Even Isometries in neutral Geometry Consider a Hilbert plane as in Hartshorne's 'Euclid and beyond' (axiomatic geometry), and its group of isometries f or 'rigid motion' generated by line reflections. Call f 'even' if it is the product of an even number of reflections, 'odd' otherwise.
Can f be even and odd at the same time?
In the standard euclidean or hyperbolic model space over the reals the answer is well known, but there are of course many more less familiar models in the non-complete case. 
Let me add some remarks: the only complete (!) Hilbert planes are euclidean two space and the Poincare upper half plane (up to isometry). In both cases an isometry of the plane cannot be even and odd at the same time. Analytically even means oriented, odd means non-oriented. However, there are various rather unfamiliar examples of Hilbert planes in the non-complete case. Like 'semi elliptic planes' in the sense of Hartshorne's book, or archimedean examples but without limiting rays. That the real projective plane ('elliptic geometry') is non-orientable doesn't prove a thing but motivates the question. Another motivation the fact that simplifying products of reflections is a crucial step in Hilbert's theory of ends. 
Thanks in advance!
 A: I believe that the answer to your question "Can $f$ be even and odd at the same time?" is no, but the argument that I have seems more complicated than I expected it to be.  The idea is synthetically to imitate the notion of a 'local orientation' and show that it has global coherence.  Here is an outline:
Let $H$ be a Hilbert plane and let $G$ be its group of isometries.  For each point $P$ in $H$ let $G_P\subset G$ be the subgroup consisting of those isometries of $H$ that fix $P$.  
Let $R_P$ denote the set of rays in $H$ that emanate from $P$. Then $G_P$ acts transitively on $R_P$, and the stabilizer in $G_P$ of a ray $r \in R_P$ is a $2$-element subgroup consisting of the identity and the reflection in the line that is the union of $r$ and its opposite ray $-r$.
Now, let $C_P \subset G_P$ denote the set of elements that are either the identity or that fix no ray in $R_P$.  Then $C_P$ is a abelian subgroup of index $2$ in $G_P$, and its complement in $G_P$ is exactly the set of reflections in lines through $P$.  (It's easy to see that every element of $C_P$ is (not uniquely) the product of two reflections in lines through $P$.)
Now, for any two points $P$ and $Q$ in $H$, there is a canonical identification of $R_P$ with $R_Q$.  Namely, if $P=Q$, we take the identification to be the identity.  If $P\not=Q$, then consider the line $PQ$ and let $r\in R_P$ denote the ray emanating from $P$ that passes through $Q$.  This ray $r$ contains a unique ray $r'\in R_Q$ and we let $(-r)'=-r'\in R_Q$ denote the ray that corresponds to the opposite ray $-r\in R_P$.  For any other ray $s\in R_P$, we let $s'\in R_Q$ denote the ray that is on the same side of the line PQ as $s$ and for which the angle made by $r'$ and $s'$ is congruent to the angle made by $r$ and $s$.  Call this mapping $B_Q^P:R_P\to R_Q$.  Then $B_Q^P$ is a bijection and $B_P^Q$ is its inverse.  
If $O$, $P$, and $Q$ are any three points, then the composition $B_P^OB_Q^PB_O^Q:R_O\to R_O$ is the action of an element of $C_O$ on $R_O$.  (It is the identity if $O$, $P$, and $Q$ are collinear.)
Meanwhile, if $\rho:H\to H$ is a reflection in a line, then $\rho$ induces a map  on rays $\rho'_P:R_P\to R_{\rho(P)}$, and $B^{\rho(P)}_P\circ\rho'_P:R_P\to R_P$ is induced by an element of $G_P$ that is not in $C_P$.
Finally, one can now divide the elements of $G$ into two classes:  The even elements $f$ are the isometries such that $B^{f(P)}_P\circ f'_P:R_P\to R_P$
is induced by an element of $C_P$.  (This condition turns out to be independent of $P$.)  These form a subgroup $G_0\subset G$.  The odd elements, which contain all the reflections in lines, are what is left.  Moreover, it now follows that $G_0$ consists of products of an even number of line reflections, while its complement consists of all of the products of an odd number of line reflections.
The proofs of all the claims made in this outline (and they do require proof) are straightforward, though tedious.
