Equality of the monolith and AComm for a closed surface group

Prove or disprove that the monolith of Comm(π₁(Σ)) is equal to the subgroup AComm(π₁(Σ)) generated by automorphism groups of finite-index subgroups of π₁(Σ), where Σ is a closed surface of genus at least two.

Background

For finitely generated free groups, the subgroup AComm(F) is known to equal the monolith of Comm(F). The paper notes that the monolith of Comm(π₁(Σ)) is contained in AComm(π₁(Σ)), but does not establish the reverse inclusion.

References

The monolith is contained in $AComm(\pi_1(\Sigma))$ as $AComm(\pi_1(\Sigma))$ is a normal subgroup of $Comm(\pi_1(\Sigma))$ Proposition 1.4.2, Lemma 2.22, but we do not know whether there is equality.

— Free abelian quotients of commensurators  (2609.35323 - Boudec, 28 Sep 2026) in Section 3, subsection “On the abstract commensurator of surface groups”