## Proof on Order of Elements in a Group

I'm trying to figure out how to prove the following...

If $$a, b \in G$$ where G is a group, then the order of $$bab^{-1}$$ equals the order of $$a$$.

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

## 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.
 Recognitions: Homework Help Science Advisor let f:G-->G be an automorphism of G. if x has order n, prove f(x) also has order n.
 But in order to demonstrate that b . b^-1 is an automorphism, you would be basically doing the very proof shown above though?
 Recognitions: Homework Help Science Advisor doing, and understanding WHAT you are doing, are two different things.
 I can show $$(bab^{-1})^n = ba^nb^{-1}$$ But how does this show the orders are equal? Any help would be great please!
 Recognitions: Homework Help Science Advisor If n is the order of a, what is $ba^nb^{-1}$ ?

 Quote by matt grime If n is the order of a, what is $ba^nb^{-1}$ ?
are you asking what the order of $ba^nb^{-1}$ 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?
 Is there some sort of theorem for the order of a composition of more than one element?

Recognitions:
Homework Help
 Quote by Firepanda are you asking what the order of $ba^nb^{-1}$ 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?

 Quote by matt grime 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? :

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