Group theory: How to show this cycle won't break, $r_1+r_2+...+r_k=n$, with the least common multiple $lcm(r_1, r_2, ..., r_k)=n$

Question

If $\sigma=(1,2,3...,n)$ is a cycle, $n$ is odd and $n\geq 3$, is it possible that $\sigma^2=(x_1,...x_{r_1})(y_1,..,y_{r_2})...(t_1,..t_{r_k})$, which means the cycle is splitted into $k$ parts, where the numbers inside each $(...)$ are distinct, each $(...)$ has $r_1, r_2, ..., r_k$ elements, $r_1+r_2+...+r_k=n$, is it possible the least common multiple $lcm(r_1, r_2, ..., r_k)=n$?

Attempt: I want to show $\sigma^2$ is still a cycle (won't split into parts). It is easy to do a computation to show $\sigma^2=(1,3,5,...,n,2,4,6,...,n-1)$, so it is a cycle. But I am wondering if I can prove this by the method below.

I consider the cyclic group generated by $\sigma$, $|\langle \sigma \rangle|=n$, and $\langle \sigma^2 \rangle=\langle \sigma^d \rangle$, where $d=\gcd(2, n)=1$, so we have $\langle \sigma^2 \rangle=\langle \sigma \rangle$. But this is not enough to show $\sigma^2$ is still a cycle, because it might happen as described above, if the least common multiple $lcm(r_1, r_2, ..., r_k)=n$, we still have $|\langle \sigma^2 \rangle|=n$, but the cycle breaks into parts.

For example, it is easy to show $k$ can't be an even integer, otherwise, since $r_1+r_2+...+r_k=n$, at least one of $r_1, r_2, ..., r_k$ will be even, then the least common multiple $lcm(r_1, r_2, ..., r_k)$ is even, which contradicts with $lcm(r_1, r_2, ..., r_k)=n$ where $n$ is odd. But how to exclude $k=3,5,7...$ (Only $k=1$ is the desired result.)

Answer 1

On the supposition that $\tau=\sigma^2$ is broken up into $k$ separate cycles, think about what a power of $\tau$, as a function, would do to the elements $x_i$. Repeatedly applying the function $\tau^j$ to $x_1$ would always return some element $x_i$ and never any element $y_j$. Then you would have a contradiction, since $\sigma$ is a power of $\tau$, as shown by $\langle \sigma^2 \rangle=\langle \sigma \rangle$.