$G$ is a subgroup of the Euclidean group which translational subgroup $T\triangleleft G$ (the isometries which are pure translations) is a lattice with full rank in Euclidean space. A finite subgroup $H$ of $G$ can not contain pure translations (because the order of a pure translation is infinite). But can a finite subgroup contain isometries with translational components (e.g. glide reflections)?

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    $\begingroup$ What is your exact notin of "has translational component"? $\endgroup$ – Hagen von Eitzen Jan 22 '13 at 16:56
  • $\begingroup$ An isometry in the Euclidean group consists of an orthogonal transformation and a translation. If the orthogonal transformation is the identity, then the isometry is a pure translation. If the orthogonal transformation is not the identity, then this is an isometry which either has a translational component or which translational component is zero. So the question was: can an isometry with an orthogonal transformation not the identity and a translational component not zero, be an element of a finite subgroup of $G$. $\endgroup$ – Wox Jan 23 '13 at 9:55
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    $\begingroup$ Ah, in that case you seem to have a distinguished origin in your Euclidean space. In that case, translational components are possible. More precisely, for every finite group there exixts a translation $\tau$ such that all elemnts are of the form $\tau g\tau^{-1}$ with $g$ orthogonal. If you swicth your origin from $O$ to $\tau O$, there is no translational component. $\endgroup$ – Hagen von Eitzen Jan 23 '13 at 18:17

Consider the group $\mathbb{Z}\rtimes \langle \tau\rangle$ where $\mathbb{Z}$ represents the group generated by the translation $t:x\rightarrow x+1$ and $\tau$ is the reflection $x\rightarrow -x$. Then the element $t\cdot\tau$ has order 2. Then $\{1,t\cdot\tau\}$ is a finite group of isometries of $\mathbb{R}$ and the element $t\cdot\tau$ has traslational component.

  • $\begingroup$ $t\tau$ is also just a reflection ... $\endgroup$ – Hagen von Eitzen Jan 22 '13 at 16:55
  • $\begingroup$ Thanks!,so, what should be the definition for an isometry of finite order to have translational component? $\endgroup$ – Diego Jan 22 '13 at 17:01
  • $\begingroup$ Great, thanks ! $\endgroup$ – Wox Jan 23 '13 at 10:03
  • $\begingroup$ I found another example in 3D Euclidean space. Consider the isometry $g\in E(3)$ which maps $[x,y,z]\mapsto [y-x,y,-z+\frac{1}{2}]$. The order of $g$ is 2 and $\{1,g\}$ is a finite subgroup. $\endgroup$ – Wox Jan 23 '13 at 10:16

Let $f$ have finite order $n$. Consider an orbit $x, f(x), f(f(x)),\ldots, f^{(n-1)}(x)$. Then the average $$y=\frac{x+ f(x)+ f(f(x))+\ldots+ f^{(n-1)}(x)}n $$ of these points is a fixed point of $f$. I think an isometry with at least one fixed point does not qualify as having a translational component.

  • $\begingroup$ Why is the average a fixed point? $\endgroup$ – anon Jan 22 '13 at 17:07
  • $\begingroup$ @anon Because $f$ is affine and $y$ is an affine combination (sum of coefficients is $1$), so applying $f$ yields $\frac{f(x)+f(f(x))+\ldots f^{(n)}(x)}{n}$. $\endgroup$ – Hagen von Eitzen Jan 23 '13 at 18:15
  • $\begingroup$ Well, $f:x\mapsto x+1$ is an affine map on $\bf R$, but $f(a+b)\ne f(a)+f(b)$. Affine doesn't imply linear does it? One would have to assume the "translational" component $b$ in the map $x\mapsto Ax+b$ was zero in order for $f$ to be linear. $\endgroup$ – anon Jan 24 '13 at 3:27

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