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.
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1$\begingroup$ 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. $\endgroup$– Arturo MagidinApr 6, 2021 at 20:14
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$\begingroup$ Hint: $A_5$ has a subgroup of index $6$. $\endgroup$– Josh B.Apr 7, 2021 at 0:33
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2 Answers
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$A_5$ acts on its $6$ $5$-Sylow subgroups by conjugation. This action is transitive by the Sylow theorems.
answered Apr 6, 2021 at 19:04
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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.
