# Proof on Order of Elements in a Group

1. Jul 20, 2008

### 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.

2. Jul 20, 2008

### 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?

3. Jul 20, 2008

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 $$G$$.

4. Jul 20, 2008

### 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.

5. Jul 21, 2008

### mathwonk

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

6. Jul 22, 2008

### maze

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

7. Jul 23, 2008

### mathwonk

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

8. May 11, 2009

### Firepanda

I can show
$$(bab^{-1})^n = ba^nb^{-1}$$

But how does this show the orders are equal?

Any help would be great please!

9. May 12, 2009

### matt grime

If n is the order of a, what is $ba^nb^{-1}$ ?

10. May 12, 2009

### Firepanda

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?

11. May 12, 2009

### Firepanda

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

12. May 12, 2009

### 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?

13. May 12, 2009

### Firepanda

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 $$(ab)^{x} = e$$?