Fixed subgroups are not compressed in direct products of surface groups (Q6632211)

Let \(G\) be a finitely generated group, then \(\operatorname{rk}(G)\), the rank of \(G\), is the minimal number of generators of \(G\). For \(\mathcal{B} \subseteq \mathrm{End}(G)\) let \(\mathrm{Fix}(\mathcal{B})=\{ g\in G \mid \phi(g)=g, \; \forall \phi \in \mathcal{B} \}\). For free groups, a celebrated result is attributed to \textit{M. Bestvina} and \textit{M. Handel} [Ann. Math. (2) 135, No. 1, 1--51 (1992; Zbl 0757.57004)] asserts that if \(F_{n}\) is a free group of rank \(n\) then, for every \(\phi \in \Aut(F_{n})\), one has \(\mathrm{rk}(\mathrm{Fix}(\phi)) \leq n\).\N\N\textit{W. Dicks} and \textit{E. Ventura} [The group fixed by a family of injective endomorphisms of a free group. Providence, RI: AMS, American Mathematical Society (1996; Zbl 0845.20018)], introduced the notions of inertia and compression. A subgroup \(H \leq G\) is said to be inert in \(G\) if \(\mathrm{rk}(K \cap H) \leq \operatorname{rk}(K)\) for every (finitely generated) subgroup \(K \leq G\). A subgroup \(H \leq G\) is said to be compressed in \(G\) if \(\operatorname{rk}(H) \leq \operatorname{rk}(K)\) for every (finitely generated) subgroup \(K\) with \(H \leq K \leq G\). In [\textit{A. Martino} and \textit{E. Ventura}, Commun. Algebra 32, No. 10, 3921-3935 (2004; Zbl 1069.20015)] it is proved that the fixed subgroups of any family of endomorphisms are compressed in free groups.\N\NThe authors, constructing appropriate counterexamples, show that the fixed subgroups are not compressed in direct products of free and surface groups. This disproves a conjecture made by the authors and \textit{E. Ventura} [Int. J. Algebra Comput. 25, No. 5, 865--887 (2015; Zbl 1352.57003)].