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Let $f$ and $g$ be rotations of the plane about distinct points, with arbitrary nonzero angles of rotation $\theta$ and $\phi$. Does the group generated by $f$ and $g$ contain a translation?

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Here I will try to work out $fgf^{-1}g^{-1}$ explicitly.

Let $\rho_\theta$ denote rotation about the origin, and WLOG take the point of rotation of $f$ be the origin and the point of rotation of $g$ be $a$. Then $$f=\rho_\theta$$ $$g=t_a\rho_\phi t_{-a}$$ Therefore we have $$f^{-1}=\rho_{-\theta}$$ $$g^{-1}=t_a\rho_{-\phi}t_{-a}$$ and hence $$fgf^{-1}g^{-1}=\rho_{\theta}t_a\rho_\phi t_{-a}\rho_{-\theta}t_a\rho_{-\phi}t_{-a}$$ By the relations $$\rho_\theta t_a=t_{\rho_\theta(a)}\rho_\theta$$ $$t_a\rho_\theta=\rho_\theta t_{-\rho_{-\theta}(-a)} =\rho_\theta t_{\rho_{-\theta}(a)}$$ we have $$fgf^{-1}g^{-1}=\rho_\theta t_a \rho_\phi t_{-a}\rho_{-\theta}t_a\rho_{-\phi}t_{-a}$$ $$=t_{\rho_\theta(a)}\rho_\theta \rho_\phi \rho_{-\theta}t_{\rho_{-\theta}(-a)}t_a\rho_{-\phi}t_{-a}$$ $$=t_{\rho_\theta(a)} \rho_\phi t_{\rho_{-\theta}(-a)+a} \rho_{-\phi}t_{-a}$$ $$=t_{\rho_\theta (a)}t_{\rho_\phi (\rho_{-\theta}(-a)+a)}\rho_\phi \rho_{-\phi}t_{-a}$$ $$=t_{\rho_\theta (a)+\rho_{\phi-\theta}(-a)+\rho_\phi(a)-a}$$ $$=t_{\rho_\theta (a)-\rho_{\phi-\theta}(a)+\rho_\phi(a)-a}$$

$fgf^{-1}g^{-1}$ yields the translation by displacement $$\rho_\theta (a)-\rho_{\phi-\theta}(a)+\rho_\phi(a)-a=f_\theta (a)-f_{\phi-\theta}(a)+f_\phi(a)-a$$

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Let $f$ and $g$ be the two rotations. The elements that we can obtain from them are products of the four elements $f$, $g$, $f^{-1}$, $g^{-1}$. We are looking for a product that is a translation. The simplest way to analyze the situation is to use the homomorphism $M \overset{\pi}{\to} O_2$ from the group $M$ of isometries to the orthogonal group. This homomorphism drops the translation from a product $t_\alpha \rho_\theta$, and keeps the rotation, sending that element to $\overline{\rho}_\theta$, the bar being put as a reminder. The kernel of $\pi$ is the group of translations.

If, for example, $\alpha$, $\beta$, and $\gamma$ are the angles of rotation about various points of some isometries $f$, $g$, and $h$, then$$\pi(fgh) = \overline{\rho}_\alpha\overline{\rho}_\beta\overline{\rho}_\gamma = \overline{\rho}_{\alpha+\beta + \gamma}.$$A product will be a translation if and only if it is in the kernel of $\pi$, which happens when the sum of the angles is zero. This being so, we try the commutator $fgf^{-1}g^{-1}$. Here, the sum of the angles is zero, so this is a translation.

However, we had better check that it is not translation by the zero vector. To check this, we rewrite the equation $fgf^{-1}g^{-1} = 1$ as $fg = gf$. When we check that $fg \neq gf$, we are done.

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  • $\begingroup$ Just a minor comment on this completely correct answer. There exist 2-generated groups $G$ of rotations in the Euclidean 4-space, such that no point in $E^4$ is fixed by $G$ and every element of $G$ is a rotation. $\endgroup$ Oct 21 '15 at 18:50
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Here my thoughts: If $\exists m, n\in \mathbb{N}$ such that $\psi:=m\phi=n\theta$ then you can rotate with angle $\psi$ around point one and then rotate back with angle $-\psi$ (which is ok since it's an inverse) around the other point - alltogether resulting in a translation when the points do not coincide.

EDIT: For the other case the answer is yes, too: rotate around point one with angle $\alpha$, rotate around point 2 with angle $\beta$, rotate around point 1 with angle $-\alpha$, rotate around point two with angle $-\beta$

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