
#1
Jul2008, 12:35 AM

P: 136

I'm trying to figure out how to prove the following...
If [tex]a, b \in G [/tex] where G is a group, then the order of [tex]bab^{1}[/tex] equals the order of [tex]a[/tex]. I'm rather stumped because the group is not necessarily abelian and it seems like it would have to be in order to directly show that you can rearrange b and b's inverse to get rid of them. I'm confused party because I'm not sure if those properties still hold when you're working with the order of the elements. Any help is greatly appreciated, thanks. 



#2
Jul2008, 01:14 AM

P: 1,076

What is (bab^{1})^{n}? And given b, and k in a group G when is it true that bkb^{1} is equal to the identity?




#3
Jul2008, 03:48 AM

P: 2

As a hint, note that [tex] (bab^{1})^2 = (bab^{1})(bab) = ba(b^{1}b)ab^{1} = baeab^{1} = baab^{1} = ba^2b^{1}[/tex], where [tex]e[/tex] is the identity in [tex]G[/tex].




#4
Jul2008, 02:08 PM

P: 136

Proof on Order of Elements in a Group
Oh ok, it just hit me it makes perfect sense. I wasn't thinking hard enough about the actual definition of order. Thank you both for your help.




#5
Jul2108, 09:39 PM

Sci Advisor
HW Helper
P: 9,428

let f:G>G be an automorphism of G. if x has order n, prove f(x) also has order n.




#6
Jul2208, 08:20 PM

P: 655

But in order to demonstrate that b . b^1 is an automorphism, you would be basically doing the very proof shown above though?




#7
Jul2308, 10:16 PM

Sci Advisor
HW Helper
P: 9,428

doing, and understanding WHAT you are doing, are two different things.




#8
May1109, 08:57 PM

P: 432

I can show
[tex](bab^{1})^n = ba^nb^{1}[/tex] But how does this show the orders are equal? Any help would be great please! 



#9
May1209, 01:14 AM

Sci Advisor
HW Helper
P: 9,398

If n is the order of a, what is [itex]ba^nb^{1}[/itex] ?




#10
May1209, 05:43 AM

P: 432

Woudl I be right in saying that it's n, because no matter what n is the b and b^1 stay the same? 



#11
May1209, 10:20 AM

P: 432

Is there some sort of theorem for the order of a composition of more than one element?




#12
May1209, 01:01 PM

Sci Advisor
HW Helper
P: 9,398

No, I'm asking you what it is. Let me try to make it even more clear: if n is the order of a, what is a^n? Now, what is ba^nb^1? 



#13
May1209, 02:26 PM

P: 432

Would you be able to help me here also? : http://www.physicsforums.com/showthread.php?t=181745 is saying [tex](ab)^{x} = 1.[/tex] the same as saying [tex](ab)^{x} = e[/tex]? :) 


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