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

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.)

• Here is an alternative argument: since $\sigma$ has odd order, we have $\langle \sigma \rangle = \langle \sigma^2 \rangle$, so $\langle \sigma^2 \rangle$ acts transitively on $\{1,2,\ldots,n\}$, and hence $\sigma^2$ must be a single cycle. Commented Mar 5, 2023 at 8:12
• Thank you, can we define cycle as: $\sigma\in S_A$, $\sigma$ is a cycle if and only if $\sigma$ is transitive ?@DerekHolt Commented Mar 5, 2023 at 21:54
• Yes, or more precisely if and only if $\langle \sigma \rangle$ is transitive. Commented Mar 6, 2023 at 7:58
• Right, thank you! Commented Mar 7, 2023 at 1:41

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 $$\langle \sigma^2 \rangle=\langle \sigma \rangle$$.
• Yes, this can give a contradiction, but is it possible to show $L\neq n$? where $L=lcm(r_1, r_2,...,r_k)$ (we know $\tau^L=e$) Commented Mar 5, 2023 at 5:50