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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?

learner
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1 Answers1

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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.
AnatolyVorobey
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  • I'm sorry... what is a glide reflection? – learner Nov 28 '14 at 23:32
  • a glide reflection is when you flip the plane around a line (reflection) and then slide it some distance along the same line (translation). Any isometry of $R^2$ must be one of 4 types: translation, rotation, reflection and glide reflection. http://en.wikipedia.org/wiki/Euclidean_plane_isometry#Classification_of_Euclidean_plane_isometries – AnatolyVorobey Nov 29 '14 at 11:08
  • In this case, wouldn't it also be okay to say (in 2.) that $Symm(X)$ cannot contain glide reflections by finiteness (like in 1.)? If you have one glide reflection $G$ then $G^n \neq G^{n-1}$ so if you had $G$ then the group would be infinite? – learner Dec 04 '14 at 03:26
  • it amounts to the same thing. The reason reflections can exist in $Symm(X)$ but glide reflections cannot is that a rotation when applied to itself again and again produces only a finite number of variants (that is, 2), but a glide reflection produces an infinite number; but that reason it does so is precisely that it has a translation "built in" in addition to reflection. – AnatolyVorobey Dec 04 '14 at 09:57
  • But... why is a glide reflection a fourth type of element? It is a combination of a reflection and a translation. Shouldn't the wikipedia article you link to say that there are 3 types of isometries of $\mathbb R^2$? –  Dec 05 '14 at 04:44
  • note it says "four types of isometries", not "four basic types of isometries out of which all others are made by combining" or something like this. A glide reflection is an isometry, and it's not one of the other three types, so that's why it's there. The surprising thing is that these 4 types are really it: if you combine them in various ways you just get them back. For example, a combination of two rotations around different centers is just another rotation (or a translation in one particular case). – AnatolyVorobey Dec 05 '14 at 22:36