# Finite subgroups of motoins always fix some point on the plane.

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.

• I have proved it on my own at least. Thank you very much. – Dbchatto67 Dec 24 '18 at 16:01
• I have proved that any rigid transformations take center of gravity to the center of gravity. – Dbchatto67 Dec 24 '18 at 16:03
• 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 $s' = \frac {1} {n} (m(s_1) + m(s_2) + \cdots + m(s_n))$. – Dbchatto67 Dec 24 '18 at 16:09
• 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. – Dbchatto67 Dec 24 '18 at 16:11