Elementary isomorphism between $\operatorname{PSL}(2,5)$ and $A_5$

Question

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.

Answer 1

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.

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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!

Answer 2

The order of $G=PSL(2,5)$ is 60. A Sylow 2-subgroup $P$ is generated by the cosets represented by $$A=\pmatrix{2&0\cr0&3\cr}\quad\text{and}\quad B=\pmatrix{0&2\cr3&0\cr}.$$ It is easy to check that $P$ is normalized by
$$ C=\pmatrix{1&2\cr1&3\cr} $$ (conjugation by $C$ cycles the involutions in a 3-cycle $B\mapsto A\mapsto AB=BA \mapsto B$). The order of $C$ is three. As $P$ is not normal in $G$, we can deduce that $N(P)$ has order 12. Thus there are 5 Sylow 2-subgroups. Therefore we get a homomorphism from $G$ to $S_5$ from the conjugation action of $G$ on the 2-Sylow subgroups. If you believe that $G$ is simple, then rest follows as in Geoff Robinson's comment. Even without using that fact at this point it would suffice to prove that this homomorphism is injective. Or equivalently that the intersection of the conjugates of $N(P)$ is trivial.

The other Sylow 2-subgroups can be gotten by conjugating $P$ with powers of $$ D=\pmatrix{1&1\cr0&1\cr}. $$

Answer 3

I know two good descriptions, which I'll put in two answers. Here is the algebraic one:

The group $PGL(2,5)$ acts on $\mathbb{P}^1(\mathbb{F}_5)$, a set which has $6$ elements. There are $15$ permutations of $\mathbb{P}^1(\mathbb{F}_5)$ which have order $2$ and no fixed points -- for example, $0 \leftrightarrow \infty$, $1 \leftrightarrow 2$, $3 \leftrightarrow 4$. Of those permutations, $10$ are induced by elements of $PGL(2,5)$. For example, the above map is $z \mapsto 2/z$. Conjugation by $PGL(2,5)$ permutes the other $5$ involutions, and this permutation gives an isomorphism $PGL(2,5) \cong S_5$.

Answer 4

Geometric description: Both are isomorphic to the group of symmetries of the icosahedron/dodecahedron. To show this, I will five color the faces of the icosahedron, and label the faces of the dodecahedron with $\{ 0,1,2,3,4,\infty \}$, so the symmetry group acts on the $5$ colors by $A_5$ and acts on the $6$ elements of $\mathbb{P}^1(\mathbb{F}_5)$ by $PSL(2,5)$. Cutout out the two images below and tape them together to see the construction: