TITLE: $G$, group. $a,b\in G$. Show that $|a|=|a^{-1}|; |ab|=|ba|$, and $|a|=|cac^{-1}|, \forall c\in G$.
QUESTION [5 upvotes]: Isn't it obvious for $|a|=|a^{-1}|$, since $\langle a\rangle = \langle a^{-1} \rangle$?
For $|ab|=|ba|$, I think we should go like this: $e=(ab)^n\Rightarrow e=(ab)(ab)^{n-1}\Rightarrow (ab)^{-1}=(ab)^{n-1}\Rightarrow b^{-1}a^{-1}=(ab)^{n-1}\Rightarrow a^{-1}=b(ab)^{n-1}$ But I get stuck here, since I don't know whether to get from here to $e=(ba)^n$.
For $|a|=|cac^{-1}|$, do I have to show that $\langle a\rangle = \langle cac^{-1}\rangle$?

REPLY [2 votes]: For the second one:
Notice that as $(ab)^n=1$. Then we have $(ab)^{n-1}=(ab)^{-1}=b^{-1}a^{-1}$.
Now consider $(ba)^n$:
$$(ba)^n=b(ab)^{n-1}a=bb^{-1}a^{-1}a=e.$$
Hence $|ba| \Big\lvert |ab|$. 
Now think, from the last computations, why one must have equality.