Exercise 9.3.4 (Order of permutation cycle)

Question

Explain why, in the last paragraph, $\sigma^r$ has order $p$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
By definition we know that the order of a permutation $ho$ is the least positive integer $r$ such that $ho^r$ is the identity permuation.
Hence for our cycle $\sigma = (a_0 \ a_1 \ ... \ a_{p-1})$ its order can never be less than $p$ since for $r \in \{1,...,p-1\}$ $\sigma^r(a_0) = a_r$.
Similarly, $\sigma^p = e$ (the identity permutation).
Hence since $\sigma^p = e$ and $\sigma^r 
eq e$ for $r \in\{1,...,p-1\}$, then $\sigma^r = p$.

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

\begin{proof} 
Let $k$ be any integer, and assume that $(\sigma^r)^k = \mathrm{id}$. Then $\sigma^{rk} =   \mathrm{id}$. Since $\sigma$ has order $p$, $p \mid rk$. Here $1 \leq r \leq p-1$, where $p$ is prime, so that $\mathrm{gcd}(r,p) = 1$. Therefore $p \mid k$. Conversely, if $p \mid k$, then $p \mid rk$, thus $(\sigma^r)^k = \mathrm{id}$:
$$\forall k \in \Z,\ (\sigma^r)^k = \mathrm{id} \iff p \mid rk \iff p \mid k.$$
This shows that $\sigma^r$ has order $p$.
\end{proof}

Exercise 9.3.4 (Order of permutation cycle)

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Exercise 9.3.4 (Order of permutation cycle)

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