Let $G$ be a group, let $a,b\in G$ with $ab=ba$, and let $|a|=m,~|b|=n$.

Then, $|ab|\mid\textrm{lcm}(m,n)=mn/d$, where $d=\gcd(m,n)$.

I have confusing in the following situation:

Let $\sigma,\tau$ be 'disjoint permutations' in certain symmetric group of orders $m$ and $n$, respectively, where 'disjoint' means that $\sigma(i)\neq i\Rightarrow\tau(i)=i$ and $\tau(j)\neq j\Rightarrow\sigma(j)\neq j$.

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

Then, what is the order of an element $\sigma\circ\tau$? Is it just $|\sigma\circ\tau|=\textrm{lcm}(m,n)$?

I think that $\gcd(m,n)$ need not be equal to $1$.

But, my attempts as follows:

Set $l=\textrm{lcm}(m,n)$. Then, since $\sigma\circ\tau=\tau\circ\sigma$, $(\sigma\circ\tau)^{l}=\textrm{id}$.

Now, assume that $(\sigma\circ\tau)^{l'}=\textrm{id}$ with $0<l'\le l$.

Then, since $\sigma\circ\tau=\tau\circ\sigma$, $\sigma^{l'}=(\tau^{l'})^{-1}$, and which implies that $\sigma^{l'}=(\tau^{l'})^{-1}=\textrm{id}$ since $\sigma$ and $\tau$ are disjoint.

Thus, $m\mid l'$ and $n\mid l'$, furthermore, $l=\textrm{lcm}(m,n)\mid l'$.

Therefore, since $0<l\le l'$, $l=l'$, and hence, the element $\sigma\circ\tau$ is of order $l=\textrm{lcm}(m,n)$.

My attempts seems to be false. Where i made a mistake?

Can someone point me out? Thank you!

  • 1
    $\begingroup$ Maybe it is a mistake in definition of disjoint permutations? I think there should be $\sigma(j) = j$ $\endgroup$ – Mikhail Goltvanitsa May 14 at 11:19

There is no mistake in your reasoning. The fact that permutations are disjoint is much stronger than the fact that permutations commute. So, in general for commuting permutations $a, b$ we have $$ \operatorname{ord} (a\cdot b) | \operatorname{lcm} (\operatorname{ord} (a), \operatorname{ord} (b)) $$ But in the case where $a$ and $b$ are disjoint we have $$ \operatorname{ord} (a\cdot b) = \operatorname{lcm} (\operatorname{ord} (a), \operatorname{ord} (b)). $$

Also we can give an exmaple where for commuting but non-disjoint permutations the last equality is not true. We have $$ \operatorname{ord}(a\cdot a^{-1}) = 1 $$ for every permutation $a$, but of course if $a$ has non-identity order equation $ \operatorname{ord} (a\cdot b) = \operatorname{lcm} (\operatorname{ord} (a), \operatorname{ord} (b)) $ is not valid.

  • $\begingroup$ I'm really appreciating for your answer. It is exactly what i want. Thank you again. $\endgroup$ – Primavera May 14 at 16:38
  • $\begingroup$ You are welcome! $\endgroup$ – Mikhail Goltvanitsa May 14 at 17:41

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