There are finitely many $n$-dimensional wallpaper groups There are $7$ Frieze groups, $17$ wallpaper groups and $230$ space groups, which are the 1,2,3-dimensional cases of isometries on $\Bbb R^n$ (I think? Frieze groups seem to require $[0,1]\times\Bbb R$ instead of just $\Bbb R$). What the linked pages don't get into, though, is why we would expect the list to be finite; I assume the folks who set out to classify these things already knew this going in or else the enumeration attempt would seem pointless.
(Correct me if I'm wrong, but) the general definition of an $n$-dimensional space group is an infinite discrete subgroup of the Euclidean group $E(n)$ of isometries of $\Bbb R^n$, presumably topologized by compact-open, in order to make sense of the "discrete" moniker.
The question is: Given the above definition, show that there are finitely many $n$-dimensional space groups up to isomorphism.
 A: There are two definitions useful in this context:


*

*A crystallographic group of the euclidean $n$-space is a discrete subgroup $\Gamma$ of the isometry group $Isom(E^n)$ of $E^n$, such that $E^n/\Gamma$ is compact. 

*A discrete subgroup $Isom(E^n)$. 
You also have to decide on the notion of equivalence between such subgroups: Conjugation in $Isom(E^n)$, conjugation in $Aff(E^n)$, topological conjugation, an abstract isomorphism. I will use the abstract isomorphism as the equivalence relation. 
There are several important theorems in this context, all due to Bieberbach. 
Theorem 1. For each discrete subgroup $\Gamma< Isom(E^n)$ there exists a $\Gamma$-invariant affine subspace $A\subset E^n$, such that $A/\Gamma$ is compact. 
Theorem 2. For each $n$, up to affine conjugation, there are only finitely many crystallographic groups of $Isom(E^n)$.
On the other hand, 
Lemma. Up to an isomorphism there are infinitely many infinite discrete subgroups of  $Isom(E^3)$. 
Proof: Consider finite cyclic groups of rotations of $E^3$ fixing a line $L$. Now, take the direct product of such groups with an infinite cyclic group of translations of $E^3$ along $L$. qed 
However, in view of Theorems 1 and 2, the following finiteness theorem holds:
Theorem 3. For each $n$ there are finitely many discrete isometry subgroups of $Isom(E^n)$, $\{\Gamma_1,...,\Gamma_k\}$, such that every 
discrete subgroup $\Gamma$ of $Isom(E^n)$ contains a normal subgroup of finite index $\Phi< \Gamma$, such that $\Gamma/\Phi\cong \Gamma_i$ for some $i$. 
Note: One can even reduce the finiteness claim here to finite normal abelian subgroups of $\Gamma$.
Read:
J.Wolf, "Spaces of constant curvature".  
