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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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    Hint: Compute $a^3ba^{-3}$ in two different ways.2012-10-23
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    Or try computing $a^3b$ in one way.2012-10-23
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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