I'm trying to understand a proof from Yahoo Answers
It does not look like part (a) is done very clearly. Can someone explain it?
Thanks
I'm trying to understand a proof from Yahoo Answers
It does not look like part (a) is done very clearly. Can someone explain it?
Thanks
What you want to prove is: Let $\mu=(x_1,...,x_k)\in S_n$ be a cycle and let $\sigma\in S_n$ be any permutation then $\tau:=\sigma\mu\sigma^{-1}=(\sigma(x_1),...,\sigma(x_k))$.
What we actually want to show is:
I. For all $1\leq i
For I: compute: $\tau(\sigma(x_i))=\sigma\mu\sigma^{-1}(\sigma(x_i))=\sigma\mu(x_i)=\sigma(x_{i+1})$. Similarly, for $\tau\sigma(x_k)$.
For II: Take any $y\neq \sigma(x_i)$ for all $1\leq i\leq k$. Now,
$$\tau(y)=\sigma\mu\sigma^{-1}(y)\overset{(*)}{=}\sigma\sigma^{-1}(y)=y$$
$(*)$ since $y\neq \sigma(x_i)$, we have $\sigma^{-1}(y)\neq x_i$, and hence $\mu\sigma^{-1}(y)=\sigma^{-1}(y)$ ($\mu$ acts non-rivially only on $x_1,..,x_k$)