Prove the binary icosahedral group is isomorphic to ${\rm SL}(2,5)$

Question

I am having difficulty proving that the binary icosahedral group $2I$ is isomorphic to ${\rm SL}(2,5)$.

The binary icosahedral group $2I$ is a finite subgroup of $H^1$, where $H^1=\{q\in\mathbb{H}\mid |q|=1\}$ is the multiplicative group of unit quaternions. Below are some notations from Voight's book "Graduate Texts in Mathematics, 288 Quaternion Algebras (2021)".

Let $\tau=\frac{1+\sqrt{5}}{2}$, $R=\mathbb{Z}[\tau]$ and $F=\mathbb{Q}(\sqrt{5})$, then $R$ is the ring of algebraic integers in $F$.

Let $\zeta = \frac{\tau + \tau^{-1}i + j}{2}$, the $R$-algebra $$O = R + Ri + R\zeta + Ri\zeta. $$ is a maximal order in $\left(\frac{-1,-1}{F}\right)$, and $2I$ is the set of units of $O$: $2I=O^\times$.

I want to prove that $2I\cong {\rm SL}(2, 5)$. I tried adopting the following strategy, but I found there were some details in the argument that I couldn't fill in.

1. Firstly reduce $O$ by the ideal $\sqrt{5}O$, and get $O/\sqrt{5}O$.

2. $O/\sqrt{5}O$ is a ring over $\mathbb{F}_5$. I think it lies in some quaternion algebra $A=\left(\frac{a,b}{\mathbb{F}_5}\right)$ which splits, hence $O/\sqrt{5}O$ is mapped to a ring lies inside ${\rm Mat}_2(\mathbb{F}_5)$.

3. The units $O^\times$ maps injectively into ${\rm SL}(2,5)$, and since both of them have order 120, they are isomprphic.

In step 2, I guess $A$ can be chosen as $\left(\frac{-1,-1}{\mathbb{F}_5}\right)$ where both the -1's come from $i,j$ in $\mathbb{H}$, but I'm not sure of this. And I have no idea of how to prove 3.

Of course, my proof strategy might be wrong. I would be very grateful if you could help clarify my doubts!

Answer 1

Your strategy looks fine to me, let's try to make it work. First, working $\bmod \sqrt{5}$ we can invert $2$ and $\tau \equiv 3 \bmod \sqrt{5}$, so we get $\zeta \equiv -1 + i + 3j \bmod \sqrt{5}$. So taking linear combinations gets us $j$ and then multiplying by $i$ gets us $k$, which gives

$$\mathcal{O}/\sqrt{5} \cong \left( \frac{-1, -1}{\mathbb{F}_5} \right)$$

as expected. Since $5 \equiv 1 \bmod 4$ this quaternion algebra splits, also as expected, so is isomorphic to $M_2(\mathbb{F}_5)$, giving us an isomorphism

$$(\mathcal{O}/\sqrt{5})^{\times} \cong GL_2(\mathbb{F}_5).$$

So the remaining questions are why the induced map $\mathcal{O}^{\times} \to GL_2(\mathbb{F}_5)$ is injective and has image contained in $SL_2(\mathbb{F}_5)$. There's probably a better way to do the second one but it suffices to observe that the binary icosahedral group has trivial abelianization.

For the first one we need to show that none of the unit icosians are congruent to the identity $\bmod \sqrt{5}$, or equivalently, that the only unit icosian congruent to $1 \bmod \sqrt{5}$ is $1$. Such an element has the form

$$z = 1 + \sqrt{5} \, w$$

where $w \in \mathcal{O}$, and has quaternion norm $1$.

At this point the cleanest way I can see to proceed is actually to use a different norm. Namely, as described on Wikipedia, if the quaternion norm is $a + b \sqrt{5}$ we can define the Euclidean norm to be $a + b$. Apparently (I have not checked this) this turns out to be a positive-definite integer-valued norm, and for $z$ the Euclidean norm is equal to $1$. But then if $w \neq 0$ then the Euclidean norm of $w$ is also at least $1$, so the Euclidean norm of $\sqrt{5} w$ is at least $5$ (this requires a small calculation: if the quaternion norm of $w$ is $a + b \sqrt{5}$ then the quaternion norm of $\sqrt{5} w$ is $5a + 5b \sqrt{5}$, so the Euclidean norm is also multiplied by $5$), which contradicts that the Euclidean norm of $z - 1$ is at most $2$ by the triangle inequality. So we must have $w = 0$ as desired.