I am trying to figure out if $A_5$ has a subgroup of order $6$. Rather than a yes/no answer I would prefer if someone could show me how they find their way to the answer. Below is my false attempt at a solution. Ideally I could get some tips on how to fix my attempt, but if it seems a dead end alternate solutions are really appreciated:).
I began by counting the number of elements of each cycle type in S5 so that I could deduce the conjugacy classes and class equation for S5. Next I checked centralizers to determine which were conjugacy classes in A5 and I got the class equation for $A_5$ as $60 = 1 + 12 + 12 + 15 + 20$.
Using the orbit-stabilizer theorem, we can deduce from the order of the last conjugacy class that there is a subgroup of order $3$, namely the centralizer of any element in this conjugacy class, for example $C((12345))={e, (12345), (15432)}$. It is easy to then find a subgroup of order $2$ such as $<(12)(34)>= {e, (12)(34)}$. Since the intersection of these two groups is ${e}$, my hope was that $<(12)(34)>(12345)$ would be a subgroup of order $6$, unfortunately not all the elements commute (If the elements of two subgroups $H$,$K$ commute and their intersection is ${e}$ then $HK$ is a subgroup with order equal to the product of the orders) so I'm left at a dead end.
Best Answer
If $\pi\in S_5$ is a permutaion of $\{1,2,3\}$, then either $\pi$ or $\pi\circ (4\,5)$ is an even permutation, i.e., $\in A_5$. This allows us to embed $S_3\to A_5$.
Or have fun with geometry: $A_5$ is the symmetry group of the dodekahedron. It is possible to select $4$ of its $20$ vertices that make up a regular tetrahedron $T_1$. A rotation of the dodekahedron by $72^\circ$ turns $T_1$ into another such tetrahedron $T_2$. Then the subgroup of $A_5$ that fixes $T_1\cup T_2$ turns out to be of isomorphic to $S_3$ (why?)