Exercise 1.3.1 (Cycle decompositions 1)

Question

Let $\sigma$ be the permuation
$$1\mapsto 3 \qquad 2 \mapsto 4\qquad 3 \mapsto 5 \qquad 4 \mapsto 2 \qquad 5 \mapsto 1$$
and let $	au$ be the permutation
$$1\mapsto 5 \qquad 2 \mapsto 3\qquad 3 \mapsto 2 \qquad 4 \mapsto 4 \qquad 5 \mapsto 1$$
Find the cycle decompositions of each of the following permutations: $\sigma,\ 	au,\ \sigma^2, \sigma 	au, 	au \sigma$, and $	au^2 \sigma$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Following Cauchy, we write
\begin{align*}
\sigma &= \begin{pmatrix} 1 & 2 & 3 & 4 & 5\ 3 & 4 & 5 & 2 & 1\end{pmatrix},\
	au &= \begin{pmatrix} 1 & 2 & 3 & 4 & 5\ 5 & 3 & 2 & 4 & 1\end{pmatrix}.
\end{align*}
Using the Cycle Decomposition Algorithm, we obtain
\begin{align*}
\sigma &= (1 \ 3 \ 5 ) (2 4),\
	au &= (1\ 5)(2\ 3).
\end{align*}
Then (the composition being from right to left),
\begin{align*}
\sigma^2 &=  \begin{pmatrix} 1 & 2 & 3 & 4 & 5\ 5 & 2 & 1 & 4 & 3\end{pmatrix} = (1\ 5\ 3 ),\
\sigma 	au &= \begin{pmatrix} 1 & 2 & 3 & 4 & 5\ 1 & 5 & 4 & 2 & 3 \end{pmatrix} = (2 \ 5 \ 3 \ 4),\
	au \sigma &= \begin{pmatrix} 1 & 2 & 3 & 4 & 5\ 2 & 4 & 1 & 3 & 5\end{pmatrix} = (1\ 2 \ 4 \ 3 ),\
\end{align*}
Since $	au$ is the commutative product of $(1\ 5)$ and $(2 \ 3)$, $	au^2 = (1\ 5)^2(2\ 3)^2 = e$, so 
$$	au^2 \sigma = \sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5\ 3 & 4 & 5 & 2 & 1\end{pmatrix}.$$
(Similarly, as a verification, $\sigma^2 = (1 \ 3 \ 5 )^2 (2 4)^2 = (1 \ 3 \ 5 )^2 = (1\ 5 \ 3)$.)
\end{proof}

Exercise 1.3.1 (Cycle decompositions 1)

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Exercise 1.3.1 (Cycle decompositions 1)

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