Permutation Product

Question

Here is a problem from Algebra by Michael Artin:

$p = \left( {\begin{array}{*{20}{c}} 3&4&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&5 \end{array}} \right),q = \left( {\begin{array}{*{20}{c}} 1&4&5&2 \end{array}} \right)$

$pq = \left( {\begin{array}{*{20}{c}} 2&3&4 \end{array}} \right),qp = \left( {\begin{array}{*{20}{c}} 1&3&5 \end{array}} \right)$

Although $pq \ne qp$, both of them turn out to be 3-cycles. Is this an accident?

I tried some more examples and it seemed to be not an accident. But I don't know how to prove this.

Any help will be appreciated.

Answer 1

Two permutations have the same cyclic structure (i.e. they decompose into the product of the same number disjoint cycles of the same length) if and only if they are conjugated. So for any $p,q\in S_n$ you have $qp=q(pq)q^{-1}$ and hence they are conjugated. So no, it is not a coincidence.