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

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.

• Note that you don't need a $6$-cycle to act transitively on a six element set. By analogy, note that the Klein $4$-group in $A_4$ acts transitively on $\{1,2,3,4\}$, without containing a $4$-cycle. Apr 6, 2021 at 20:14
• Hint: $A_5$ has a subgroup of index $6$. Apr 7, 2021 at 0:33

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

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.