If $\alpha$ is a k-cycle with k odd then there exists a cycle $\beta$ such that $\beta^2$=$\alpha$

Question

If $\alpha$ is a k-cycle with k odd then there exists a cycle $\beta$ such that $\beta^2$=$\alpha$.
I know that $\alpha$ is even since it has odd length but I don´t know how to continue.

Answer 1

By hypothesis, you have $\alpha^k = e$ where $e$ is the identity element of the symmetric group $\mathfrak S_n$.

As $k$ is odd, $\frac{1-k}{2}$ is an integer. Denote $\beta = \alpha^{\frac{1-k}{2}}$. We have

$$\beta^2= \left(\alpha^{\frac{1-k}{2}}\right)^2 = \alpha^{1-k} = \alpha.$$

Because $k + 2 \frac{1-k}{2} = 2$ and $k$ is odd, $k$ and $\frac{1-k}{2}$ are coprimes. It is known that for a cycle of length $l$, $\sigma^r$ is also a cycle of length $l$ when $l,r$ are corpimes.

This conclude our proof: $\beta = \alpha^{\frac{1-k}{2}}$ is a $k$-cycle satisfying $\beta^2=\alpha$.