Groups of glide reflections and translations I am interested in the following question, completing a proof of Stillwell:
Let $\Gamma$ be a discontinuous, fixed point free group made of of glide reflections and
translations, let $g$ be a glide reflection of minimal length in r$\Gamma$, and let $h$
be an element of minimal length not in the direction of $g$. Show that $g, h$
must have perpendicular directions (e.g., by finding shorter elements when
the directions of g, h are not perpendicular). Deduce that $g, h$ generate $\Gamma$.
I am a bit disturbed because I do not find a proper proof of this fact anywhere. There is this very interesting answer on MSE, but it does not seem totally answering the question. In this particular answer (that seems really elegant, or at least less "classically geometrical"), I have the following precise issue


*

*what about the case of $g$ being a translation and not a glide reflection? (which does not seem to be considered by the answer)

*there is a part of the argument I do not get: the linear part of $[f, g]$ is not the identity, thus $[f, g]$ is not a translation. But why would it necessarily be a rotation?is there anything to do with orientation preserving transformation?


Thanks for all your clarification! (and in addition to clarifying this more linear algebraic proof, I would be happy to know if there is an answer in the spirit Stilwell suggest, only considering minimal elements in some sense, if it is not too complicated).
 A: I will be identifying the Euclidean plane $E^2$ with the complex plane ${\mathbb C}$. 
Accordingly, for a complex number $a\in {\mathbb C}$, $T_a$ will denote the translation $T_a(z)=z+a$. 
Suppose that $f, g$ are orientation-reversing isometries of $E^2$. Then their composition is always orientation-preserving. Ditto for the compositions of an even number of orientation-reversing isometries. In particular, the commutator of any pair of isometries is orientation-preserving. By the classification of isometries of $E^2$, an orientation-preserving isometry is either a rotation or a translation. Here is how to see this algebraically. Consider an isometry 
$$
f(z)=az+b, |a|=1. 
$$ 
Let's solve the equation $f(z)=z$:
$$
az+b=z, z(a-1)=-b, z= b(1-a)^{-1},
$$
unless $a=1$. Thus, an orientation-preserving isometry has a fixed point unless $a=1$, i.e. unless $f$ is a translation. To see that if $f$ has a fixed point $z_0$ then it is a rotation, make a change of coordinates $w=z-z_0$ so that $w=0$ is a fixed point. In the new coordinate system, $f$ becomes the rotation
$$
f(w)=aw. 
$$ 
Now, I will discuss orientation-reversing isometries. Recall that a glide-reflection $g$ in the Euclidean plane is the composition of a translation along a line $L$ and a reflection fixing $L$. The line $L$ is the axis of $g$, it is uniquely determined by $g$. Alternatively the axis $L_g$ can be described  as the minimum set of the "displacement  function" 
$$
z\mapsto dist(z, g(z)). 
$$
Even though this is not needed for the solution, here are few facts about glide-reflections: All glide-reflections can be written in the form
$$
z\mapsto a \bar{z} +b, 
$$
where $|a|=1$. The complex number $a=e^{i\theta}$ determines the direction of the axis $L_g$ as follows. The equation $a\bar{z}=z$ has exactly two solutions with absolute value $1$: $z= e^{i\theta/2}$ and $z=e^{i(\theta +2\pi)/2}$. The axis $L_g$ is parallel to the line through these two solutions. If two glide-reflections 
$$
g_k: z\mapsto a_k \bar{z} +b_k, k=1,2, 
$$
have different linear parts $z\mapsto a_k \bar{z}$ then the product $g_1 g_2^{-1}$ is a nontrivial rotation (with some center) by the angle $arg(a_1 a_2^{-1})$.  Hence, if the product $g_1 g_2^{-1}$ is a translation then $g_1, g_2$ have parallel axes. 
(This is the content of my answer here.) 
Suppose  that $G$ is a subgroup of the group of isometries of Euclidean plane. Then $G=G_+\sqcup G_-$, where $G_+$ is the subgroup of $G$ consisting of orientation-preserving isometries and $G_-$ is the subset of $G$ consisting of orientation-reversing isometries. The subgroup $G_+$ has index 2 in $G$. 
Suppose now that $G$ is discrete and  acts freely on the Euclidean plane, i.e. $G$ contains no reflections and nontrivial rotations. 
Theorem. One of the following holds:


*

*$G$ is cyclic, generated by a translation or by a glide-reflection. 

*$G\cong {\mathbb Z}^2$ and is generated by two translations $T_a, T_b$ where $a, b$ are linearly independent over ${\mathbb R}$. 

*$G$ is generated by a translation $h=T_a$ and a glide-reflection $g$ such that the axis of $g$ is orthogonal to $a$ regarded as a vector in ${\mathbb R}^2$. More geometrically, $L_g$ is orthogonal to any line invariant under the translation $h$.  
Proof. I will consider the most interesting case, when $G_-$  is non-empty, i.e $G$ does not preserve the orientation on the plane. 
Since $G$ contains no nontrivial rotations, axes of glide-reflections in $G_-$ are all parallel to each other (compositions of glide-reflections are translations in our case). Without loss of generality (by conjugating $G$ by a rotation) we can assume that these axes are parallel to the real axis in the complex plane. Thus, every $g\in G_-$ 
determines a real number $\tau(g)\in {\mathbb R}$:
$$
Re( g(z)- z)= \tau(g), \forall z\in {\mathbb C}. 
$$
The function $\tau: G_-\to {\mathbb R}$ satisfies:
$$
\tau(g^{-1})=-\tau(g), g_1\circ g_2= T_{\tau(g_1) + \tau(g_2)}. 
$$ 
Since the group $G$ is discrete, so is $G_+$, hence, the set of values of $\tau: G_-\to {\mathbb R}$ has no accumulation points, i.e. 
it forms a discrete and closed subset of ${\mathbb R}$. In particular, there exists $g_1\in G_-$ such that  $\tau(g_1)>0$ realizes the minimum of the subset
$$
\{\tau(g): g\in G_-\} \cap (0,\infty). 
$$
Without loss of generality, by conjugating $G$ via some element of the group of complex-affine transformations $z\mapsto az+b$, $a\in {\mathbb C}^\times, b\in {\mathbb C}$, we can (and will) assume that $L_{g_1}$ is the real axis and $\tau(g_1)=1$. (This is just the matter of notation convenience.) The square $h_1=g_1\circ g_1$ is the translation $T_2$. 
Recall that for a subset $S$ of a group $G$, the notation $\langle S \rangle$ stands for the subgroup of $G$ generated by $S$. Since $h_1(z)=z+2$, every coset in $G_+/\langle h_1\rangle$ has a representative $h=T_a$ such that $Re(a)\in [-1,1]$. 
Lemma. The number $a$ as above is imaginary: $Re(a)=0$. In other words, $h$ is a vertical translation. 
Proof. Replacing $h$ with its inverse, we can assume that $0\le a\le 1$. 
Next, $g_1 h g_{1}^{-1}= T_{\bar{a}}$ and, hence, $h g_1 h g_{1}^{-1}= T_{b}$, $b=2Re(a)\in {\mathbb R}$. 
If $Re(a)\in (0,1)$ then either $g=T_b g_1^{-1}$ or $g= T_{-b}g_1$ is a glide-reflection which satisfies 
$$
0< \tau(g) <1
$$
which contradicts our choice of $g_1$ attaining the minimal positive value of $\tau$. Suppose that $Re(a)=1$. Then 
$$
\tau( T_{-b}g_1)=0,
$$
i.e. $T_{-b}g_1$ is a reflection (not a glide-reflection) contradicting the assumption that $G$ acts freely on the plane. Thus, $Re(a)=0$. qed 
The subgroup of $G_+$ consisting of vertical translations is discrete, hence, it is either trivial or is infinite cyclic. In the former case, 
$G_+= \langle h_1\rangle$ and, hence, $g_1$ generates $G$. 
Assume, therefore, that  the subgroup of vertical translations of $G_+$ is infinite cyclic. It is then generated by some translation 
$$
h_2=T_a, Re(a)=0. 
$$ 
Since every coset of $G_+/\langle h_1\rangle$ is represented by a vertical translation, i.e. an element of $\langle h_2\rangle$, we conclude that $G_+$ is generated by $h_1, h_2$, horizontal and vertical translations. Since $G=G_+\cup g_1 G_+$ (as $G_+$ has index two in $G$), we obtain that $g_1$ and $h_2$ generate $G$. qed 
A: 
I would be happy to know if there is an answer in the spirit Stilwell suggest, only considering minimal elements in some sense, if it is not too complicated

Honestly, given the ground already covered in GoS, I think Stillwell envisioned a simple proof. Noting that we can identify $\mathbb{E}^2$ with $\mathbb{C}$, WLOG write $g(z)=\overline{z}+1$, then for any $h\in\Gamma$ not in the direction of $g$ $$h(z)=\begin{cases} z+2n+imb ,     & \quad \text{if } h \text{ is a translation}\\ \overline{z}+2n+1+imb , & \quad \text{if } h \text{ is a glide reflection}\end{cases}.$$ The key observation is that, multplying $h$ on the left by $g^{-a}$, where $a=2n$ or $a=2n+1$, we can find a translation $t$ perpendicular to $g$ and smaller than or equal to $h$.
Proof:  Let $h\in\Gamma$ be a translation $$h(z)=z+u,$$ not in the real direction, with $u=a+ib$.
We will prove if $h$ is minimal then $a=0$ , otherwise $h$ is generated by $g$ and a translation $t$ perpendicular to $g$ and smaller than $h$.
If $a=0$ then by discontinuity of $\Gamma$ either $h$ is minimal in the imaginary direction or $$h(z)=z+ib=z+inc=t^n(z).$$
Assume $a\neq 0$. For the rest of the proof we will also assume $b$ is minimal.
Let $j=hg^{-1}hg$, then $j\in\Gamma$ and $$j(z)=z+\overline{u}+u=z+2a.$$ Thus $|a|\geq 1/2$, since $g$ is minimal in the real direction. Further, arguing from discontinuity again, either $a$ is an integer or $|a|=1/2$.

Case 1: $|a|=1/2$,

then $$g^{\pm 1}j(z)=\overline{z},$$ which is a contradiction, since $\Gamma$ acts freely on $\mathbb{C}$ and so cannot contain reflections.

Case 2: $a\in\mathbb{Z}-\{0\}$,

then $$t(z)=g^{-a}h(z)=\begin{cases} z+ib ,     & \quad \text{if } a \text{ is even}\\ \overline{z}+ib , & \quad \text{if } a \text{ is odd}\end{cases}.$$ By the classification of isometries $\overline{z} +ib$ is equal to a product of three reflections in three parallel lines, which in turn is equal to a reflection. Thus $a$ is even, $t$ is a translation of $\Gamma$ smaller than $h$, and $h$ is on the lattice formed by $t$ and $g$.
If $h$ is instead a glide reflection, then the line of reflection is parallel to $g$, since otherwise $gh$ would equal a rotation, which is a contradiction. Thus  $h=gt$ for some translation $t\perp g$ if $h$ is minimal. If $h$ is not minimal we can repeat the same basic arguments as above (i.e. $h(z)=\overline{z}+a+ib$ for some $a=2n+1$) to conclude the result. $\qquad\square$
Note: Since powers of $g$ and $t$  completely exhaust translations in the real and imaginary directions, and we have shown any $h$ not on these axes must be on the lattice formed by $g$ and $t$, it follows that $\Gamma$ is a lattice generated by a glide reflection $g$ and a perpendicular translation $t$ or a glide reflection $g$ and a glide reflection $gt$.
