How can it be proven that a cycle of length k is an even permutation if and only if k is odd?

Question

How can it be proven that a cycle of length k is an even permutation if and only if k is odd? I know it can be done using the fact that a permutation which exchanges two elements but leaves the rest unchanged is an odd permutation.

Answer 1

A cycle of length $k$ is the product of $k-1$ inversions. To see this, write the cycle as

$$(a b c d e) = (d e) \circ (c e) \circ (b e) \circ (a e)$$

Since the parity of a permutation is determined by the parity of the number of inversions it contains, the desired result follows.

Answer 2

Let $c=(\alpha_1\;\alpha_2\;\cdots \alpha_k)$ a cycle of length $k$ so

$$c=(\alpha_1\;\alpha_2)(\alpha_2\;\alpha_3)\cdots(\alpha_{k-1}\;\alpha_k)$$ is the composition of $k-1$ transpositions and recall that the signature is a morphism of groups hence

$$\epsilon(c)=\prod_{i=1}^{k-1}\epsilon((\alpha_i\;\alpha_{i+1}))=(-1)^{k-1}$$ and the result follows easily.