Exercise 2.3.14 (Powers of a cycle)

Question

Let $\sigma = (1\ 2 \ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12)$. For each of the following integers $a$ compute $\sigma^a$:

$a= 13,\ 65,\ 626,\ 1195,\ -6,\ -81,\ -570$ and $-1211$.

Richard Ganaye · Accepted Answer

\begin{proof} 
$\sigma$ is a cycle of order $12$, so $|\sigma| = 12$. Then for all integers $a$, 
$$\sigma^a = \sigma^r,\qquad	ext{where } a = 12 q + r,\ 0\leq r < 12.$$
The remainders are
\begin{center}
\begin{tabular}{|c|cccccccc|}
\hline
 $a$ &13& 65& 626& 1195& -6& -81& -570 & -1211\
\hline
 $r$ &1& 5& 2& 7& 6& 3& 6& 1 \
 $\sigma^a$ &$\sigma^{1}$& $\sigma^{5}$& $\sigma^{2}$& $\sigma^{7}$& $\sigma^{6}$& $\sigma^{3}$& $\sigma^{6}$& $\sigma^{1}$\
\hline
\end{tabular}
\end{center}
So 
\begin{align*}
\sigma^{13} = \sigma &= (1\ 2 \ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12),\
\sigma^{65} = \sigma^5&=(1\ 6\ 11\ 4\ 9\ 2\ 7\ 12\ 5\ 10\ 3\ 8),\
\sigma^{626} = \sigma^2 &= (1\ 3\ 5\ 7\ 9\ 11)(2\ 4\ 6\ 8\ 10\ 12),\
\sigma^{1195} = \sigma^7 &= (1\ 8\ 3\ 10\ 5\ 12\ 7\ 2\ 9\ 4\ 11\ 6),\
\sigma^{-6} = \sigma^6 &= (1\ 7)(2\ 8)(3\ 9)(4\ 10)(5\ 11)(6\ 12),\
\sigma^{-81} = \sigma^3 &= (1\ 4\ 7\ 10)(2\ 5\ 8\ 11)(3\ 6\ 9\ 12),\
\sigma^{-570} = \sigma^6 &= (1\ 7)(2\ 8)(3\ 9)(4\ 10)(5\ 11)(6\ 12),\
\sigma^{-1211} = \sigma &=  (1\ 2 \ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12).
\end{align*}
\end{proof}

With Sagemath:
\begin{verbatim}
sage: G = SymmetricGroup(12)
sage: s = G([(1,2,3,4,5,6,7,8,9,10,11,12)])
sage: s^5
(1,6,11,4,9,2,7,12,5,10,3,8)
...
\end{verbatim}

Exercise 2.3.14 (Powers of a cycle)

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$a$	13	65	626	1195	-6	-81	-570	-1211

$r$	1	5	2	7	6	3	6	1
$σ^{a}$	$σ^{1}$	$σ^{5}$	$σ^{2}$	$σ^{7}$	$σ^{6}$	$σ^{3}$	$σ^{6}$	$σ^{1}$

Exercise 2.3.14 (Powers of a cycle)

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