# Show cycle permutations meets certain properties

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?

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.