Let $H=\langle (3\quad 4\quad 5),(1\quad 2\quad3)(7\quad8\quad9)\rangle \le S_9$ be a subgroup in $S_9$. Find number of orbits and their order.
First I noticed $\mathrm{orb}(6)=\{6\}$. I also think that $\mathrm{orb}(3)=\{1,2,3,4,5\}$ but I can't explain how (and if so, then also $\mathrm{orb}(7)=\{7,8,9\}$). Another problem is I don't know to explain myself why a permutation send one of $\{1,2,3,4,5\}$ to $\{7,8,9\}$ does not exist. How can I explain both problems?
This is not a complete answer, but rather a hint on what you need to understand. I assume (though you should specify) that the set upon which your $S_9$ is acting is $\{1,2,3,4,5,6,7,8,9\}$. If this is the case, then the permutation $\sigma=(3,4,5)$ takes $3$ to $4$, takes $4$ to $5$, takes $5$ to $3$, and leaves $1,2,6,7,8,9$ fixed. In the same manner you can see what the permutation $\tau=(1,2,3)(7,8,9)$ does to each element of the set $\{1,\dots,9\}$. In order to compute the orbits of this group action you need also to consider composing the two permutations that generate your subgroup, for example, $(\tau\circ\sigma)(5)=1$. Methods for doing that involve Burnside's lemma and the orbit-stabilizer theorem.