# Frieze group p2mg

I'm looking at Frieze groups. Wikipedia went to the point to depict how the basic pattern looks like.

I'm trying to understand the next claim about $$p2mg$$ group:

(TRVG) Vertical reflection lines, Glide reflections, Translations and 180° Rotations: The translations here arise from the glide reflections, so this group is generated by a glide reflection and either a rotation or a vertical reflection.

This image, I suppose, come to explain how the basic pattern of $$p2mg$$ looks like.

From the quote above, it seems that I can apply generators to the basic pattern and get (maybe a translated) basic pattern back.

In class, we denoted the basic glide reflection $$\gamma : (x, y) \mapsto (x + \frac{1}{2}, -y)$$. (Basic pattern is centered at $$(0, 0)$$ and have sizes of $$1/2 \times 1/2$$).

So using this definition I can see how glide reflection acts on the pattern, and it is mapped to itself. So far so good.

I then, can easily see why a vertical reflection $$v : (x, y) \mapsto (-x, y)$$ also maps the basic pattern to itself.

What am I failing to see is how the rotation ($$180^{\circ}$$) $$r: (x, y) \mapsto (-x, -y)$$ acts on the pattern.

It seems, that rotation acts on the left and right halves of the basic pattern independently.

For comparison I paste here the images for basic pattern of $$p2$$ (rotation + translations)

and of $$p11g$$ (Glide reflections only)

In both I can see how generators acts on the pattern (with translation just mapping to the (one of the) next pattern(s)).

So, my question is

How can we see that either used by Wikipedia indeed describes the same group of isometries?

Cause it seems, that Glide Reflection + Vertical Reflection is not the same thing as Glide Reflection + Rotation.

• The centres of 180 rotation symmetry are shown by the green dots. The lines of reflection symmetry are shown by the blue lines. Note how those dots and lines alternate, and never coincide. This is because combining a rotation around a point with a reflection about a vertical line through that point gives a reflection through a horizontal line through the point (symmetry which this frieze does not have). Your algebraic formulation of reflection is for reflection through the y-axis, and similarly your rotation is around the origin. You need to shift one or the other so that they don't coincide. Jan 14, 2020 at 12:53

I will consider the following pattern, with the origin marked by a star *:

 A     C     E     G
/\    /\    /\    /\
o  o  o  o  *  1  2  3  4
\/    \/    \/    \/
B     D     F     H


The negative numbers $$-4$$, $$-3$$, $$-2$$, $$-1$$ are not shown, just place them correspondingly in the o places, using a reflection w.r.t. the origin $$0=*$$.

• Let us denote by $$R$$ the $$180^\circ$$-rotation around the origin $$0$$. It moves in the same time $$E\to D$$, $$1\to -1$$, $$F\to C$$, $$2\to-2$$, and so on. It can be seen as a reflection w.r.t. origin, since $$R(x,y)=(-x,-y)$$. In particular, it preserves orientation.
• Let us denote by $$V$$ (vertical mirror) the reflection in the vertical line through $$E$$. It moves in the same time $$E\to E$$, $$1\to 0$$, $$F\to D$$, $$2\to 0$$, and so on. It changes the orientation of the plane.
• Let us denote by $$H$$ (horizontal mirror) the reflection in the horizontal line through $$0,1,2,3,\dots$$. It moves in the same time $$E\to$$(mid point of $$DF$$), and invariates $$0,1,2,3,\dots$$. It changes the orientation of the plane.
• Let us denote by $$T$$ the translation in the direction of the horizontal axis by $$1/2$$. So to move in the same time $$E\to G$$, $$1\to 3$$ we use $$TTTT=T^4$$. (So $$T$$ is not a symmetry of the pattern, but $$T^4$$ is.)

Wikipedia claims on https://en.wikipedia.org/wiki/Frieze_group "The translations here arise from the glide reflections, so this group is generated by a glide reflection and either a rotation or a vertical reflection." And indeed:

• Compozing the above with itself we get $$(HTT)(HTT)=TTTT$$. (It preserves the orientation and moves $$0\to 2$$, $$2\to 4$$, etc. So the translations $$T^4$$ are obtained in this way.
• Let us start with $$HTT$$ and $$R$$. Then \begin{aligned} (HTT)R\ 0 &= HTT\ 0 = 1\ ,\\ (HTT)R\ 1 &= HTT\ -1 = 0\ ,\qquad\text{ and in general}\\ (HTT)R\ k &= HTT\ -k = -k+1\ ,\ k\in\Bbb Z\\[2mm] (HTT)R\ E &= HTT\ D = E\ ,\\ (HTT)R\ C &= HTT\ F = G\ ,\\ \end{aligned} so $$(HTT)R$$ is $$V$$. The relation $$(HTT)R=V$$ shows that starting with the two elements $$HTT$$, $$R$$ we can obtain $$V$$ as $$(HTT)R$$, and that starting with the two elements $$HTT$$, $$V$$ we can obtain $$R$$ as $$R=(HTT^{-1})V\ .$$ This means $$\langle\ HTT\ ,\ R\ \rangle = \langle\ HTT\ ,\ V\ \rangle \ .$$
• So basically you define a vertical mirror at $n+\frac{1}{2}$ points. Back to the Wikipedia pattern image - how do you place the pattern around the origin, so it is either preserved or translated by the rotation? Jan 14, 2020 at 14:33
• The pattern on the wiki page has a long line, the Ox axis, this is not drawn above to have a light picture, it would be the line $Ox$ through all points $0,1,2,3,\dots$ and the total pattern has a $\Lambda$ (series) pattern above the $Ox$, and a $V$ (series) pattern below it, with a slight shift. I picked one vertical mirror $V$. (At one point, corresponding to the vertical line through the point $E(1/2, 1/2)$ in the picture.) The rotation $G$ of $Ox$ by $180^\circ$ is again $Ox$ with reversed orientation. The $\Lambda$ at NE w.r.t. the origin $0$ if moved to the $V$ at SE w.r.t. the origin. Jan 14, 2020 at 15:15