Let $G$ be a group with order $2m$ where $m$ is odd. Consider the left action $\lambda_g:G\to G$.
It appears that if $g$ has odd order iff $\lambda_g$ has odd order iff $\lambda_g$ is an even permutation.
Is this true and can we generalize this observation?
On each orbit, $\lambda_g$ is cyclic of odd length thus has even sign. Also $\lambda_g$ decomposes into $|\langle g\rangle/G|$ number of disjoint cycles. Therefore the sign of $\lambda_g$ is the sum of signs of the disjoint cycles, which is even.
For the converse, one can use proof by contradiction. If $g$ has even order, then $\lambda_g$ has odd sign on each orbit. There are $|\langle g\rangle /G|$ number of cycles, which is odd by assumption that $|G| = 2m$. Therefore the sign of $\lambda_g$ on $G$ is an odd sum of odd integers, which is odd. Contradiction.