TITLE: Order of $b$ when $ba=ab^3$ and $\operatorname{ord}(a)=4$.
QUESTION [5 upvotes]: Let $G$ be a non-abelian group and let $a,b \in G$ such that

$\operatorname{ord}(a) = 4$, and
$ba=ab^3$

Given that $\operatorname{ord}(b)$ is an odd prime number, compute its value.

So, $\operatorname{ord}(b)$ may be $3,5,7,11,13,...$ but I have no idea how I can compute it precisely.
I will be grateful for any help.

REPLY [3 votes]: Also, there is a bit similar way to @Daniel attempt. We have $ba=ab^3$ so we have $$a^{-1}ba=b^3\to b=ab^3a^{-1}=(aba^{-1})^3$$ so:
 $$b=(aba^{-1})^3=\left(a(aba^{-1})^3a^{-1}\right)^3=(a^2b^3a^{-2})^3=(a^2ba^{-2})^{3^2}=...=(a^4ba^{-4})^{3^4}=b^{81}$$