Parity of K-Cycle

Theorem

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

Then $\operatorname{sgn} \left({\pi}\right) = \begin{cases} 1 & : k \ \mathrm {odd} \\ -1 & : k \ \mathrm {even} \end{cases}$.

Thus:

$\operatorname{sgn} \left({\pi}\right) = \left({-1}\right)^{k-1}$

or equivalently:

$\operatorname{sgn} \left({\pi}\right) = \left({-1}\right)^{k+1}$

Proof

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

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

Thus $\pi$ is even iff $k$ is odd.

$\blacksquare$

Sources

• John F. Humphreys: A Course in Group Theory (1996): $\S 9$: Corollary $9.18$