# the order of a product two elements in symmetric group.

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.

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=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!

• Maybe it is a mistake in definition of disjoint permutations? I think there should be $\sigma(j) = j$ – 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.