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Looking for a very fast/"smart" way to compute this number (it was a question asked on an hour-long exam I recently took, so listing everything out for each element in $S_5$ was not an option since I wanted to do the other questions :P). Is there a key observation I am missing that enables us to find this number without much excessive computation, and if so, what is it?

(Note: this is for an introductory class in Abstract Algebra.)

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    $\begingroup$ If $H$ is a subgroup of $G$, then the orbits under right or left multiplication are just the left or right cosets. So the number of orbits is the index of $H$ in $G$. $\endgroup$ – verret Nov 16 '18 at 0:17
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2 orbits. $A_5$ is a subgroup of index 2. Firstly, $A_5$ is an orbit itself, since for any $g,h\in A_5$, there is some $k\in A_5$ such that $kg=h$. The map $A_5\rightarrow S_5$ given by $g\mapsto g.(12)$ gives the other orbit.

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    $\begingroup$ Btw, if you want a "big idea" (that is possibly overkill for this problem) you could use the orbit stabilizer theorem or orbit counting theorem. O-S theorem, applied with $G=A_5$ and $X=S_5$ gives orbit sizes of 60. $\endgroup$ – user25959 Nov 15 '18 at 21:59

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