Consider the dihedral group, that is, group of symmetries of a regular $n$-gon. Note that a rotation either leaves all points fixed (identity roation) or moves all points except the rotation axis.
Also note that a reflection fixes all the points on the line of reflection.
Using this I can argue why rotation composed with rotation is again a rotation: there is exactly one point that's fixed if we compose two rotations and that the axis of rotation so the composition of two rotations is again a rotation.
Similarly it is clear why two reflections composed are a rotation: each reflection fixes a line and the points that are fixed by the composition is the intersection of the two lines which is a point.
Now I tried to do the same to argue why a reflection followed by rotation (or vice versa) is a reflection. But the problem is that say, we have a non-identity rotation it moves all points except the centre of rotation. Then applying the reflection I don't see why this should fix a line.
How can I argue geometrically why rotation followed by reflection is
reflection?
Best Answer
Locate the line of reflection, and call it $L$. From the center of rotation, draw two lines such that the angle between the two lines is the angle of rotation, and such that $L$ bisects that angle.
When the rotation is first applied, it will carry one of the two lines you have drawn onto the other, and then when the reflection is applied, it will carry the line back to where it started. Whichever of those two lines it is, that is the line of reflection for the composition of the rotation with reflection.
You can reduce the second case to this case by noting that if $\sigma$ is a rotation and $\tau$ is a reflection, then you know that $\tau\sigma^{-1}$ is a reflection, so $\tau\sigma^{-1}=(\tau\sigma^{-1})^{-1}=\sigma \tau$. Since you've already understood where the line of reflection is for $\tau\sigma^{-1}$, that is exactly the line of rotation for $\sigma\tau$.