<s>[INST] Order of $b$ when $ba=ab^3$ and $\operatorname{ord}(a)=4$. 
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.
 [/INST] You can find out a multiple of the order of $b$ by shifting $a^4 = e$ past $b$, one $a$ after the other:
$$\begin{align}
b &= b\cdot e = b(a^4) = (ba)(a^3) = a(b^3a)a^2\\
&= a (b^2ab^3)a^2 = a(bab^6)a^2 = a^2 (b^9) a^2\\
&= a^3(b^{27}a) = a^4 b^{81}\\
&= b^{81}
\end{align}$$
So we have $b^{81} = b \iff b^{80} = e$, hence the order of $b$ divides $80 = 2^4\cdot 5$. The order of $b$ is an odd prime dividing $80$, hence $5$.
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