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I'm trying to understand a proof from Yahoo Answers of the following.

(a) Show that if $μ = (x_1, x_2, \dots, x_k) ∈ S_n$ is a $k$-cycle and $\sigma \in S_n$ is any permutation, then $\sigma μ \sigma^{−1}$ is the $k$-cycle $\sigma \mu \sigma^{−1} = (\sigma(x_1),\sigma(x_2), \dots ,\sigma(x_k))$.

(b) Using the above, find a necessary and sufficient condition for two permutations in $S_n$ to be conjugate to each other.

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

Thanks

  • 0
    probably the inverse2012-02-12

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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, $\tau(\sigma(x_i))=\sigma(x_{i+1})$ and $\tau(\sigma(x_k))=\sigma(x_1)$.
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-trivially only on $x_1,..,x_k$)