TITLE: Order infinite or finite
QUESTION [2 upvotes]: If $b$ has infinite order and $a$ has finite order what is the order of $bab^{-1}$? 
Sorry for the formatting. I'm using the app.
Thanks in advance.

REPLY [0 votes]: A misunderstanding you seem to have (sorry if I'm wrong) is that you seem to think exponents distribute over products. This is not true unless the group is abelian. So for example
$$(bab^{-1})^n\neq b^na^nb^{-n}$$
However, in this case the exponent does distribute, but it's an accident. It's because you get the identity either way. 
An equation is worth a thousand words:
$$(bab^{-1})^2=bab^{-1}bab^{-1}=baab^{-1}=ba^{2}b^{-1}$$
I'm sure you see that if you keep doing this the exponent of $a$ is the only thing that changes. This explains why an element has the same order as its conjugates.
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