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Consider the following isometries in the Euclidean group of distance preserving maps of the plane $\mbox{Iso}(\mathbb R^2)$ which is generated by the following:

  1. Rotations $\rho_\theta$ about the origin by an angle of $\theta$.
  2. Reflection $r$ about $x$-axis.
  3. Translation $t_a$ by a vector $a$.

I wish to prove that the group $O_2=\{f\in\mbox{Iso}(\mathbb R^2):f(0)=0\}$ is generated by $\{r,\rho_\theta:\theta\in\mathbb R\}$. What puzzles me is as to what prohibits a term such as $t_a\circ f_1\circ\cdots\circ f_n\circ t_b$ from being in $O_2$.

It is not that $O_2$ is abelian that we may use the commutative property to rewrite this term as a translation followed by some rotations/reflections following which we may obtain a translation in $O_2$ which is clearly an absurdity.

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    $\begingroup$ There are "such terms" in $O_2$, but any such term can also be written without translations. (Say $T$ is linear. Then $T(x+p)-q=Tx$ if $q=Tp$.) $\endgroup$ May 9 '16 at 16:36
  • $\begingroup$ The more interesting question is: Why is $Iso(R^2)$ generated by 1, 2, 3? $\endgroup$ May 9 '16 at 18:45
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It is not that a composition of the form $t_a\circ f_1\circ\cdots\circ f_n\circ t_b$ could not possibly be in $O_2$, but in that case it is equal to $f_1\circ\cdots\circ f_n$. This is because the subgroup of translations is a normal subgroup, so that for every operation $f$ and every translation $t_a$ there is some vector $a'$ such that $t_a\circ f=f\circ t_{a'}$. Then using such transformations you can "chase" the original translation $t_a$ all the way to the right (changing it to a different translation at each step), where it combines ultimately with $t_b$ to a single translation. Moreover since the resulting translation is now the only one that can possibly move the origin, it must be be the zero translation if the whole composition is to fix the origin.

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