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Let $G$ be a group and $a,b \in G$ with $a \ne 1$ and $b \ne 1$. Assuming $ab^2 = b^3a$ and $a^2=1$ I need to prove that the order of $b$ is $5$.

I have proved by contradiction that it can't be 2 or 3 but I don't know how to prove that it must be 5 and it can't be 4.

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  • $\begingroup$ Why can't it be 6 or 7, for example? Do you anything about the order of $G$ itself? $\endgroup$ – user307169 Mar 26 '16 at 13:44
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Note that $$b^5 = b^3a^2b^2=(b^3a)(ab^2)=(b^3a)(b^3a)=b^3(ab^2)ba=b^3(b^3a)ba=b^6aba,$$ multiplying by $b^{-5}$ from the left we get $1=baba$. Multiplying by $a$ from the right we get $a=baba^2=bab$. Thus $$1=a^2=(bab)^2=b(ab^2)ab=b(b^3a)ab=b^4a^2b=b^4\cdot 1 \cdot b = b^5.$$

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  • $\begingroup$ How would you proof the order is not 4? $\endgroup$ – Rodrigo Mar 26 '16 at 14:10
  • $\begingroup$ @Rodrigo Because 4 does not divide 5. $\endgroup$ – Ted Shifrin Mar 26 '16 at 14:12
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    $\begingroup$ Well there is a general fact that if $x^n=1$ then the order of $x$ divides $n$. In our case $b^5=1$ and so the order of $b$ divides $5$ and so is equal to either $1$ or $5$. However $b\neq 1$ so the order of $b$ must equal $5$. $\endgroup$ – timon92 Mar 26 '16 at 14:13
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Consider that $$ab^2a=b^3$$ And so: $$b^2=ab^3a=ab^2ba=ab^2aaba=b^3aba$$ From here: $$b^{-1}=aba$$ Thus $$b^{-2}=ab^2a=b^3$$ which means: $$b^5=1$$

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  • $\begingroup$ How would you proof the order is not 4? $\endgroup$ – Rodrigo Mar 26 '16 at 14:10
  • $\begingroup$ Since $b^5=1$ and $5$ is a prime number. $\endgroup$ – hamid kamali Mar 26 '16 at 14:13

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