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Let $\sigma$ and $\tau$ be distinct(means relative prime) permutations of order $m$ and $n$, respectively(need not be $\gcd(m,n)=1$) in certain symmetric group.

Now, suppose that $\sigma\circ\tau=\tau\circ\sigma$.

Then, which one is true?:

(i) the order of $\sigma\circ\tau$ is $\textrm{lcm}(m,n)$.

(ii) the order of $\sigma\circ\tau$ divides $mn$.

(iii) the order of $\sigma\circ\tau$ divides $\textrm{lcm}(m,n)$.

Can anyone help me? Thank you!

AnonyMath
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  • What do you mean by relative prime permutations? – Groups May 14 '19 at 08:40
  • What have you tried? Did you have a look at this site already? – Dietrich Burde May 14 '19 at 08:40
  • In case you mean "coprime orders" see this duplicate. There a re a lot of good answers here concerning this question. – Dietrich Burde May 14 '19 at 08:43
  • @DietrichBurde, No, it's not coprime orders. I've tried as follows: Let $d=\gcd(m,n)$, and let $l=\textrm{lcm}(m,n)=mn/d$. Then, $(\sigma\circ\tau)^{l}=\textrm{id}$ since $\sigma\circ\tau=\tau\circ\sigma$. So, the order of $\sigma\circ\tau$ divides $l$. Now, assume that $(\sigma\circ\tau)^{c}=\textrm{id}$ with $0<c\le l$. Then, $\sigma^{c}=(\tau^{c})^{-1}$, and further, $\sigma^{c}=\textrm{id}=(\tau^{c})^{-1}$ since $\sigma$ and $\tau$ are distinct. Thus, $m\mid c$ and $n\mid c$, and therefore, $l=\textrm{lcm}(m,n)\mid c$. Hence, $l=c$, i.e., the order of $\sigma\circ\tau$ is $l$. Right? – AnonyMath May 14 '19 at 09:26
  • @DietrichBurde, In my attempt, I think I made a mistake, but I don't know where I made the mistake. Can I ask you for a correction? – AnonyMath May 14 '19 at 09:27
  • @DietrichBurde Is this a duplicate of the linked question? It’s similar, but I don’t think it’s a dup. – Santana Afton May 14 '19 at 10:46

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