Question

Ifis the C -cycle$\left(a_1{a}_2\cdots a_k\right)$ and if$\sigma \in S_n$. prove that$\sigma \tau {\sigma }^{-1}=\left(\sigma \left(a_1\right)\sigma \left(a_2\right)\dots \sigma \left(a_k\right)\right)$.

Short answer

It is proved that $\sigma \tau {\sigma }^{-1}=\left(\sigma \left(a_1\right)\sigma \left(a_2\right)\dots \sigma \left(a_k\right)\right)$

Step-by-step solution

Consider

Assume that$\tau =\left(a_1{a}_2\dots a_k\right)\in S_n$ be a K-cycle and$\sigma \in S_n$ be a permutation.

Prove the result

Consider$c\ne \left\{\sigma \left(a_1\right),\sigma \left(a_2\right),\dots ,\sigma \left(a_k\right)\right\}$and $1⩽i⩽k$, we can say that:

$\begin{array}{rcl}\sigma \tau {\sigma }^{-1}\sigma \left(a_i\right)& =& \sigma \tau \left(a_i\right)\\ & =& \sigma \left(a_{i+1}\mathrm{mod}k\right)\\ \sigma \tau {\sigma }^{-1}\left(c\right)& =& \sigma {\sigma }^{-1}\left(c\right)\\ & =& c\\ & & \end{array}$

This implies that $\sigma \tau {\sigma }^{-1}=\left(\sigma \left(a_1\right)\sigma \left(a_2\right)\dots \sigma \left(a_k\right)\right)$.

Hence,it is proved that$\sigma \tau {\sigma }^{-1}=\left(\sigma \left(a_1\right)\sigma \left(a_2\right)\dots \sigma \left(a_k\right)\right)$