Permutation Powers And Cycles

Every permutation can be written as the composition of disjoint cycles: $\iota = \sigma_1\sigma_2\cdots\sigma_m$ . Furthermore, disjoint cycles commute. Thus, $\iota^k = \sigma_1^k\sigma_2^k\cdots\sigma_m^k$ Note that the product of two disjoint cycles can never be the identity permutation. So if $\iota^k=I$ , then $\sigma_i^k=I$ for all $1\leq i\leq m$ . Using the notation given, the length of $\sigma_i$ is $\ell_i$ . The length of a cycle is also its order; if $\sigma_i^k = I$ then $\ell_i\mid k$ . Thus $k = \mathrm{LCM}(\ell_1,\ell_2,\ldots,\ell_m)$ and the answer is True .

• Bonus #1: It need not be true that $k\geq n$ . For example, if $\iota$ is a transposition of two elements then $k=2$ (for any $n$ ).
• Bonus #2: We already know that $k = \mathrm{LCM}(\ell_1,\ell_2,\ldots,\ell_m)$ is the smallest positive integer for which $\iota^k = I$ . But $\iota^k = \iota^{k-1}\circ\iota = \iota\circ\iota^{k-1}$ . Thus, $p=k-1$ for any $k>1$ (if $k=1$ then $\iota = I$ and no such $p$ exists).