Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 1 / 10E (Groups, Rings and Modules)

Use the following standard facts: $A_5$ has order $60$ and is a nonabelian simple group; the coset action on the cosets of an index-$j$ subgroup is a transitive homomorphism into $S_j$. If $H<A_5$ has index $j>1$, that action is nontrivial. Its kernel is a normal subgroup, hence is trivial by simplicity. It would therefore embed $A_5$ in $S_j$. For $j=2,3,4$ this is impossible since $j!<60$. There are no subgroups of indices 2, 3 or 4.

An index-5 subgroup has order $12$ by Lagrange's theorem. Consider its natural action on the five letters. It cannot be transitive, since an orbit of size 5 would divide its order by the orbit-stabilizer theorem. If it had no fixed letter, its orbits would have sizes 2 and 3. It would then be a subgroup of the permutations preserving that partition, namely $S_2\times S_3$. Exactly half of those 12 permutations are even, so $(S_2\times S_3)\cap A_5$ has order 6, too small to contain $H$.

Thus $H$ fixes a letter $i$. The stabilizer of $i$ in $A_5$ is the copy of $A_4$ consisting of even permutations of the other four letters; it has order 12. Containment and equality of orders force $H$ to equal that stabilizer. Conversely each such stabilizer has index 5. ConsequentlyThese five subgroups are distinct: for $i\ne j$, a three-cycle on letters other than $i$ can move $j$, so it belongs to the stabilizer of $i$ but not that of $j$.