Question

Let alpha = (a1, a2, a3,...,ae) be a cycle of length L. Prove that alpha^2 is a cycle if and only if L is odd.

Answer 1

Question 1. Let $\alpha = (a_{1}, a_{2}, a_{3}, \ldots, a_{L})$ be a cycle of length $L$. Prove that $\alpha^{2}$ is a cycle if and only if $L$ is odd.

Solution. Suppose $L = 2n$, $n \in \mathbb{N}$. Then

$$a_{1} \xrightarrow{\alpha^{2}} a_{3} \xrightarrow{\alpha^{2}} a_{5} \xrightarrow{\alpha^{2}} \ldots \xrightarrow{\alpha^{2}} a_{2n-3} \xrightarrow{\alpha^{2}} a_{2n-1} \xrightarrow{\alpha^{2}} a_{1} \xrightarrow{\alpha^{2}} \ldots$$ $$a_{2} \xrightarrow{\alpha^{2}} a_{4} \xrightarrow{\alpha^{2}} a_{6} \xrightarrow{\alpha^{2}} \ldots \xrightarrow{\alpha^{2}} a_{2n-2} \xrightarrow{\alpha^{2}} a_{2n} \xrightarrow{\alpha^{2}} a_{2} \xrightarrow{\alpha^{2}} \ldots$$

Therefore, $\alpha^2$ is the product of two independent cycles:

$$\alpha^{2} = (a_{1}, a_{3}, \ldots, a_{2n-1})(a_{2}, a_{4}, \ldots, a_{2n}).$$

Now suppose $L = 2n - 1$, $n \in \mathbb{N}$. Then

$$a_{1} \xrightarrow{\alpha^{2}} a_{3} \xrightarrow{\alpha^{2}} \ldots \xrightarrow{\alpha^{2}} a_{2n-1} \xrightarrow{\alpha^{2}} a_{2} \xrightarrow{\alpha^{2}} a_{4} \xrightarrow{\alpha^{2}} \ldots \xrightarrow{\alpha^{2}} a_{2n-2} \xrightarrow{\alpha^{2}} a_{1} \xrightarrow{\alpha^{2}} \ldots$$

So, $\alpha^2$ is the cycle

$$\alpha^{2} = (a_{1}, a_{3}, \ldots, a_{2n-1}, a_{2}, a_{4}, \ldots, a_{2n-2}).$$

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