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If $|g|=15$ and $|h|=16$, find $|\left \langle g \right \rangle \cap \left \langle h \right \rangle|$.

Would the order be $lcm(15,16)$ because $\left \langle a^{lcm(m,n)} \right \rangle = \left \langle a^{m} \right \rangle \cap \left \langle a^{n} \right \rangle$? I am not very good with this kind of question.

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    $\begingroup$ Try to think about the group and its elements rather than looking ro a formula. What are the possible orders for an element in that intersection. $\endgroup$ – Ethan Bolker Mar 13 '18 at 23:22
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Since $\langle g\rangle\cap \langle h\rangle \leq \langle g\rangle, \langle h\rangle$, we have $|\langle g\rangle\cap \langle h\rangle|$ divides $|\langle g\rangle|,|\langle h\rangle|$. So $|\langle g\rangle\cap \langle h\rangle|$ divides $\gcd (15,16)=1$. Hence $|\langle g\rangle\cap \langle h\rangle|=1$.

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    $\begingroup$ Wow! What a very, very clear answer... you truly made everything make sense. Thank you very much. $\endgroup$ – numericalorange Mar 14 '18 at 23:20
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    $\begingroup$ @numericalorange Also, do we have the assumption that $g,h$ are elements of a group? $\endgroup$ – Delong Mar 15 '18 at 0:46
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    $\begingroup$ Yes, I apologize for forgetting to mention this! $\endgroup$ – numericalorange Mar 15 '18 at 18:50

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