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Let $V$ an euclidean space, $X$ a finite set of non-zero vertors in V and $\mathcal{H}$ be the set of hyperplanes of the form $a^{\perp}$ for some $a\in X$. Let $W$ be the group generated by reflecions $s_{H}$ for $H\in \mathcal{H}$.

If the set $X$ is invariant under the action of $W$ then is finite. My question is : Does the reciprocal is true ???

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Is your question «if $W$ is finite, is $X$ invariant under the action of $W$?»? –  Mariano Suárez-Alvarez Nov 2 '12 at 22:51
Obviously NO, take two vectors in the plane. If $X$ is the set of all unit normal vectors to all hyperplanes of reflections in $W$ then obviously YES. –  Anton Petrunin Nov 2 '12 at 23:38
Thanks Mrs Andres. –  Rajkarov Nov 2 '12 at 23:42
To elaborate on Anton's remark: If you take two vectors $a_1,a_2$ in the plane so that the angle between $a_1,a_2$ is not in ${\mathbb Q}\pi$, then the product of reflections in $a_1^\perp, a_2^\perp$ has infinite order. There are more interesting counter-examples in ${\mathbb R}^3$, where you can find three vectors $a_1,a_2,a_3$ so that each $\angle(a_i, a_j)$ is of the form $r_{ij}\pi$, $r_{ij}\in {\mathbb Q}$, but the reflection group is still infinite. To get a better idea of what is going on, read any book on Coxeter groups. –  Misha Nov 3 '12 at 5:01

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