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.
Group Theory – Elementary Isomorphism Between $\operatorname{PSL}(2,5)$ and $A_5$
exceptional-isomorphismsfinite-groupsgroup-theory
Best Answer
One way to see this, which can be found in Galois's letter to Chevalier that he wrote on the night before his death, is that $G = PSL(2,5)$ contains a maximal subgroup $H$ of index $5$. The action of $G$ on $G/H$ is faithful (because $G$ is simple), and so we get an embedding of $G$ into $S_5$. Since $S_5$ doesn't contain many subgroups of order 60, we are done.
Galois more generally considers the action of $PSL(2,p)$ on the fibres of the modular curve $X_0(p)$ over $X_0(1)$ (which have $p+1$ points generically), and from this point of view sees that $PSL(2,p)$ can appear as the Galios group of a degree $p+1$ equation (the equation cutting out the fibre over a typical $j$-invariant in $X_0(1)$). He asks whether we can replace this degree $p+1$ equation by a degree $p$ one, and observes that this is possible for $p = 5,7,11$ (i.e. these are the primes for which $PSL(2,p)$ has an index $p$ subgroup).
It's quite amazing to see just how much Galois understood!