Question

Let $\mathbf{\sigma }\mathbf{=}\left(a_1{a}_2,,,,,,,,a_k\right)$and $\mathbf{r}\mathbf{=}\left(b_1{b}_2.........b_r\right)$ be disjoint cycles in S_n. Prove that $\mathbf{\sigma \tau }\mathbf{=}\mathbf{\tau \sigma }$[Hint: you must show that$\mathbf{\sigma \tau }$ and$\mathbf{\tau \sigma }$ agree as a function on each i in $\left\{1,2,...n\right\}$. Consider three cases: i is one of the a's, i is one of the b's, i is neither.]

Short answer

It is proved that$\mathrm{\sigma \zeta }=\mathrm{\zeta \sigma }$.

Step-by-step solution

Leveraging hint from the question

We have to show $\mathbf{\sigma \zeta }$and $\mathbf{\zeta \sigma }$agree as a function on each i in $\left\{1,2,.....n\right\}$Consider three cases: i is one of the a's, i is one of the b's, i is neither.

Proving that σζ=ζσ

According to the hint, taking $\mathrm{i}\in \left\{1,2,....\mathrm{n}\right\}$

Now, we must compare $\mathrm{\sigma \zeta }\left(\mathrm{i}\right)=\mathrm{\zeta \sigma }\left(\mathrm{i}\right)$, for different i values.

Suppose $\mathrm{i}={\mathrm{a}}_{\mathrm{i}}\in \left({\mathrm{a}}_1{\mathrm{a}}_2........{\mathrm{a}}_{\mathrm{k}}\right)$

Therefore,

$\mathrm{\sigma }\left({\mathrm{a}}_{\mathrm{i}}\right)=\left\{\begin{array}{cc}{\mathrm{a}}_{\mathrm{j}+1}& \mathrm{i}⩽\mathrm{j}<\mathrm{k}\\ {\mathrm{a}}_1& \mathrm{j}=\mathrm{k}\end{array}\right\}$

So,

$\mathrm{\zeta }\left(\mathrm{\sigma }\left({\mathrm{a}}_{\mathrm{i}}\right)\right)=\mathrm{\zeta }\left(\left\{\begin{array}{cc}{\mathrm{a}}_{\mathrm{j}+1}& 1⩽\mathrm{j}<\mathrm{k}\\ {\mathrm{a}}_1& \mathrm{j}=\mathrm{k}\end{array}\right\}\right)$

Since we know that $\mathrm{\sigma }$and $\mathrm{\zeta }$are two disjoint cycles, so ${\mathrm{a}}_{\mathrm{j}}$cannot be an element in $\mathrm{\zeta }=\left({\mathrm{b}}_1{\mathrm{b}}_2...{\mathrm{b}}_{\mathrm{r}}\right)$.

Hence, for all values of $\mathrm{i}={\mathrm{a}}_{\mathrm{j}}$, $\zeta \left({\mathrm{a}}_{\mathrm{i}}\right)=a_j$

Therefore, from the above equation, we get,

$\mathrm{\zeta }\left(\mathrm{\sigma }\left({\mathrm{a}}_{\mathrm{j}}\right)\right)$=$\left\{\begin{array}{cc}{\mathrm{a}}_{\mathrm{j}+1}& \mathrm{i}⩽\mathrm{j}<\mathrm{k}\\ {\mathrm{a}}_1& \mathrm{j}=\mathrm{k}\end{array}\right\}$

$=\mathrm{\sigma }\left({\mathrm{a}}_{\mathrm{i}}\right)$

Since we know, $\zeta \left({\mathrm{a}}_{\mathrm{j}}\right)=a_j$

$\zeta \left(\mathrm{\sigma }\left({\mathrm{a}}_{\mathrm{j}}\right)\right)=\sigma \left(\zeta \left(a_j\right)\right)$

This implies $\mathrm{\sigma \zeta }=\mathrm{\zeta \sigma }$.

Hence, it is proved that $\mathrm{\sigma \zeta }=\mathrm{\zeta \sigma }$.