Let \(\sigma = (4\, 2\, 1)(6\, 1\, 3\, 2) \in S_6\) in cycle notation.
Write \(\sigma\) as a product of disjoint cycles.
Compute the order of \(\sigma\). What is the general theorem about the order of cycles?
Determine if \(\sigma\) is even or odd. What is the general theorem?
Solution
Permutations are composed from right to left.
1. \(\sigma=(1\,3)(2\,6\,4)\).
Applying \((6\,1\,3\,2)\) and then \((4\,2\,1)\): \(1\mapsto3\mapsto3\), \(3\mapsto2\mapsto1\), \(2\mapsto6\mapsto6\), \(6\mapsto1\mapsto4\), \(4\mapsto4\mapsto2\), \(5\mapsto5\). So \(\sigma\) swaps \(1\) and \(3\), cycles \(2\mapsto6\mapsto4\mapsto2\), and fixes \(5\).
2. If \(\pi=c_1\cdots c_k\) is a product of disjoint cycles of lengths \(\ell_1,\dots,\ell_k\), then \(\operatorname{ord}(\pi)=\operatorname{lcm}(\ell_1,\dots,\ell_k)\); hence \(\operatorname{ord}(\sigma)=\operatorname{lcm}(2,3)=\boxed{6}\).
Disjoint cycles commute, so \(\pi^N=c_1^N\cdots c_k^N\), and the \(c_i^N\) have disjoint supports. Thus \(\pi^N=1\) exactly when every \(c_i^N=1\), that is, when \(\ell_i\mid N\) for every \(i\).
3. With \(\pi\) as in step 2, \(\operatorname{sgn}(\pi)=(-1)^{\sum_i(\ell_i-1)}\); hence \(\operatorname{sgn}(\sigma)=(-1)^{1+2}=-1\) and \(\sigma\) is odd.
A \(k\)-cycle is the product \((a_1\,a_k)(a_1\,a_{k-1})\cdots(a_1\,a_2)\) of \(k-1\) transpositions, so it has sign \((-1)^{k-1}\), and \(\operatorname{sgn}\) is a homomorphism.