Symmetries of a dodecahedron vs $\mathrm{GL}(2,\mathbb C)$

Question

We know the following facts:
i) The symmetry group of a dodecahedron is $A_5$.
ii) Rotations in 3D space can be expressed by $\mathrm{SO}(3)\cong\mathrm{PSU}(2)$, which is a subgroup of $\mathrm{GL}(2,\mathbb C)$.
iii) $A_4$ is not isomorphic to a subgroup of $\mathrm{GL}(2,\mathbb C)$.

Proof of iii)
As $A_4\cong G:=\langle a, b | a^3 = b^2 = 1, aba = ba^2 b \rangle$, assume there is a monomorphism $\tau:G\to\mathrm{GL}(2,\mathbb C)$, then let $A=\tau(a)$ and $B=\tau(b)$. Since $B^2=I$, the only possible non-diagonal choices for $B$ is $\left(\begin{matrix}x&y\\(1-x^2)/y&-x\end{matrix}\right)$. Similiarily, $$A=\left(\begin{matrix}z&w\\(-\overline\omega-\omega z-z^2)/w&-z-\omega\end{matrix}\right)\text{ where }\omega^3=1$$ Solving $ABA=BA^2B$ (using software) gives no result.

Combining i) and ii), we can see $A_4<A_5<\mathrm{SO}(3)<\mathrm{GL}(2,\mathbb C)$, which is contradicting with iii).
Question Where does the mistake occur?
P.S. I'm sure iii) is correct as I saw an alternative argument of it.

Answer 1

Because $PSU(2)$ is not a subgroup of $GL(2,\mathbb C)$. It is a quotient of a subgroup of $GL(2,\mathbb C)$.