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

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


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


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 $(12)(25)=8$.

Can anyone correct me please!!!!

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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
up vote 3 down vote accepted

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.

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ohh that means it is a transposition right? – Timoci Lagilevu Feb 6 '13 at 0:48
no, (12)(25) is not a transposition. It is a product of two transpositions. It can also be re-written in cycle form as (125), a single 3-cycle. – Ittay Weiss Feb 6 '13 at 1:39

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 $\sigma$: $$\mathcal{O}_3=\{\sigma^0(3),\sigma^1(3),\sigma^2(3),\ldots\}=\{3\}$$ $$\mathcal{O}_4=\{\sigma^0(4),\sigma^1(4),\sigma^2(4),\ldots\}=\{4\}$$

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thankx alexander – Timoci Lagilevu Feb 6 '13 at 0:46

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