Exercise 1.3.15 (Order of an element in $S_n$)

Question

Prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its decomposition. [Use Exercise 10 and Exercise 24 of Section 1.]

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $\sigma$ be a permutation in $S_n$, and $\sigma = 	au_1	au_2 \cdots 	au_m$ its cycle decomposition. Since the cycles of this decomposition commute, by Exercise 1.24, we obtain, for all integers $k$, $\sigma^k = 	au_1^k 	au_2^k\cdots 	au_m^k$. Let $l_i$ denote the length of $	au_i$. By Exercise 10, $|	au_i| = l_i$. Since the $	au_i$ are disjoint cycles, for all integers $k$,
\begin{align*}
\sigma^k = e &\iff  	au_1^k 	au_2^k\cdots 	au_m^k = e\
&\iff 	au_1^k = 	au_2^k= \cdots  =	au_m^k = e\
&\iff l_1 \mid k,\ l_2 \mid k,\ \ldots,\ l_l \mid k\
&\iff \mathrm{l.c.m}(l_1,l_2,\ldots,l_m) \mid k,
\end{align*}
by definition of the l.c.m.

Therefore
$$|\sigma| = \mathrm{l.c.m}(l_1,l_2,\ldots,l_m).$$
So the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its decomposition.
\end{proof}

Exercise 1.3.15 (Order of an element in $S_n$)

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Exercise 1.3.15 (Order of an element in $S_n$)

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