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Let $a$ and $b$ be elements of a group $G$. If $|a| = 12, |b| = 22$ and $\langle a \rangle \: \cap \: \langle b\rangle \ne e$, prove that $a^6 = b^{11}$.

Any ideas where to start would be helpful.

Thanks.

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2 Answers 2

I sketch a solution; you should be able to fill in the details.

The intersection of the groups generated by $a$ and $b$ must have order dividing each, by Lagrange's theorem. Since this order is not 1, looking at the prime factorization tells us it must be 2. Hence it contains a single non-identity element $c$ with order $2$. They only such element in the group generated by $a$ is $a^6$, and the only such element in the group generated by $b$ is $b^{11}$. So these are equal.

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First observe that orders of $a$ and $b$ are even and so there exists element $d$ such that $d^{2}=e$ in $\langle a\rangle$ and $\langle b\rangle$. Therefore $a^{6}=b^{11}=d$, such that $d^{2}=e$.

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LaTeX tip: < and > are not the same thing as \langle and \rangle. –  Zev Chonoles Feb 21 '13 at 5:08
    
Thanks Zev. i didn't know about those commands before. Thanks again –  Phani Raj Feb 21 '13 at 5:13

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