Question: > 
> **Theorem.** If $x$ is a cycle of odd length then $x^2$ is also a cycle.
> 
> 
> 

*This is the main idea behind the proof.* 

When you look at $(a\_1a\_2a\_3...a\_n)(a\_1a\_2a\_3...a\_n)$, you can see that $a\_1$ goes to $a\_3$, $a\_3$ to $a\_5$, etc... continuing to skip every other element, and because there's an odd number of elements, this skipping goes through every element.

*This is how I began formalizing the previous idea using generators.*

$\langle x \rangle = \mathbb{Z}\_{|x|}$. Because $|x|$ is odd $x^2$ generates $\langle x \rangle$. Therefore...

...but I don't know how to translate the idea of skipping.

1. How do I complete this proof?
2. Which parts of the correct completion corresponds to skipping?

Answer: 
Here's the most elementary way I can think of doing this. 

If the cycle $x$ has order $n$, then we may think of $x$ as the operation $k\mapsto k+1$ on $\mathbb Z/n\mathbb Z$ (we are considering a faithful, transitive action of the subgroup $\langle x\rangle \subset S\_m$, modelled as $\mathbb Z/n\mathbb Z$ acting on itself by translation). Thus $x^2$ corresponds to the operation $k\mapsto k+2$. In this formulation, $x^2$ is a cycle if and only if the action of $\langle x^2\rangle $ is transitive, meaning that we can get between any two elements of $\mathbb Z/n\mathbb Z$ by repeatedly adding $2$. But this is the same as saying that any $k\in \mathbb Z/n\mathbb Z$ is of the form $k=2l$ for some $l\in \mathbb Z/n\mathbb Z$ (meaning we can get from $0$ to $k$ by adding $2$ a total of $l$ times). In other words, we see that $x^2$ is a cycle $\Longleftrightarrow$ $2$ generates $\mathbb Z/n\mathbb Z$ $\Longleftrightarrow$ $\gcd(2,n)=1$.

More generally, this line of reasoning shows that if $x$ is an $n$-cycle, then $x^k$ is a cycle if and only if $\gcd(k,n)=1$. Or even more generally, if $x$ is an $n$-cycle and $g=\gcd(n,k)$, then $x^k$ is the product of $g$ cycles, eaching having length $n/g$ (this uses something about quotients of cyclic groups). 

I would imagine that you can also give an argument using Burnside's lemma.
Let $x$ be a cycle of odd length; say it is an element of $S\_N$ for some $N$, and let $X$ be the support of $x$; that is,
$$ X = \mathrm{supp}(x) = \{i\mid 1\leq i\leq N\text{ and }x(i)\neq i\}$$
the elements of $\{1,\ldots,N\}$ that are "moved" by $x$. 

Let $G=\langle x\rangle$. Then $G$ is a group of order the length of $x$, that is, $|G|$ is odd. Also, $G$ acts on $X$, by letting $h\in G$ act as $h\cdot a = h(a)$ (note that $h$ is a permutation of $X$).

The **Orbit-Stabilizer Theorem** says that if $X$ is a set, $G$ is a group acting on $X$, then for every $a\in X$, there is a bijection between the $G$-orbit of $a$ and the cosets of the stabilizer of $a$ in $G$; that is, if we let 
$$\begin{align\*}
G\cdot a &= \{ g\cdot a\mid g\in G\}\\
G\_a &= \{g\in G\mid g(a)=a\},
\end{align\*}$$
then
$$|G\cdot a | = [G:G\_a]$$
in the sense of cardinality. The bijection is given by taking a coset $gG\_a$ and sending it to the element $g\cdot a$ of $G\cdot a$. It is not hard to verify this is well defined and gives the desired bijection.

Note that for $G$, we have that $G\cdot a = X$ for all $a\in X$, since $x$ is a cycle. So $[G:G\_a] = |X| = |G|$ for all $a\in X$.

Now let $a\in X$, and we want to verify that the orbit of $a$ under $\langle x^2\rangle$ is all of $X$. Because $x$ has odd length, $|x|$ is odd, so $|x^2| = |x|/\gcd(2,|x|) = |x|$, hence $\langle x^2\rangle = \langle x\rangle= G$; so we have:
$$|X|=[G:G\_a] = [\langle x^2\rangle:\langle x^2\rangle\_a] = \langle x^2\rangle\cdot a \leq |X|.$$
Thus, we have equality throughout, and in particular $\langle x^2\rangle\cdot a = X$. That is, the orbit of $a$ is all of $X$, so the action of $x^2$ is a cycle.