Proof on Order of Elements in a Group

[jeffreydk]

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.

[d_leet]

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]

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.

[jeffreydk]

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]

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

[maze]

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

[mathwonk]

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