transitive action from $A_5$ on a set of $6$ elements

Question

Can we find a transitive action from the alternating group $A_5$ on the set $X$ with 6 elements?

I think we can't because the $|X|=6$ which greater than $5$. It also because $A_4$ does not have a 6-cycle.

Answer 1

$A_5$ acts on its $6$ $5$-Sylow subgroups by conjugation. This action is transitive by the Sylow theorems.

Answer 2

Background: There is an exceptional outer automorphism $\phi$ of $S_6$. If we apply it to the usual copy of $A_5\subset S_6$, we end up with an exotic embedding $A_5\to S_6$ which is a transitive action.

One way to see the action geometrically is to view $A_5$ as the rotational symmetry group of the icosahedron. One can show this symmetry group, traditionally denoted $I$, has size

$$ |G|=12\cdot5=30\cdot2=20\cdot3 $$

by applying the orbit-stabilizer theorem with vertices, edges, or faces. There are also $30/(3\cdot2)=5$ compounds of three perpendicular golden rectangles, since there are $3\cdot2=6$ edges per compound and every edge uniquely determines a compound. Example of a compound:

This automatically means $I$ embeds in $S_5$ as an index $2$ subgroup, which must be none other than $A_5$. Then $I\cong A_5$ acts transitively on the $12/2=6$ antipodal pairs of opposite vertices.

Algebraically, we can see it as the action of $A_5$ on its six $5$-Sylow subgroups, as Lukas notes.