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If $\langle a \rangle \cap \langle b \rangle=\{e\}$ then the group contains one element with least common multiple of m,n.

I have constructed the element $ab$ will have order $ab=LCM(m,n)$. $$ LCM(m,n)= fm= gn:1\le f\le n,1\le f\le m $$ $$ab^{LCM(m,n)}=(a^{m})^f(b^n)^g= e$$

Is this proof correct I used commutative nature to get that exponent distributes but I never seem to use the fact that intersection is just identity. Where have I gone wrong?

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    $\begingroup$ Titles are there to describe the question, not to hold the initial sentence. $\endgroup$ – celtschk Sep 4 '18 at 10:04
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To finish it off:

$(ab)^r=e\implies a^r=b^{-r}\implies a^r=e\implies m\mid r$. Similarly, $n\mid r$. $\therefore r\ge\operatorname{lcm}(m,n)$.

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No, the proof is not correct. To be more precise, it is not complete. You proved that $(ab)^{\operatorname{lcm}(m,n)}=e$. This only proves that $o(ab)\mid\operatorname{lcm}(m,n)$, not that $o(ab)=\operatorname{lcm}(m,n)$.

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  • $\begingroup$ So we will need the condition that intersection is identity to prove that LCM is the smallest such integer. $\endgroup$ – Sonal_sqrt Sep 4 '18 at 9:02
  • $\begingroup$ @PiyushDivyanakar Yes, that's it. $\endgroup$ – José Carlos Santos Sep 4 '18 at 9:03
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If $k:=\text{lcm}(m,n)$ then $(ab)^k=a^kb^k=ee=e$

This proves that the order of $ab$ divides $k$, but here we are not ready yet. It must also be proved that the order of $ab$ is not smaller than $k$.

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