Farah and Solecki showed the following in [Borel subgroups of Polish groups](https://doi.org/10.1016/j.aim.2005.07.009):

**Theorem:** Every Polish group $G$ admits Borel subgroups of arbitrarily high Borel rank.

However, the construction is far from uniform. I want to know if the procedure can be made uniform, i.e. if we have a reasonably absolute transfinite operation which, on countable ordinal $\alpha$, spits out a Borel subgroup of $G$ which is not $\Pi_\alpha$. There are various morally equivalent ways to make this precise, for example:

**Question:** Work in $\text{ZF} + \text{AD}$ (axiom of determinacy). Suppose $G$ is a Polish group. Is there is an injection from $\omega_1$ to the set of Borel subgroups of $G$?

If such an injection existed, then after pruning, we can arrange the Borel ranks of subgroups to be strictly increasing. This is because under AD, Hjorth proved in [An absoluteness principle for Borel sets](https://doi.org/10.2307/2586857) there is no injection from $\omega_1$ to $\Pi_\alpha$ subsets of $\mathbb{R}$, for any fixed $\alpha < \omega_1$.

Some remarks:

 - The answer might depend on $G$. I would accept a positive or negative answer for any specific (uncountable) $G$.
 - $\omega_1$ is reducible to the set of Borel subsets of $\mathbb{R}$: e.g. let $X_\alpha$ be the set of well-founded linear orders of rank less than $\alpha$.
 - If we looked at subgroups up to conjugacy, rather than just subgroups, the answer is "Yes" for $S_\infty$ (in fact we just need closed subgroups). But otherwise it may depend on properties of the group $G$; for instance if $G$ is abelian then conjugacy is trivial, so we reduce to the original question.
 - We can ask the same question for various adjectives for Borel subgroups; for instance we can restrict attention to Polishable groups, or to non-Archimedian Polishable groups.