What are the finite subgroups of isometries of a flat triangular torus? Let $\mathbb{Z}^2$ act on $\mathbb{R}^2$ as follows: $(1,0)$ acts by translating the plane by $(1,0)$, and $(0,1)$ acts by translating plane by $(1/2, \sqrt{3}/2)$. 
Now consider the torus $\mathbb{R}^2 / \mathbb{Z}^2$. What are the finite subgroups of the group of isometries? It seems like such a classification should be possible, from the classification of wallpaper groups.
I see that the following groups are possible:
(1) Any group $G \times H$ where $G$ and $H$ are cyclic or dihedral,
(2) or any subgroup of $D_{12}$, the dihedral group of order $12$.
Are there any other finite groups of isometries of this torus?
 A: One way to classify the ways that a finite group can act on the regular hexagonal torus $H$ is to first lift the action to the plane, then name the so-called wallpaper group $W$, and then determine whether (and in what ways) the fundamental group of $H$ can arise as a subgroup of $W$.  Using Dror Bar-Natan's illustrated list of the 17 wallpaper groups, the 3 that possess a 90 degree rotation are impossible in this context.  I think that the others are all possible, but there are subtle difference in group structure because of the distinction between a reflection and a glide reflection.  I.e., $W$ always has a translational subgroup $T$, which is a normal subgroup, and a rotational quotient group $R$, but it can be a non-split extension of $R$ by $T$, not just a semi-direct product.
Your list of candidates is not quite complete.  In the case that the orbifold quotient (of either the torus $T$ or the whole plane) is a Klein bottle, it means that you have glide reflections but no reflections.  In this case the finite group acting on $H$ can be a non-split central extension of a dihedral group by a cyclic group.  Besides, even after listing the finite groups that can act on $H$, a particular group in the list can act in more than one way, with subtle differences.   For instance I think that there are 5 ways that the group with 2 elements acts on $H$, because the quotient can be a Klein bottle (in two ways), an annulus (again in two ways), or a sphere with four cone points of order two (in one way).
