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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.

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It may depend if you want a geometric construction or an algebraic one. Since a Sylow $2$-subgroup of ${\rm PSL}(2,5)$ is a Klein $4$-group, Burnside's normal $p$-complement theorem guarantees that a Sylow $2$-normalizer has order $12,$ so that ${\rm PSL}(2,5)$ has $5$ Sylow $2$-subgroups, yielding a homomorphism from ${\rm PSL}(2,5)$ to $S_5$ with trivial kernel and image of order $60.$ – Geoff Robinson Dec 23 '11 at 20:42
(somewhat) related:… – Grigory M Dec 23 '11 at 20:48
Since those Sylow 2-subgroups have order 4, presumably there might be a relatively simple description of representative matrices that generate these groups? That would be explicit enough to satisfy me. – Barry Smith Dec 23 '11 at 20:50
I would then also be happy with an extremely explicit construction of this covering. – Barry Smith Dec 23 '11 at 20:56
up vote 16 down vote accepted

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!

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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:

By the way, there is also an algebraic way to see that $PSL(2,5)$ is isomorphic to the symmetries of the dodecahedron. The dodecahedral symmetry group embeds in $SO(3)$, using entries in $\mathbb{Z}[\tau]$ where $\tau$ is the golden ratio $\frac{1 + \sqrt{5}}{2}$. There is a ring map $\mathbb{Z}[\tau] \mapsto \mathbb{F}_5$ with $\tau \mapsto 3$. Applying that to these matrices, we get a representation of the dodecahedral group with entries in $SO(3, \mathbb{F}_5)$. Since $\mathbb{F}_5$ contains square roots of $-1$ (namely, $\pm 2$), we have $SO(3, \mathbb{F}_5) \cong PSL_2(\mathbb{F}_5)$. But the isomorphism $A_5 \cong PSL_2(\mathbb{F}_5)$ seems harder to see by pure algebra.

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Alternatively, there is a way (two, actually) to partition the vertices of a dodecahedron in groups of four in such a way that each part consists of the vertices of a regular tetrahedron. It is not difficult to see that a motion of the solid cannot give a transposition of two of the tetrahedra. – Mariano Suárez-Alvarez Mar 12 '12 at 19:16
Yup. The colors here are vertices of the tetrahedra Mariano speaks of. I've always thought images of the tetrahedra were too cluttered to understand and that this might be clearer. It looked better in my head than on paper, though. – David Speyer Mar 12 '12 at 19:20
Google will find modular origami instructions to build the five tetrahedra. It takes a little patience to fold the pieces and quite a bit of ingenuity to assemble, but the result is quite extraordinary. Drop by at my office if you are ever in the neighborhood and I'll let you play with mine :) – Mariano Suárez-Alvarez Mar 12 '12 at 19:24
Feel free to edit in a photo of your desktoys! – David Speyer Mar 12 '12 at 19:27

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 $PGL(2,5) \cong S_5$.

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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 $$ D=\pmatrix{1&1\cr0&1\cr}. $$

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I wasn't really assuming ${\rm PSL}(2,5)$ simple in my comment, it's clear that the homomorphism into $S_5$ is injective by a direct argument of the kind you outlined. – Geoff Robinson Dec 24 '11 at 1:45

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