# Unusual presentation of a cyclic group.

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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What have you tried? If this is homework, tag it as such. –  lhf Oct 1 '12 at 14:35

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

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

Can you take it from here?

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The last implication on the top row isn't correct. Should be $a^{-1}b^{-n} = a$. –  Brandon Carter Oct 1 '12 at 14:29
It's already been taken care of. thanks –  DonAntonio Oct 1 '12 at 14:32

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

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Could you say more about how this determinates the order of $a$ ? –  Ragib Zaman Oct 1 '12 at 15:05
@Ragib: I want to leave some work for the OP to do. –  Brandon Carter Oct 1 '12 at 15:13