Hi,
I dont understand the step that goes: Using associativity, we get (ab)^{x} = a(ba)^{x1}b = 1. Could someone elaborate, thanks! 



#9
Nov108, 01:50 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101





#10
Nov108, 02:05 PM

P: 10

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
Nov108, 02:07 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

(ab)^{x} = a(ba)^{x1}b 



#12
Nov108, 02:14 PM

P: 10

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




#13
Nov108, 02:17 PM

P: 10

Hey, now i can!




#14
May1209, 10:45 AM

P: 432

is saying [tex](ab)^{x} = 1.[/tex] the same as saying [tex](ab)^{x} = e[/tex]?




#15
May1209, 07:05 PM

P: 432





#16
May1209, 11:00 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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 [itex]
(ab)^{x} = e [/itex] can't even make sense. 



#17
Nov1611, 07:12 PM

P: 376

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? 


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