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Let

$$\rho=\left(\begin{matrix}1&2&3&4&5&6&7&8&9\\3&2&5&6&1&4&8&9&7\end{matrix}\right)$$

$$\sigma=\left(\begin{matrix}1&2&3&4&5&6&7&8&9\\3&1&2&7&6&5&8&4&9\end{matrix}\right)$$

$$\tau=\left(\begin{matrix}1&2&3&4&5&6&7&8&9\\2&1&3&5&6&7&8&9&4\end{matrix}\right)$$

of the symmetric group $S_9$ of all permutations of the set $S=\{1,2,3,4,5,6,7,8,9\}$.

  1. Find $\rho,\sigma,\tau$ as a prdocut of disjoint cycles and as a product of transpositions.

  2. Show that $\tau\in A_9,$ but $\rho,\sigma\notin A_9$.

  3. Express the product $\rho\sigma^2$ as the product of disjoint cycles.

  4. Show that $\rho$ is conjugate to $\sigma$ but not $\tau$.

  5. Show a permutation $\alpha\in S_9$ such that $\alpha\rho\alpha^{-1}=\sigma$

  1. Disjoint Cycles

$$\rho=(135)(46)(789)$$ $$\sigma=(132)(478)(56)$$ $$\tau=(12)(456789)$$

  1. Transpositions

$$\rho=(15)(13)(46)(79)(78)$$ $$\sigma=(12)(13)(48)(47)(56)$$ $$\tau=(12)(49)(48)(47)(46)(45)$$

  1. The even permutations of the group $S_n$ are elements of the group $A_n$. An even permutation is when there is an even number of transpositions. We can clearly see $\tau$ is the only element of the three defined that is even and in $A_9$.

  2. Multiplying from right to left of

$$\rho\sigma^2=\left((135)(46)(789)\right)\left((132)(478)(56)\right)\left((132)(478)(56)\right)$$

$$\rho\sigma^2=(125)(4976)$$

  1. To be conjugate we require that two elements contain the same number of $2$-cycles, $3$-cycles etc, as disjoint cycles, then these two elements are considered conjugate. We can see $\rho$ and $\sigma$ contain the same number of $2$ and $3$ cycles.

  2. How would I do this?

Have I made any mistakes?

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Hint for question 5.

For a cycle $(a_1 \ a_2 \dots a_m) \in \mathcal S_n$ and $\alpha \in \mathcal S_n$ you have

$$\alpha (a_1 \ a_2 \dots a_m) \alpha^{-1} = (\alpha(a_1) \ \alpha(a_2) \dots \alpha(a_m))$$

Also notice that $\sigma \mapsto \alpha \sigma \alpha^{-1}$ is an (inner) automorphism. Therefore for $\sigma_1, \dots \sigma_p \in \mathcal S_n$ you have

$$\alpha \sigma_1\sigma_2 \dots \sigma_m \alpha^{-1} = \left(\alpha \sigma_1 \alpha^{-1}\right)\left(\alpha \sigma_2 \alpha^{-1}\right) \dots \left(\alpha \sigma_m \alpha^{-1}\right)$$

Now the good thing (!!) is that both the given $\sigma$ and $\rho$ are composed of two $3$-cycles and one $2$-cycle.

With all those ingredients, you should be able to cook the solution!

Note: this is a general result. Two permutations decomposed with the same types of cycles (number of cycles with associated orders) are conjugated.

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