Elements in finite symmetry groups share a common fixed point

Question

I am reading Tapp's intro to matrix groups for undergraduates. On page 46 he states the following theorem:

For $X\subseteq \mathbb R^2$ if $Symm(X)$ is finite then it is isomorphic to $D_m$ or $\mathbb Z_m$ for some $m$.

Following it he writes:

The proof involves two steps. First, when $Symm(X)$ is finite its elements must share a common fixed point.

But he does not elaborate and it is not obvious to me. Why is this clear?

Answer 1

Neat, I read this book recently and liked it!

I think Tapp doesn't mean to signal that this should be obvious, he just cites the theorem. I found a detailed proof in Aarts' "Plane and Solid Geometry", p. 99. It proceeds by careful examination of available isometries in $R^2$ (translations, rotations, reflections and glide reflections). Here's a sketch:

1. $Symm(X)$ cannot contain translations (by finiteness).
2. $Symm(X)$ cannot contain a glide reflection, because one such is enough to construct a translation.
3. If it contains two rotations $R_1, R_2$ around different centers, show that you can combine them to obtain a translation. Conclusion: rotations, if any, are around the same center $O$.
4. If it contains a reflection and a rotation, the reflection line must pass through $O$, otherwise you can construct a glide reflection.
5. Putting it all together: if it has more than one reflection, it has at least one rotation as a product of two of them, and apply 3 and 4. If it has just one reflection and some rotations, apply 3 and 4. If just one reflection and no rotations, trivial.