N_3 and N_4 periodic and pseudo Anosov auto-homeomorphisms It is well know that the genus three non orientable surface, N3, has only periodic and reducible auto-homeomorphisms, meanwhile the surface N4 is the first non orientable surface with pseudo Anosov maps. Also, recently profesor B.Szepietowski gave the MCG presentation of N4, from where, I calculated that there are seven periodic mapping classes. The question is: are there more?
Curiously, the torus and N3 show also seven periodic mapping classes each but we would like to understand better why N3 loses pseudo Anosov maps which contrast with the fact that  N3 is got from the 2-torus via an one-point blow up...
 A: Just to lend some context to the above question: the mapping class group of the two-torus  is naturally isomorphic to GL(2, Z).  If we restrict to orientation preserving homeomorphism the mapping class group is SL(2, Z).  The periodic mapping classes (isotopy classes of homeomorphisms) are exactly those with trace less than two in absolute value.  (Hmm, and +/- Id, I guess!)  Now we need to count the number of conjugacy classes of periodic elements.  There should be a cool algebraic way to do this.  (Perhaps it would help to give a purely algebraic proof that the order of torsion is at most 6?)
I think that there is a geometric way to do this: every periodic element occurs as the symmetry of some flat torus (= parallelogram with opposite sides identified).  All tori have have the hyperelliptic symmetry, corresponding to rotation by 180 degrees about any point. (These maps lie in the mapping class of the negative identity.)  Other symmetries:
Rombic tori have a reflection symmetry as do rectangular tori.
The square torus has a rotation by 90 degrees.
The hexagonal torus has a rotation by 60 degrees.
So I count:
1. the identity, Id 
2. the hyperelliptic = -Id = rotation by 180
3. rotation by 90
4. rotation by 60
5. rotation by 120
6. the reflection [[-1,0],[0,1]] (reflection in an axis) and
7. the reflection [[0,1],[1,0]] (exchange axes).
You can prove that the last two are distinct algebraically.  Perhaps the lack of 45 degree rotation is a geometric proof.
Now, we could perform similar geometric tricks to obtain symmetries of $N_4$ and get at least all of the rotations... [Edit: For example, it is possible to build a copy of $\rm{Sym}_4$ by placing the cross-caps at the vertices of a tetrahedron.]
