Unusual presentation of a cyclic group.

Question

Show that the group with presentation $$\langle a,b| aba^{-1}=b^n, b=(ba)^2\rangle$$ is a cyclic group generated by $a$ and determine its order.

Answer 1

$$b=(ba)^2=baba=\Longrightarrow aba=1=aba^{-1}b^{-n}\Longrightarrow a=a^{-1}b^{-n}\Longrightarrow$$

$$1=aba=(a^{-1}b^{-n})ba=a^{-1}b^{-n+1}a\Longrightarrow b^{n-1}=1\Longrightarrow b^n=b=a^{-2}$$

Can you take it from here?

Answer 2

Hint: $b = (ba)^2$ implies $b = baba \Rightarrow aba = e \Rightarrow b = a^{-2}$.