Question

Use the following steps to prove Theorem 7.24: Every permutation$\tau$ in$S_n$ is a product of disjoint cycles.

1. Let be any elements of $\left\{1,2,\dots ,n\right\}$such that $\mathit{\tau }\left(a_1\right)\mathbf{\ne }{\mathit{a}}_{\mathbf{1}}$. Let ${\mathit{a}}_{\mathbf{2}}\mathbf{=}\mathit{\tau }\left(a_1\right)\mathbf{,}{\mathit{a}}_{\mathbf{3}}\mathbf{=}\mathit{\tau }\left(a_2\right)\mathbf{,}{\mathit{a}}_{\mathbf{4}}\mathbf{=}\mathit{\tau }\left(a_3\right)$, and so on. Let K be the first index such that$\mathit{\tau }\left(a_k\right)$ is one of ${\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{k}\mathbf{-}\mathbf{1}}$.Prove that$\mathit{\tau }\left(a_k\right)\mathbf{=}{\mathit{a}}_{\mathbf{1}}$ . Conclude that$\tau$ has the same effect on${\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{k}}$ as the cycles$\left(a_1{a}_2\dots a_k\right)$ .

Short answer

It is proved that$\tau \left(a_k\right)=a_1$ .

Step-by-step solution

Assume

Assume that $\tau \left(a_k\right)=a_i$for some$1<i<k$ , this implies that$\tau \left(a_k\right)=\tau \left(a_{i-1}\right)$ .

Prove the result

Since $a_k=a_{i-1}$but $\tau \left(a_{i-1}\right)=a_i\in \left\{a_1,\dots ,a_{k-1}\right\}$, which contradicts the assumed minimality of K.

Therefore, $\tau \left(a_k\right)=a_1$.

Hence, it is proved that $\tau \left(a_k\right)=a_1$.