Let $\sigma$ be a product $\sigma_{1}\sigma_{2}...\sigma_{n}$ of disjoint cycles. Prove that the permutation $\sigma$ is even if and only if an even number of $\sigma_{i}$'s are of even length.
Attempt at proof:
Let the permutation $\sigma$ be even. Then $\sigma=\tau_{1}\tau_{2}...\tau_{m}$ where each $\tau_{i}$ is a transposition, $m$ is even, and $i=1,2,...,m$. It then follows that $\sigma_{1}\sigma_{2}...\sigma_{n}=\tau_{1}\tau_{2}...\tau_{m}$. Since any permutation can be written as a product of transpositions, one way for the preceding equation to be true is if there are an even number of $\sigma_{i}$'s that can each be written as a product of an odd number of transpositions, thus making each $\sigma_{i}$ even. (My question is as follows: Can there be an even number of $\sigma_{i}$'s each of even length?)
Now let an even number of $\sigma_{i}$'s be of even length. Since each of the $\sigma_{i}$'s are of even length, they are each odd cycles and can be written as a product of an odd number of transpositions. Since there are an even number of these $\sigma_{i}$'s, their product $\sigma$ can be written as an even number of transpositions.
Thanks in advance!