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Let $G$ be a group, $a$ and $b$ are non-identity elements of $G$, $ab=b^2a$. If the subgroup of $G$ generated by $a$ has order 3, what about the order of the subgroup of $G$ generated by $b$?

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    The problem came up again, http://math.stackexchange.com/questions/231072/what-can-ab-b2a-and-a-3-imply-about-the-order-of-b-when-b-neq-e/231078#comment513196_231078 --- my answer there expands on Thomas' hint here.2012-11-06

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We have $ ab = bba $ multiplying gives $ aaab = aabba $ $ b = aabba $ $ baa = aabb $ Now we van work: $ baa = a(ab)b = a(bba)b = (ab)b(ab) = (bba)b(bba) = b^2(ab)bba = b^2(bba)bba $ $ = b^4(ab)ba = b^4(bba)ba = b^6(ab)a = b^6(bba)a = b^8aa $ so $baa = b^8aa \Rightarrow b^7 = e $

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    So... not pretty, following Thomas Andrews' hint might be easier, but both ways should give an answer.2012-10-23
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$b=a^2b^2a\Rightarrow b^n=a^2b^{2n}a$ i.e., $b^{2n}=ab^na^2 $ then, $b^8=ab^4a^2=aab^2a^4=aaaba^6=b$ whence, $b^7=e$ futhermore, since 7 is a prime and $b$ is not a identity, we can not have a positive interger $k$ less than 7 that $b^k=e$
hence, 7 is the order of b. (This answer seems easier.)

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    Great solution. It does take some insight to try 7.2015-10-22