# number of orbits of $A_5$ acting by left multiplication in $S_5$

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.)

• 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 $G$. – verret Nov 16 '18 at 0:17

## 1 Answer

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.

• 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. – user25959 Nov 15 '18 at 21:59