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Proof on Order of Elements in a Group

by jeffreydk
Tags: elements, order, proof
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jeffreydk
#1
Jul20-08, 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.
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d_leet
#2
Jul20-08, 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?
bad poster
#3
Jul20-08, 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].

jeffreydk
#4
Jul20-08, 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.
mathwonk
#5
Jul21-08, 09:39 PM
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let f:G-->G be an automorphism of G. if x has order n, prove f(x) also has order n.
maze
#6
Jul22-08, 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?
mathwonk
#7
Jul23-08, 10:16 PM
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doing, and understanding WHAT you are doing, are two different things.
Firepanda
#8
May11-09, 08:57 PM
P: 431
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!
matt grime
#9
May12-09, 01:14 AM
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If n is the order of a, what is [itex]ba^nb^{-1}[/itex] ?
Firepanda
#10
May12-09, 05:43 AM
P: 431
Quote Quote by matt grime View Post
If n is the order of a, what is [itex]ba^nb^{-1}[/itex] ?
are you asking what the order of [itex]ba^nb^{-1}[/itex] is?

Woudl I be right in saying that it's n, because no matter what n is the b and b^-1 stay the same?
Firepanda
#11
May12-09, 10:20 AM
P: 431
Is there some sort of theorem for the order of a composition of more than one element?
matt grime
#12
May12-09, 01:01 PM
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Quote Quote by Firepanda View Post
are you asking what the order of [itex]ba^nb^{-1}[/itex] is?

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?
Firepanda
#13
May12-09, 02:26 PM
P: 431
Quote Quote by matt grime View Post
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?
Ah i gotcha! Kk it all made sense now.

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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