Algebraic argument for why any $A_5$ in $S_6$ can be extended to an $S_5$ in $S_6$

Question

It is well known that $S_5$ is a subgroup of $S_6$ in a way that acts transitively on $6$ points, a surprising fact related to the outer automorphism of $S_6$. One can see this using the icosahedron: its rotational symmetry group is isomorphic to $A_5$, and acts transitively on the $6$ axes of the icosahedron that go through opposite vertices. One can show that this exotic $A_5$ in $S_6$ extends to an $S_5$ in $S_6$ (see this excellent page by John Baez), using some geometric observations.

Is there a purely algebraic way to see that any $A_5$ including into $S_6$ extends to an inclusion of $S_5$ in $S_6$, that avoids working directly with the icosahedron? If such an argument were slick enough, I might prefer it to the way Baez gives (not that I don't love the icosahedron!)

Answer 1

Suppose we're given a transitive action of $A_5$ on a set $X$ of size 6. Then the point stabilizers have order 10, which forces them to be the usual copies of $D_{10}$ in $A_5$; equivalently, these are the normalizers of the six Sylow 5-subgroups. But this implies that $X$ is isomorphic as an $A_5$-set to the set $Syl_5(A_5)$ consisting of Sylow 5-subgroups of $A_5$ under conjugation. Once you know this, it's clear that the $A_5$-action extends to an $S_5$-action: $A_5$ and $S_5$ have the same Sylow 5-subgroups, so $S_5$ also acts on them by conjugation.

Answer 2

$A_5 \cong {\rm PSL}(2,5) < {\rm PGL}(2,5) \cong S_5$.

I suppose that doesn't show that the $A_5$ that is isomorphic to ${\rm PSL}(2,5)$ is the same as (i.e. conjugate to in $S_5$) the $A_5$ that is the symmetry group of the icosahedron, but your could not really prove that without thinking about the icosahedron!

Added: You can avoid thinking about the icosahedron by proving directly that, up to conjugacy in $S_6$, $A_6$ has a unique transitive subgroup isomorphic to $A_5$. Here is a sketch proof.

$A_5$ is generated by $x,y$ with $x$, $y$, and $xy$ having order $2$, $3$ and $5$, respectively.

In the image in $A_6$, we can assume that $x=(1,2)(3,4)$. For $\langle x,y \rangle$ to be transitive, $y$ must consist of two $3$-cycles.

The essentially different possibilities for $y$ (by which I mean up to conjugation in $S_6$) are

(i) $y=(1,3,5)(2,4,6)$, $xy$ has order $3$;

(ii) $y=(1,3,5)(2,6,4)$, $xy$ has order $4$;

(iii) $y=(1,2,3)(4,5,6)$, $xy$ has order $5$.