In my quest to understand interpretability and Shelah's $(-)^{\text{eq}}$ construction, I stumbled upon the following remark in page 29 of Marker's Model Theory:
Suppose that we have a definable group $G$ and a definable normal subgroup $H$. We might want to look at the group $G/H$. It is possible that $G/H$ does not correspond to a definable group in our structure.
After looking around for some time, I have not found a concrete example which realises the latter sentence of the statement above, and I am struggling to come up with such an example myself. So far I've been able to understand most of the theory surrouding interpretability and the $(-)^{\text{eq}}$ construction together with the provided examples, but it seems that any possible example of a quotient of definable groups not being definable is trying to hide from me.
I'm pretty sure such an example doesn't have to by anything fancy, and my bet is that such quotient not being definable will have to do about its universe not being a subset of any power of the universe of the original definable group.
Any idea is much appreciated!
Edit: I just realised that instead of asking for a particular example, a more interesting question would be:
Under which conditions the quotient of definable groups is definable?
This seems (at least to my newbie model-theoretic eyes) a highly non-trivial question, and I'm afraid that even if an answer is given to this I will probably not be able to follow it. However, if anoyone could shed some light on this last question it would be nice too, of course.
Erik Walsberg told me the following example:
Consider the $p$-adic field $\mathbb{Q}_p$. The ring of $p$-adic integers $\mathbb{Z}_p$ is definable in $\mathbb{Q}_p$, so the multiplicative group of units $\mathbb{Z}_p^\times$ is a definable subgroup of the multiplicative group $\mathbb{Q}_p^\times$. The quotient $\mathbb{Q}_p^\times/\mathbb{Z}_p^\times$ is isomorphic to $\mathbb{Z}$ (this is the value group of $\mathbb{Q}_p$, viewed as a valued field). But no group isomorphic to $\mathbb{Z}$ is definable in $\mathbb{Q}_p$, since every definable subset of $\mathbb{Q}_p$ is finite or has cardinality $2^{\aleph_0}$.