An $m$-cycle is odd iff $m$ is even

Show that an $m$-cycle is an odd permutation if and only if $m$ is an even number.

Solution

If \[ \sigma=(a_1\ a_2\ \dots\ a_m), \] then \[ \sigma=(a_1\ a_m)(a_1\ a_{m-1})\cdots(a_1\ a_2), \] a product of exactly $m-1$ transpositions. Hence \[ \operatorname{sgn}(\sigma)=(-1)^{m-1}. \] Therefore $\sigma$ is odd exactly when $m-1$ is odd, equivalently when $m$ is even.