Exercise 1.3.14 (Elements of order $p$ in $S_n$)

Question

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 needs not be the case if $p$ is not a prime.

Richard Ganaye · Accepted Answer

\begin{proof} 
If $\sigma = 	au_1	au_2\cdots 	au_n$ is a non void product of commuting $p$-cycle, then $\sigma 
e e$ and
$$\sigma^p = 	au_1^p	au_2^p\cdots 	au_n^p =e,$$
then the order $|\sigma|
e 1$ of $\sigma$ divides $p$, so $|\sigma| = p$.

\bigskip
Conversely, let $\sigma = 	au_1	au_2\ldots 	au_n$ be the cycle decomposition of $\sigma$. Then $	au_1, 	au_2,\ldots ,	au_n$ are disjoint cycles and commute. Then
$$\sigma^p = 	au_1^p 	au_2^p\cdots 	au_n^p = e.$$
Since the cycles $	au_1, 	au_2,\ldots ,	au_n$ are disjoint, we obtain
$$	au_1^p = 	au_2^p = \cdots = 	au_n^p = e.$$
We write $	au_1 = (a_1\ a_2 \ \ldots \ a_m)$, where $m \geq 2$. Since $	au_1^p = e$, and $	au_1$ is a $m$-cycle, then $m \mid p$. Since $p$ is prime, and $m\geq 2$, then $m = p$, so $	au_1$ is a $p$-cycle. Similarly, $	au_2,\ldots, 	au_n$ are $p$-cycles.

In conclusion, if $p$ is prime, an element has order $p$ in $S_n$ if and only if its cycle decomposition is a product of commuting $p$-cycles.

\bigskip
If $p$ is not prime, consider the counterexample $	au = (1\ 2)(3 \ 4 \ 5)$.

For every integer $k$,
$$	au^k = e \iff (1\ 2)^k = (3\ 4 \ 5)^k = e \iff 2 \mid k 	ext { and } 3 \mid k \iff 6 \mid k,$$
 so the order of $	au$ is $6$, but the cycle decomposition $	au = (1\ 2)(3 \ 4 \ 5)$ is not a product of $6$-cycles.
\end{proof}

Exercise 1.3.14 (Elements of order $p$ in $S_n$)

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Exercise 1.3.14 (Elements of order $p$ in $S_n$)

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