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I'm trying to understand a proof from Yahoo Answers

http://in.answers.yahoo.com/question/index;_ylt=ArPgiZQnQIXqbb0t61DlLusazKIX;_ylv=3?qid=20120209041852AANkOYp

It does not look like part (a) is done very clearly. Can someone explain it?

Thanks

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    Questions should be self-contained if possible.2012-02-12
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    What does $\mu - 1$ in the statement of (a) indicate?2012-02-12
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    probably the inverse2012-02-12

1 Answers 1

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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 II. If $y\neq \sigma(x_i)$ for all $1\leq i\leq k$, then $\tau(y)=y$.
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 $x_1,..,x_k$)