Classification of Euclidean plane isometries I suppose this question has already been asked here, but I cannot find it. 
Is there any simple way to prove that there are 5 possibilities for isometries in the Euclidean plane? Namely: Identity, Reflection, Rotation, Translation and  Glide reflection. 
Articles I've read (including wikipedia) simply say that it can be proved, but I am curious about how to prove that. Could you recommend me a book, etc where this subject is described? Or post the proof here?
 A: Any isometry can be fully described by its effect on three points, unless these lie on a single line. This is the case because for any given point, knowing the distances to the three defining preimage points will define three circles of corresponding radius about the image points. Two of them will intersect in two points, and the third will select one of these two points of intersection. So now you have some tool to describe all isometries.
Next you get every image point in place one after the other. You can map the first point using a translation, preserving the relative position of the second and third point in the process. Next you can map the second point using a rotation around the image of the first, since its distance to the first point is preserved. Unless the third point is already in its final position, you can use a reflection in the axis spanned by the first two image points to get it there. So taken together, any isometry can be described by a translation (perhaps by a zero displacement), followed by a rotation (perhaps by $0°$) possibly followed by a reflection.
Now all you have to do is analyze the possible ways these three steps can interact. A non-zero translation and a non-zero rotation together are simply a rotation around some other center. A rotation followed by a reflection is a reflection in some different line. A translation followed by a reflection in an axis parallel to the direction of the translation is a glide reflection. If the axis of reflection is not parallel to the direction of the translation, the combination is a reflection in a different axis. And if all steps are identity transformations, then so is the result.
A: Here's a sketch of how it could be proved using classical coordinate-free geometry:
First divide according to whether there's any fixed point of the isometry (i.e. any point that the isometry maps to itself).
If there is at least one fixed point, then it is easy to see that the the isometry must either be a rotation about that point (possibly by 0°, that is, the identity), or a rotation followed by a reflection about a line through the fixed point. And in the latter case, such a composition of a rotation an reflection always amounts to a reflection about some (usually different) line.
Suppose now there is no fixed point. We can then first prove that there exists collinear points $X$, $Y$ and $Z$ such that $X$ maps to $Y$ which maps to $Z$. To see this, select an arbitrary point $A$, and let the isometry map $A$ to $B$, $B$ to $C$, and $C$ to $D$. If $A$, $B$ and $C$ are collinear, then we're done. Otherwise, triangles $ABC$ and $BCD$ are congruent and isosceles. The centers of the circumcircles of $ABC$ and $BCD$ both lie on the perpendicular bisector of $BC$, and have the same distance to $BC$. They cannot be on the same side of $BC$, because then they would be the same point, which would then map to itself, contrary to assumptions. So they are on opposite sides of $BC$, which implies that $A$ and $D$ are on opposite sides of $BC$.
Therefore (by vertical angles) the midpoints of $AB$, $BC$, and $CD$ are collinear, and they clearly have to map to each other.
Thus in any case, an isometry without fixed points must have collinear $X\ne Y\ne Z$ such that $X\mapsto Y \mapsto Z$. Then we must also have $X\ne Z$ because otherwise the midpoint of $XY$ would be equal to the midpoint of $YZ$ which it maps to, and we're assuming no fixed points. In other words, $Y$ is between $X$ and $Z$.
Now it is easy to see that the action of the isometry of any point on the line $XYZ$ must be to translate it by a distance of $|XY|$ along the line -- because that's the only way for it to preserve its distances to both $X$ and $Y$ as they translate along the line.
Furthermore, once the action of the isometry on that line is given, there are only two posible images of each point in the plane outside the line -- one on the same side of the line, and one on the opposite side. Because the isometry is continuous, it must be the same choice of "same side" or "opposite side" for all points on one side of the line, and then, by injectivity, also for points on the other side. "Same side" means that the isometry is a translation of the plane; "opposite side" is a glide symmetry.

Excercise: Why doesn't this argument work for the hyperbolic plane? Does that mean that the hyperbolic plane admits additional conjugacy classes of isometries? (Hint: yes.) Describe them.
A: See Affine Maps Euclidean Motions and Quadrics, Springer SUMS.
