Permutations of Prime Order

Question

Here is the problem

Let $p$ be a prime. Show that an element has order $p$ in $S_n$ if and only if its cycle decomposition is a product of commuting $p$-cycles. Show by an explicit example that this need not be the case of p is not prime.

I was able to show the "only if" part easily enough. But I am having difficulty with the "if" direction. I suppose that $\sigma \in S_n$ has order $p$ yet $\sigma$ is not just the product of commuting $p$-cycles, which would mean that an $n$-cycle, with $n \neq p$, would appear in the product. But this is where I was unsure of how to proceed, so I consulted this. However, the author claims that "the cycle decomposition of $\sigma$ contains an $n$-cycle $\tau$ for some $n \neq p$, and we can write $\sigma = \tau \omega$." Why does the $n$-cycle $\tau$ appear at the right-end of $\sigma$? Why couldn't it be at the left-end, or even in the interior?

Answer 1

The cycles in a cycle decomposition of $\sigma$ (not just any product of cycles) are disjoint, hence they commute. So it doesn't matter if it's in the interior; you can always move it to wherever you need it to be.

For example: (123)(45)(678) = (123)(678)(45).

Answer 2

The order of a permutation is the least common multiple of the lengths of cycles in the decomposition, so every of these lengths divides the order. In particular, if the order is a prime number $p$, then every cycle has length $1$ or $p$.

As to the opposite direction, if $n$ is composite, then you can write $n=pq$ with $(p,q)=1$ and $p,q\neq1$. We see that the product of $(1\ldots p)$ and $(p+1\ldots p+q)$ has order $pq=n$.