TITLE: Elementary isomorphism between $\operatorname{PSL}(2,5)$ and $A_5$
QUESTION [18 upvotes]: At this Wikipedia page it is claimed that to construct an isomorphism between $\operatorname{PSL}(2,5)$ and $A_5$, "one needs to consider" $\operatorname{PSL}(2,5)$ as a Galois group of a Galois cover of modular curves and consider the action on the twelve ramified points.  While this is a beautiful construction, I wonder if this really is necessary.  Is there a construction of a map that takes a representative matrix of a class in $\operatorname{PSL}(2,5)$ and uses some relatively simple computation to produce a permutation in $S_5$ that can be shown to be even?  I don't mind if describing the construction and providing the verification that it is well-defined and does what it should takes several pages.  I'd just like to think that it's possible.

REPLY [6 votes]: Yet another way: I will show that both of them are isomorphic to the group of rotational symmetries of the dodecahedron. Let $H$ be the rotational symmetry group of the dodecahedron. 
$H$ embeds in $SO(3)$, using entries in $\mathbb{Z}[\tau]$ where $\tau$ is the golden ratio $\frac{1 + \sqrt{5}}{2}$.  Since these are orthogonal matrices, they preserve the quadratic form $x^2+y^2+z^2$.
First, consider the ideal $\langle \sqrt{5} \rangle = \langle 2 \tau-1 \rangle$ in $\mathbb{Z}[\tau]$. Reducing our matrices modulo this ideal, we get matrices in $SL_3(\mathbb{F}_5)$ preserving the quadratic form $x^2+y^2+z^2$. The conic $x^2+y^2+z^2=0$ (in homogenous coordinates) has points over $\mathbb{F}_5$, namely, the $6$ permutations of $(0:1:2)$. So the matrices act on this conic by matrices in $PGL_2(\mathbb{F}_5)$. This gives a map $H \to PGL_2(\mathbb{F}_5)$, and it is not bad to show it is an isomorphism $H \to PSL_2(\mathbb{F}_5)$.
Now, consider the ideal $\langle 2 \rangle$ in $\mathbb{Z}[\tau]$. The quotient is the field with $4$-elements, so we get a representation of $H$ by matrices in $SL_3(\mathbb{F}_4)$ preserving $x^2+y^2+z^2 = (x+y+z)^2$. Thus, these matrices preserve the plane $x+y+z=0$. We get an action on the set of $5$ lines in this plane. This gives a map $H \to S_5$, and it isn't hard to see that it is an ismorphism $H \to A_5$. Remark: We have also just shown $H \cong PGL_2(\mathbb{F}_4)$ (which is the same as $PSL_2( \mathbb{F}_4)$).
Nothing interesting seems to happen if we reduce modulo $\langle 3 \rangle$, I'm sorry to say. We get an action of $H$ on $\mathbb{P}^1(\mathbb{F}_9)$, but I can't find much to say about it.