Finding the "square root" of a permutation

Suppose $r$ is odd, and ${\rm ord}(\alpha)=r$. ($\alpha,\beta$ are cycles.)

Now, $\alpha=(a_1\cdots a_n)$.

I need to find $\beta$ that will make $$\alpha = \beta^2$$

How can I show it?

What I tried?

${\rm ord}(\alpha)=r$ so $a^{r+1}=\alpha$.

Let say that $\beta=\alpha^{\frac{r+1}{2}}$, $\frac{r+1}{2}$ is an integer, how do I prove that $\alpha^\frac{r+1}{2}$ is a cycle?

Because then $\alpha=\beta^2$...

Thank you!

Let $\alpha$ be the cycle given by:

$$(a_1~a_2~\ldots~a_n)$$

We are furthermore given that $r$ is the smallest natural number such that $\alpha^r = ()$.

Using the cycle notation, we easily determine that $\alpha^k(a_1) = a_{k+1}$ for any $k$ (where $a_{n+1} = a_1$ and so on) -- similarly for the other entries. It follows that $r = n$.

Now to find out why $\beta = \alpha^{\frac{r+1}2}$ is also a cycle. Let us contemplate what it means for $\beta$ to consist of shorter cycles.

Recall or observe that:

• The order of a cycle is the length of that cycle;
• Disjoint cycles commute.

Therefore, $\beta$ will consist of one $r$-cycle if and only if there are no $a_i$ and $k < r$ such that $\beta^k(a_i) = a_i$. (For, otherwise, $(a_i~\beta(a_i)~\beta^2(a_i)~\beta^{k-1}(a_i))$ would be a subcycle of the permutation $\beta$. Conversely, any subcycle is of that form.)

Aiming for a contradiction, suppose $\beta^k(a_i) = a_i$ for some $0 < k < r$ and $i$.

By definition of $\beta$ and $r = {\rm ord}(\alpha)$, we know that$\beta^k = \alpha^{k\frac{r+1}2}$ leaves some $a_i$ fixed if and only if $r \mid k \frac{r+1}2$ (since $\alpha$ is a cycle).

But $r$ and $\frac{r+1}2$ are coprime. For, suppose that, for some prime $p$, $p \mid r$ and $p \mid \frac{r+1}2$. Then $p \mid 2(\frac{r+1}2)-r = 1$, a contradiction.

It therefore follows that $r \mid k$. But we assumed $0<k < r$, yielding yet another contradiction. Therefore, ${\rm ord}(\beta) = r$, and, being a power of $\alpha$, it must be an $r$-cycle, as desired.