Let $K$ be a quadratic imaginary field, and $E$ an elliptic curve whose endomorphism ring is isomorphic to the full ring of integers of $K$. Let $j$ be its $j$-invariant, and $c$ an integral ideal of $K$. Consider the following tower:

$$K(j,E[c])\ \ /\ \ K(j,h(E[c]))\ \ /\ \ K(j)\ \ /\ \ K,$$

where $h$ here is any Weber function on $E$. (Note that $K(j)$ is the Hilbert class field of $K$).

We know that all these extensions are Galois, and any field has ABELIAN galois group over any smaller field, EXCEPT POSSIBLY THE BIGGEST ONE (namely, $K(j,E[c]) / K$).

Questions:

- Does the biggest one have to be abelian? Give a proof or counterexample.

My suspicion: No, it doesn't. I've been trying an example with $K = \mathbf Q(\sqrt{-15})$, $E = C/O_K$, and $c = 3$; it just requires me to factorise a quartic polynomial over $\bar{ \mathbf Q}$, which SAGE apparently can't do.

- What about if I replace $E[c]$ in the above by $E_{tors}$, the full torsion group?