Proof that there exist only 17 Wallpaper Groups (Tilings of the plane) I had a professor who once introduced us to Wallpaper Groups. There are many references that exist to understand what they are (example Wiki, Wallpaper group).
The punchline is 

$$There \,\, are \,\, exactly \,\, 17 \,\, wallpaper \,\, groups \,\,(17 \,\, ways \,\, to \,\, tile \,\, the \,\, plane)$$

My question is $2$-fold:


*

*Can someone sketch out the proof or at least give some high level ideas of why this may be true?

*Can someone refer me to a website or a textbook that develops the proof in detail?
 A: The best I have, is this (and I admit it is not very good).  The Euler characteristic of the infinite plane is 2.
The members of the wallpaper group have a notation:
632 or 4*2 or *2222  
It uses some sequence of numbers, and the symbols $*,\circ, \times$
The numbers represent rotations, the $*$ represents the presence of reflection, the $\times$ represents a glide symmetry. The $\circ$ indicates translations without reflections or rotations. 
This notation suggests an algebra.  For each digit before the star we add $\frac {n-1}{n}$.  The star adds 1.  For each digit after the star we add $\frac {n-1}{2n}$ or one half of what you otherwise would add. 
$\times$ adds 1, an $\circ$ adds 2. 
This sum must equal 2.
For the groups above: $\frac {5}{6}+\frac{2}{3} + \frac {1}{2} = 2$ and $\frac {3}{4} + 1 + \frac {1}{4} = 2$ and $1+\frac 14 + \frac 14 + \frac 14 + \frac 14=2$ 
With this algebra, we can brute force through all possible combinations of rotations, reflections, glides, etc.
https://en.wikipedia.org/wiki/Orbifold_notation
However, I don't remember the proofs that associate this algebra to groups.
A: Sketch of the proof: Let $\Gamma \le {\rm Iso}(\Bbb R^2)$ be a wallpaper group. Then $\Gamma$ has a normal subgroup isomorphic
to $\Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $\Bbb Z^2$ by conjugation. 
We obtain a faithful representation
$$
F \hookrightarrow {\rm Aut}(\Bbb Z^2)\cong GL_2(\Bbb Z).
$$
The group $GL_2(\Bbb Z)$ has exactly $13$ different conjugacy classes of finite subgroups,
called arithmetic ornament classes:
\begin{align*}
C_1 & \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\rangle,\;
C_2 \cong \left\langle \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle,\;
C_3 \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix} \right\rangle, \\
C_4 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right\rangle, \;
C_6 \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix} \right\rangle, \;
D_1 \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle, \\
D_1 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\rangle, \;
D_2 \cong  \left\langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
\right\rangle, \\
D_2 & \cong  \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
\right\rangle, \; D_3 \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}\right\rangle, \\
D_3 & \cong  \left\langle \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}
\right\rangle,\\
D_4 & \cong  \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
\right\rangle, \; D_6\cong  \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}\right\rangle.
\end{align*}
This is an easy computation. Here $C_1,C_2,C_3,C_4,C_6$ are cyclic groups and $D_1,D_2,D_3,D_4,D_6$ are dihedral groups. We use here, that the order $n$ of a subgroup must satisfy $\phi(n)=deg(\Phi_n)\mid 2$, so that $n=1,2,3,4,6$. This is called the crystallographic condition. The wallpaper groups arise from these $13$ classes by equivalence classes of extensions
$$
1\rightarrow \Bbb Z^2\rightarrow \Gamma\rightarrow F\rightarrow 1,
$$
determined by $H^2(F,\Bbb Z^2)$. 
By computing $H^2(F,\Bbb Z^2)$ in each case we obtain $18$ inequivalent extensions, because in $13$ cases the cohomology is trivial, and in three cases we get $C_2,C_2$ and $C_2\times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
This yields $17$ different groups, because two of them turn out to be isomorphic.
