Commutativity of permutations

Question

Two permutations $\tau_1$ and $\tau_2$ commute if and only if they are disjoint or there exists a permutation $\sigma$ and $k_1,k_2\in\mathbb{Z}$ such that $\tau_1=\sigma^{k_1}$ and $\tau_2=\sigma^{k_2}$

The $\Leftarrow$ implication is obvious but ¿is it true the biconditional? ¿Can you prove it or show a counterexample?

Answer 1

In $S_4$, let $p,q$ be given by \begin{align*} p&=(1\;2)\\[4pt] q&=(1\;2)(3\;4)\\[4pt] \end{align*} It easily verified that $pq=qp$.

It's clear that $p,q$ are not disjoint.

Noting that $p,q$ both have order $2$, it follows that $p,q$ can't belong to the same cyclic subgroup of $S_4$, since a cyclic group can't have two distinct elements of order $2$.

So that's one counterexample.

Here's an entire class of counterexamples . . .

Let $G$ be any finite, abelian, non-cyclic group.

Let $a,b\in G$ be any two elements such that the subgroup of $G$ generated by $a,b$ is not cyclic.

To see that such a pair $a,b$ always exists, we can let $a$ be an element of $G$ of maximum order, and let $b$ be any element of $G$ not in the cyclic subgroup generated by $a$.

Now let $n=|G|$, and let $p,q\in S_n$ be the permutations in a left regular representation of $G$ corresponding to $a,b$, respectively.

Since $G$ is abelian, we have $ab=ba$, hence $pq=qp$.

Since neither of $a,b$ is the identity in $G$, it follows that $p,q$ are fixed-point free, hence $p,q$ are not disjoint.

Finally, if $p,q$ were elements of the same cyclic subgroup of $S_n$, then since any subgroup of a cyclic group is cyclic, the subgroup of $S_n$ generated by $p,q$ would be cyclic, contradiction, since the subgroup of $G$ generated by $a,b$ is not cyclic.