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.
Unusual presentation of a cyclic group.
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group-theory
group-presentation
2 Answers
7
Hint: $b = (ba)^2$ implies $b = baba \Rightarrow aba = e \Rightarrow b = a^{-2}$.
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1Could you say more about how this determinates the order of $a$ ? – 2012-10-01
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2@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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1The last implication on the top row isn't correct. Should be $a^{-1}b^{-n} = a$. – 2012-10-01
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0It's already been taken care of. thanks – 2012-10-01