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In the group of orientation preserving isometries of the plane, can we characterize the subgroups generated by two non commuting elements ?

Denote an isometry by $$ \varphi(z) = e^{i\theta}z +b.$$ In the simple case of $\varphi_1(z)=z+1$, and $\varphi_2(z)=-z$, we get the infinite dihedral group. Can we say something about the general case ?

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If the two $\theta$-angles are not rational multiples of $\pi$, then you get all kinds of messy rotations (the set of angles of rotations will be dense). OTOH, if both the angles are multiples of $2\pi/m$, then more can be said. If the two generators have a common fixed point, then you get a finite dihedral group of a regular polygon around that common fixed point. Check that first! What happens is there isn't one? The answer likely depends on whether the (distinct) fixed points and their images fall on points of a nice lattice. Probably $m\mid 12$ is needed for something nice to happen. –  Jyrki Lahtonen Jan 4 '13 at 10:09
@JyrkiLahtonen: Thank you. The assumption of a common fixed point amounts to the assumption that the functions commute. Since the dihedral group is non-abelian, It cannot be the one generated by two such functions with a common fixed point. –  Teddy Jan 6 '13 at 11:04
More explicitly, assuming $z_0$ is a fixed point, we have $$ \varphi(z) = \varphi(z_0 + z - z_0) = e^{i\theta}z_0 + e^{i\theta}(z-z_0) + b = z_0 + e^{i\theta}(z-z_0).$$ Now having $z_0$ as a common fixed point for $\varphi_1$ and $\varphi_2$, we get two rotations around $z_0$, which surely commute, and generate a cyclic group. (assuming $\theta = q \pi$, for $q$ rational) –  Teddy Jan 6 '13 at 11:09
You're right. I was thinking about the group as a subgroup of a group generated by reflections, but there are no reflections around here. So yeah, a common fixed point implies that it is abelian. –  Jyrki Lahtonen Jan 6 '13 at 16:03

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