Show that a permutation is odd if and only if its disjoint-cycle decomposition contains an odd number of even-length cycles.
Solution
An \(m\)-cycle has sign \[ (-1)^{m-1}, \] so it is odd exactly when \(m\) is even.
Write a permutation as a product of disjoint cycles \[ \sigma=c_1\cdots c_r. \] Then \[ \operatorname{sgn}(\sigma)=\prod_{i=1}^r\operatorname{sgn}(c_i). \] Odd-length cycles contribute \(+1\), while even-length cycles contribute \(-1\). Hence \[ \operatorname{sgn}(\sigma)=(-1)^e, \] where \(e\) is the number of even-length cycles. Therefore \(\sigma\) is odd exactly when \(e\) is odd.