# Finding the number of orbits

How many orbits are there of $(12)(25)$ in $S_{5}$?

Considering the permutation $(12)$, it has $4$ orbits and is as follows:

$\{\{1,2\},\{3\},\{4\},\{5\}\}$

and (25) also has 4 orbits and is also listed below:

$\{\{1\},\{2,5\},\{3\},\{4\}\}$.

To find the number of orbits of $(12)(25)$.

Need to add the number of orbits of $(12)$ which is $4$ together with the number of orbits of $(25)$ and is also $4$.

That is, $4+4=8$.

Therefore the number of orbits of $(12)(25)=8$.

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You mean orbits of $(12)(25)$ under some action of $S_5$ on itself, or the orbit of some $n\in \{1,\ldots,5\}$ under the action of $\langle (12)(25) \rangle$? – Alexander Gruber Feb 6 '13 at 0:01
@Timoci Lagilavu, I edited your post to have the right latex codes. You might want to have a look for future posts. – Ittay Weiss Feb 6 '13 at 0:05
thankx alot ittay weiss – Timoci Lagilevu Feb 6 '13 at 0:46

You can't compute the number of orbits of each cycle independently and then just add them. It doesn't work that way. You need to compute the orbits of the permutation $(12)(25)$. The orbits are $\{\{1,2,5\},\{3\}, \{4\}\}$ so there are three orbits.
Just compute the orbits for that permutation. Letting $\sigma=(12)(25)$, $$\mathcal{O}_1=\{\sigma^0(1),\sigma^1(1),\sigma^2(1),\ldots\}=\{1,2,5\}$$ Since sharing an orbit is an equivalence relation, we then know that $\mathcal{O}_1=\mathcal{O}_2=\mathcal{O}_5$, so now all we need are $\mathcal{O}_3$ and $\mathcal{O}_4$. But these are easy, because they are obviously fixed by $\sigma$: $$\mathcal{O}_3=\{\sigma^0(3),\sigma^1(3),\sigma^2(3),\ldots\}=\{3\}$$ $$\mathcal{O}_4=\{\sigma^0(4),\sigma^1(4),\sigma^2(4),\ldots\}=\{4\}$$