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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.

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

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    @Ragib: I want to leave some work for the OP to do.2012-10-01
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$b=(ba)^2=baba=\Longrightarrow aba=1=aba^{-1}b^{-n}\Longrightarrow a=a^{-1}b^{-n}\Longrightarrow$

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

Can you take it from here?

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    It's already been taken care of. thanks2012-10-01