Exercise 1.3.5 (Order of $(1\ 12 \ 8 \ 10 \ 4)(2\ 13)(5\ 11 \ 7)(6\ 9)$)

Question

Find the order of $(1\ 12 \ 8 \ 10 \ 4)(2\ 13)(5\ 11 \ 7)(6\ 9)$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Put $\sigma = (1\ 12 \ 8 \ 10 \ 4)(2\ 13)(5\ 11 \ 7)(6\ 9)$.

Since these four cycles of the cycle decomposition of $\sigma$ are disjoint, for all $k \in \mathbb{Z}$,
\begin{align*}
\sigma^k = e & \iff (1\ 12 \ 8 \ 10 \ 4)^k = (2\ 13)^k = (5\ 11 \ 7)^k = (6\ 9)^k = e\
&\iff k\mid 5 	ext{ and } k\mid 2 	ext{ and }k\mid 3 	ext{ and }k\mid 2\
&\iff k \mid 30. 
\end{align*}
Therefore the order of $\sigma$ is $30$:
$$|\sigma| = 30.$$
\end{proof}

Exercise 1.3.5 (Order of $(1\ 12 \ 8 \ 10 \ 4)(2\ 13)(5\ 11 \ 7)(6\ 9)$)

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Exercise 1.3.5 (Order of $(1\ 12 \ 8 \ 10 \ 4)(2\ 13)(5\ 11 \ 7)(6\ 9)$)

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