Let me first introduce two definitions:
For a structure $\mathcal{M}$ in a language $\mathscr{L}$ and a subset $X \subseteq M^n$, the fully induced structure on $X$ is a structure $\mathcal{X}$ with the universe $X$ in a language consisting of following relational symbols: For every formula $\varphi (\overline{x}_1, \dots, \overline{x}_m)$, where $\overline{x}_1, \overline{x}_n, \dots, \overline{x}_m$ are $n$-tulpes of variables (all pairwise distinct), there is a rel. symbol $R_{\varphi}(y_1, \dots, y_m)$ interpreted in $\mathcal{X}$ as $\varphi(M) \cap X^m$.
Consider a (commutative) group $G$ to be an $\mathscr{L}$-strure, where $\mathscr{L}$ is an extension of the language of groups. $G$ is called weakly normal provided that every definable set $X \subseteq G^n$ is a finite Boolean combination of cosets of definable subgroups of $G^n$. (Definability is meant wrt the language $\mathscr{L}$).
These definitions are taken from the survey article "Diophantine geometry from model theory" by Thomas Scanlon.
In the article the author claims that
If G is a weakly normal group, then the induced structure on any subgroup is also weakly normal.
(Page 8, 2nd paragraph; he actually writes "is also weakly minimal," but from the context it is obvious that it is meant "weakly normal".)
From the context it seems most likely that the language on the subgroup considered is the induced one from the original group.
My question is simply
How does one see that this is true?
It is certainly true provided that the subgroup (let's call it $H$ inside a weakly normal group $G$) is definable (in $G$): whenever $X=\varphi(H)\subseteq H^m$, one can produce formula $\psi$ in the language of the group $G$ from $\varphi$ by replacing every predicate $R_{\sigma}(x_1, \dots, x_k)$ simply by the formula $\sigma(x_1, \dots, x_k)$, replacing quantifiers $\forall x$ by $\forall x \in H$ and finally replacing $\exists x$ by $\exists x \in H$. Then $\psi(G)\cap H=\varphi(H)$ and one uses the weak normality of $G$.
However, I have trouble seeing how does one make similar process to work without definability of $H$ or other extra assumptions (e.g. QE for the group $G$). Yet from what follows, $H$ not being definable seems to be pretty much standard situation.
Thanks in advance for any help.