# 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:

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