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Let $G$ be a finite subgroup of the group of motoins $M$ on the plane. Then there exists a point on the plane which is left fixed by every element of $G$.

The proof of this is sketched by our instructor as follows $:$

Let $s$ be an element on the plane. Let $S$ be the set of all images of $s$ under the action of the elements of $G$ on $s$. Then every element of $G$ permutes the elements of $S$. To see this let us take $f \in G,s' \in S$. If we can show that $f(s') \in S$ then we are through. Since $s' \in S$ $\exists$ $g \in G$ such that $g(s) = s'$. Since $f,g \in G$ so $fg \in G$. Therefore $fg (s) \in G$ i.e. $f(s') \in S,$ as claimed. Since $G$ is finite, $S$ is also finite. Let $S = \{s_1,s_2, \cdots , s_n \}.$

After that our instructor asserted that the element $s^{*} = \frac {1} {n} (s_1 + s_2 + \cdots + s_n)$ is fixed by every element of $G$. This is the stage where I am struggling. Why $s^{*}$ is fixed by every element of $G$? If the element is rotation about origin or reflection about a line passing through origin then I have understood that the claim is true because they are linear operators which act linearly on $s^{*}$ and permute the $s_i$'s. Since vector addition is commutative we get $s^{*}$ back by acting them on $s^{*}$. Also clearly $G$ doesn't contain any proper translation since they are of infinite order. So what are the elements of $G$ other than rotations about origin and translations about a line through origin? I know that the elements of the group of motoins $M$ are either of the form $t_{a} \rho_{\theta}$ (which are precisely rotations about some point fixed by it ) or of the form $t_{a} \rho_{\theta}r$ (which are precisely glide reflections). So according to me the elements of $G$ is either of the form $t_{a} \rho_{\theta}$ or of the form $\rho_{\theta} r$. Do they all fix $s^{*}$? Please help me in this regard.

Thank you very much.

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  • $\begingroup$ I have proved it on my own at least. Thank you very much. $\endgroup$ – Dbchatto67 Dec 24 '18 at 16:01
  • $\begingroup$ I have proved that any rigid transformations take center of gravity to the center of gravity. $\endgroup$ – Dbchatto67 Dec 24 '18 at 16:03
  • $\begingroup$ To be explicit let $m \in M$ and let $S=\{s_1,s_2, \cdots , s_n \}$ be any finite collection of points on the plane. Let $s = \frac {1} {n} (s_1+ s_2 + \cdots + s_n)$ then $m(s) = s'$ where $s' = \frac {1} {n} (m(s_1) + m(s_2) + \cdots + m(s_n))$. $\endgroup$ – Dbchatto67 Dec 24 '18 at 16:09
  • $\begingroup$ In this case $m \in G$ and it permutes the elements of $S$. Since vector addition is commutative so we have $s'=s$, as claimed. $\endgroup$ – Dbchatto67 Dec 24 '18 at 16:11

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