Is there an algorithm to determine all the Coxeter subgroups of a given Coxeter group? If we only want the Coxeter subgroups of finite index does that make the question easier? If we only want a Coxeter subgroup of a given type does that make the question easier? Are there special cases of these questions which can be answered?
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3$\begingroup$ Be careful about the formulation: being a "Coxeter group" requires fixing a set of involutive generators. However, a finite symmetric group $S_n$ will typically contain a lot of smaller Coxeter groups whose generators have nothing to do with those of $S_n$ itself, since every finite group has some embedding in a symmetric group. In another direction, a subgroup of a Coxeter group generated by a finite set of "reflections" (conjugates of the given generators) will be a Coxeter group relative to this new set of involutions (Deodhar, Dyer). Many possibilities. $\endgroup$ – Jim Humphreys Aug 13 '15 at 18:56

$\begingroup$ By Coxeter subgroup I mean that you specify involution generators with presentation given by the relations of a Coxeter diagram. $\endgroup$ – i. m. soloveichik Aug 13 '15 at 19:01

$\begingroup$ But if the given Coxeter group is finite then the answer to the question is clearly yes. Also the subgroups of finite index that are isomorphic to a Coxeter group is recursively enumerable. I would guess that the answer to the general questionis no. $\endgroup$ – Derek Holt Aug 14 '15 at 8:40

$\begingroup$ Related question: given two finite Coxeter graphs, determine whether the associated Coxeter groups are abstractly commensurable (that is, admit isomorphic subgroup of finite index) $\endgroup$ – YCor Aug 16 '15 at 10:17