Problem

Find the order of \((1\ 12\ 8\ 10\ 4)(2\ 13)(5\ 11\ 7)(6\ 9)\).

Solution

Let \(\tau\) be the given cycle decomposition. Note the cycles appearing in \(\tau\) are disjoint.

A cycle of size \(n\) permutes an element \(e\) to \(e\) itself if and only if the cycle is applied for a multiple of \(n\) times.

\(\tau\) consists of cycles of size \(5, 2, 3\), and \(2\). Therefore, the smallest number \(n\) such that \(\tau^n\) makes each of the cycles act as an identity function is the least common multiple of \(5, 2\), and \(3\).

The order of \(\tau\) is \(30\).

$$\tag*{$\blacksquare$}$$