# Subgroup membership problem in simple groups

Let $$G$$ be a finitely presented simple group. By Kuznetsov (1958), $$G$$ has decidable word problem. However, by Scott [1], $$G$$ may have undecidable conjugacy problem. Is anything known about other decision problems for $$G$$, in generality? I am particularly interested in the question of the subgroup membership problem, i.e. the problem of, given a set of words $$w_1, \dots, w_k$$ and a word $$w$$, deciding whether or not $$w$$ belongs to the subgroup $$\langle w_1, \dots, w_k \rangle$$ of $$G$$. This is undecidable already for the direct product $$H \times H$$ of any group $$H$$ admitting a finitely presented quotient with undecidable word problem (in particular, we can take $$H$$ to be a free group of rank $$2$$), but such direct products are of course very far from being simple.

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[1] Scott, Elizabeth A., A finitely presented simple group with unsolvable conjugacy problem, J. Algebra 90, 333-353 (1984). ZBL0544.20029.

As another example, the problem of computing the order of an element of the finitely presented simple Brin–Thompson group $$2V$$ is undecidable by
After some digging, I was able to find that the answer to my question exists: the problem can be undecidable. Rattaggi, in an unpublished manuscript (available here), proved that there exists a finitely presented simple group $$G$$ which contains $$F_2 \times F_2$$ as a subgroup. This group $$G$$ arises from constructions related to the Burger-Mozes groups. In particular, $$G$$ is incoherent (and, importantly for my question), has undecidable subgroup membership problem.
• Great. But you don't need Rataggi: Thompson's group $V$ contains $V\times V$ (trivial fact), and $V$ contains a non-abelian free subgroup $F$ (very easy too), so $V$ contains $F\times F$.