Here is a proof of Bieberbach's theorem:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.460.1436&rep=rep1&type=pdf
I can follow the computing details, but I have no intuitive sense of Lemma B in the proof.
Definition & Notation:
(1) The isometry group of standard $\mathbb{R}^n$ is the Euclidean group $E(n)$, which is the semi-product of $O(n)$ and $T(n)=(\mathbb{R}^n,+)$ and is endowed with product topology. Every element of $E(n)$ is represented by some $(A,a)$ where $A\in O(n)$ and $a\in \mathbb{R}^n$, which maps $x \in \mathbb{R}^n$ to $Ax+a$.
(2) A subgroup $G$ of $E(n)$ is called $crystallographic$ if $G$ is discrete (in subspace topology of $E(n)$) and the orbit space $\mathbb{R}^n/G$ is compact.
(3) For any $A\in O(n)$ we define $m(A)=\|A-id\|=\sup_{0\neq x\in \mathbb{R}^n}\frac{|Ax-x|}{|x|}=\max_{|x|=1}|Ax-x|.$
Then lemma B says that given a crystallographic group $\Gamma$, any element $(A,a)\in\Gamma$ satisfying $m(A)\leq \frac{1}{2}$ must be a translation, i.e. $A=id.$
I can check this result, but I have no intuitive sense of it. How can there exist a constant $\frac{1}{2}$, which is independent of $n$ and $\Gamma$, such that Lemma B holds? Amazing! Is there anybody who can help me?