Question

Use the following steps to prove Theorem 7.24: Every permutation $\mathit{\tau }$in ${\mathit{S}}_{\mathbf{n}}$is a product of disjoint cycles.

(b) Let${\mathit{b}}_{\mathit{1}}$ be any elements of$\left\{1,2,\dots ,n\right\}$ other that${\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{k}}$ that is not mapped to itself by $\tau$. Let ${\mathit{b}}_{\mathbf{2}}\mathbf{=}\mathit{\tau }\left(b_1\right)\mathbf{,}{\mathit{b}}_{\mathbf{3}}\mathbf{=}\mathit{\tau }\left(b_2\right)$, and so on. Show that$\mathit{\tau }\left(b_i\right)$ is never one of ${\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{k}}$. Repeat the argument in part (a) to find$\mathit{a}{\mathit{b}}_{\mathbf{r}}$ such that$\mathit{\tau }\left(b_r\right)\mathbf{=}{\mathit{b}}_{\mathbf{1}}$ and $\mathit{\tau }$agrees with the cycles $\left(b_1{b}_2\dots b_r\right)$on the $\mathit{b}\mathbf{\text{'}}\mathit{s}$.

Short answer

It is proved that $\tau \left(b_i\right)$is never one of $a_1,\cdots ,a_k$.

Step-by-step solution

Assume

Assume that $\tau \left(b_i\right)\in \left\{a_1,a_2,...,a_k\right\}$for some $a_j$, this implies that

$\begin{array}{rcl}{b}_1& =& {\tau }^{-i+1}\left(b_i\right)\\ & =& {\tau }^{-i}\left(a_j\right)\\ & =& a_{j-i\mathrm{mod}k}\\ & & \end{array}$

which is a contraction since $b_1$is assumed to not be contained in$\left\{a_1,...,a_k\right\}$

Prove the result

Repeating the argument in part (a), we get a disjoint cycle $\left(b_1...b_r\right)$that agrees with $\tau$on the$b_i$ . Hence, it is proved that$\tau \left(b_i\right)$ is never one of $a_1,\cdots ,a_k$