Find $o(b)$ if $aba^{-1}=b^2$ and given that $a^5=e$ If in a group $G$, $a^5=e$, $aba^{-1}=b^2$ for some $a,b\in{G}$. Find $o(b)$. 
I wrote $aba^{-1}=b^2$ as $ab=b^2a$. Then $(ab)^5=(b^2a)^5$ but then I am stucked up.
 A: Given $a^5=e$ 
$aba^{-1}=b^2$
Squaring both sides 
$ab^2a^{-1}=b^4$
Again re substituting value of $b^2$ in  above we get  
$a^2ba^{-2}=b^4$ Similarly we get   
$a^3ba^{-3}=b^8$
$a^4ba^{-4}=b^{16}$
$a^5ba^{-5}=b^{32}$ This leads to order of b as 31.
A: Hint:
$$
a^2ba^{-2}=a(aba^{-1})a^{-1}=ab^2a^{-1}=aba^{-1}aba^{-1}=(aba^{-1})(aba^{-1})=b^2b^2=b^4.
$$
Repeat until you have calculated $a^5ba^{-5}$. Then use other given facts.

The OP indicated that they solved the question. Here's a spoilerized solution:

 After a few more iterations we get that $a^5ba^{-5}=b^{2^5}=b^{32}.$ Given that $a^5=e$ the l.h.s. is equal to $b$. Thus $b=b^{32}$, or $b^{31}=1$. As $31$ is a prime, we can conclude that either $o(b)=31$ or $o(b)=1$.

A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 37 on p.49 in Herstein's book.

If $a=e$, then $b=ebe^{-1}=aba^{-1}=b^2$.
So, $b=e$ must hold.
And $a=b=e$ satisfies the conditions $a^5=e$ and $aba^{-1}=b^2$.


So I think the answer of this problem is the following:
If $a=e$, then $o(b)=1$.
If $a\ne e$, then $o(b)=31$.

