TITLE: Coxeter subgroups of Coxeter groups
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: The embedding problem among Coxeter groups (i.e. deciding, given two Coxeter groups, whether or not one is (abstractly) isomorphic to a subgroup of the other) is widely open, even in the right-angled case. And this problem seems much simpler than determining all the Coxeter groups that embed into a given Coxeter group.
Nevertheless, there are a few (very very) particular cases where an answer to your problem can be found. For instance, the following statement, dealing with right-angled Coxeter groups, can be found in Morphisms between right-angled Coxeter groups and the embedding problem in dimension two.
Proposition 1. Let $\Gamma$ be a finite simplicial graph and let $C_n$ denote a cycle of length $n \geq 5$. Then $C(\Gamma)$ is isomorphic to a subgroup of $C(C_n)$ if and only if $\Gamma$ is either a disjoint union of segments or a single cycle of length divisible by $n -4$.
Observe that a Coxeter subgroup of a right-angled Coxeter group is necessarily a right-angled Coxeter group, so the proposition provides an answer to your problem for a specific family of Coxeter groups.
Another result from the same article, and which might be of interest for the discusion, is:
Proposition 2. Let $R, S$ be two finite trees. Then $C(R)$ is isomorphic to a subgroup of $C(S)$ if and only if there exists a graph morphism $\varphi : R \to S$ that sends a vertex of degree $2$ to a vertex of degree $\geq 2$ and a vertex of degree $\geq 3$ to a vertex of degree $\geq 3$.

Fixing $R$ and $S$, the existence of such a graph morphism can be checked algorithmically, so, fixing $S$, the set of Coxeter subgroups in $C(S)$ is enumerable.