Parity of K-Cycle

Theorem

Let $\pi$ be a $k$-cycle.

Then:

$\map \sgn \pi = \begin{cases} 1 & : k \ \text {odd} \\ -1 & : k \ \text {even} \end{cases}$


Thus:

$\map \sgn \pi = \paren {-1}^{k - 1}$

or equivalently:

$\map \sgn \pi = \paren {-1}^{k + 1}$


Proof

From Transposition is of Odd Parity, any transposition is of odd parity.

From K-Cycle can be Factored into Transpositions, we see that a $k$-cycle is the product of $k - 1$ transpositions.


Thus $\pi$ is even if and only if $k$ is odd.

$\blacksquare$


Sources

  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 81 \alpha$
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $9$: Permutations: Corollary $9.18$
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