# Classifying finite subgroups of a group

This is a survey article related to:subgroups
View other survey articles about subgroups

This article explores the interesting question: given an infinite group, how do we classify all finite subgroups of the group? we'll mainly be looking at linear groups and groups that arise in geometric situations; for instance, fundamental groups or isometry groups of metric spaces and Riemannian manifolds.

## General tactics: pass to easier subgroups, quotients and covers

### Pass to and from subgroups

Suppose $H$ is a subgroup in a group $G$. Then we have a natural map from the finite subgroups of $G$, to the finite subgroups of $H$, given by intersecting with $H$: $K \mapsto K \cap H$

This map is surjective, because any finite subgroup of $H$ is also a finite subgroup of $G$. Thus:

• Going from $G$ to $H$: To classify the finite subgroups of $H$, it suffices to first classify the finite subgroups of $G$, and then check which of them lie inside $H$.
• Going from $H$ to $G$: if we know all the finite subgroups of $H$, then we can, for each finite subgroup $L \le H$, try to determine all the possibilities for a finite subgroup $K$ of $G$ such that $K \cap H = L$. This may not always be easy, but in some cases, it is not hard. For instance, if $H$ is a subgroup of index two in $G$, then any pre-image of $L$ that is not itself $L$, is generated by $L$ and exactly one element in $G \setminus H$. A bit of manipulation can put strong restrictions on what such an element must look like. More generally, the problem is tractable if $H$ is a subgroup of finite index.

### Pass to and from quotients by finite normal subgroups

Suppose $N$ is a normal subgroup of $G$, and $H = G/N$ is the quotient group with the quotient map $p:G \to H$. We have a natural map from finite subgroups of $G$ to subgroups of $H$: $K \mapsto p(K)$

This map is surjective, because for any finite subgroup $L$ of $H$, we can take the full inverse image $p^{-1}(L)$. Thus:

• Going from $G$ to $H$: To classify the finite subgroups of $H$, it suffices to classify the finite subgroups of $G$, and then take the image under $p$
• Going from $H$ to $G$: If we know all the finite subgroups of $H$, then for each finite subgroup $L \le H$, try to determine all the possible finite $K$ such that $p(K) = L$. This is not always easy, but it can be achievable in some cases, particularly the case where $N$ is a finite normal subgroup.

## Finding finite-dominating subgroups

### What's a finite-dominating subgroup?

A subgroup $H$ of a group $G$ is a finite-dominating subgroup if given any finite subgroup $K$ of $G$, $K$ is a subconjugate subgroup of $H$: in other words, there exists $g \in G$ such that $gKg^{-1} \le H$.

If $H$ is a finite-dominating subgroup in $G$, then we can reduce the problem of classifying finite subgroups of $G$, to the problem of classifying finite subgroups of $H$. Here's how:

• Classify all finite subgroups of $H$
• Then, the finite subgroups of $G$ are simply all the subgroups that can be expressed as conjugate subgroups to the finite subgroups of $H$.