# Let $R$ be any rotation and $P$ any reflection then $R \circ P$ and $P \circ R$ are both glide reflections

Let $R$ be any rotation and $P$ any reflection then $R \circ P$ and $P \circ R$ are both glide reflections

I am having trouble showing $P \circ R$ is a glide reflection, I manage to get $R \circ P$, so I shall show my work.

For $R \circ P$ :

Let $R$ be a rotation where $R=R_{A,\alpha}$ where A is the center of rotation and $\alpha$ is the angle of rotation. Let $p$ be the reflection $p=p_k$ where $k$ is a line. Now let the line which goes through $A$ be parallel to $K$ and let the line $n$ be another line through $A$ whose oriented angle from $m$ is $\frac{\alpha}{2}$

Then observe:

Since rotations can be expressed as two reflections, we shall substitute:

$R \circ P=(p_n \circ p_m) \circ p_k=p_n \circ (p_m \circ p_k)$

Since $p_m \circ p_k$ is a translation because lines m and k are parallel. As a result, a reflection composed with a translation is a glide reflection.

For $P \circ R$ :

I used the same as above, so this might sound a bit redundant:

Let $R$ be a rotation where $R=R_{A,\alpha}$ where A is the center of rotation and $\alpha$ is the angle of rotation. Let $p$ be the reflection $p=p_k$ where $k$ is a line. Now let the line which goes through $A$ be parallel to $K$ and let the line $n$ be another line through $A$ whose oriented angle from $m$ is $\frac{\alpha}{2}$

Then observe:

Since rotations can be expressed as two reflections, we shall substitute:

$P \circ R=p_k \circ (p_n \circ p_m)=(p_k \circ p_n) \circ p_m$

I'm not sure where to proceed from there because $p_k \circ p_n$ is not a translation.

Here is the pic: ## 1 Answer

Any isometry which is orientation-reversing and not a reflection is a glide reflection. Since any reflection is orientation-reversing, and any rotation is orientation-preserving, their combination is orientation-reversing as well. So essentially all you have to show is that the operation is no reflection.

But if you want to follow the approach you already have, then for the second case choose $n$ as rotated by $-\frac\alpha2$ against $m$. Then you have

$$P \circ R=p_k \circ (p_m \circ p_n)=(p_k \circ p_m) \circ p_n$$

But this doesn't free you of the need to consider the special case of a reflection, unless you treat reflections as a special case of glide reflections. Indeed, if the center of rotation lies on the axis of reflection, then the net effect will be a simple reflection, with zero glide. So you probably want to include reflections as a possibility here. Of course it depends on how you define glide reflection.

• Makes sense. For the part where you said: Any isometry which is orientation-reversing and not a reflection is a glide reflection. So essentially all you have to show is that the operation is no reflection is basically the summary of what you said below. Since, $-\frac{\alpha}{2}$ is being oriented reversing and as a result we get a translation which is not a reflection. – Mark Apr 15 '15 at 18:17
• @Mark. Every reflection is orientation-reversing: it turns a triangle with counter-clockwise orientation into one with clockwise orientation. So an odd number of reflections is orientation-reversing, an even number orientation-preserving. A rotation is orientation-preserving, so combining that with a reflection will be orientation-reversing. – MvG Apr 16 '15 at 6:38