## Characterization of elements of prime order in a symmetric group

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.

$(\Leftarrow)$ If $\sigma \in S_n$ is a product of commuting $p$-cycles, then $|\sigma| = p$ by the lemma to the previous exercise.

$(\Rightarrow)$ Let $\sigma \in S_n$ be an element of order $p$, and suppose that $\sigma$ is not aa product of commuting $p$-cycles. Then the cycle decomposition of $\sigma$ contains an $n$-cycle $\tau$ for some $n \neq p$, and we can write $\sigma = \tau \omega$ for some $\omega \in S_n$. Note that if $n > p$, then $|\tau| = n > p$ by a previous exercise. Then, also by a previous exercise, $\sigma^p = \tau^p \omega^p \neq 1$, a contradiction. Now if $1 < n < p$, then $\mathsf{gcd}(n,p) = 1$. So by a previous theorem, $\tau^p$ is an $n$-cycle, thus not 1. So $\sigma^p = \tau^p \omega^p \neq 1$, also a contradiction. So no such $\tau$ exists in the decomposition of $\sigma$, so that $\sigma$ is a product of commuting $p$-cycles.

Consider $\sigma = (1\ 2\ 3)(4\ 5)$. Now $\sigma$ is an element of order 6 in $S_n$ for $n \geq 5$, but has cycle shape $(3,2)$. Thus the result does not hold (for $n \geq 5$) if $p$ is not prime.

The result holds vacuously in $S_1$, $S_2$, and $S_3$, since the possible cycle shapes in these groups are $(1)$, $(2)$, and $(3)$. (There are no elements of composite order.)

In $S_4$, the result holds for composite $p$ in an uninteresting way: the cycle shapes in $S_4$ are $(1)$, $(2)$, $(2,2)$, $(3)$, and $(4)$. The only elements of composite order have cycle shape $(4)$, and are thus length 1 products of commuting 4-cycles.