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I am reading the book Abstract Algebra by Dummit and Foote.

In the section about the group $D_{2n}$ (of order $2n$) the authors claim that knowing where two adjacent vertices move to, completely determine the entire motion.

Based on this, the book gets the conclusion that $|D_{2n}|\leq2n$ and by showing existence of $2n$ different motions it concludes $|D_{2n}|=2n$.

My question is about the claim maid: Why does knowing where two adjacent vertices of a regular $n$-gon move to, completely determine the entire motion ?

I am looking for a convincing argument, not neccaseraly a formal proof, I just want to get the idea.

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  • $\begingroup$ There's the additional constraint that the $n$-gon is rigidly mapped onto itself. Thus all vertices are mapped to vertices, inner points to inner points, not-vertex boundary points to non-vertex boundary points. Then there aren't many possibilities left. $\endgroup$ – Daniel Fischer Oct 23 '13 at 18:38
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An Euclidean movement is determined by the effect on three (noncollinear) points. If $A\mapsto A'$, $B\mapsto B'$, $C\mapsto C'$, we can consider the movement composed of

  1. the translation along $\vec{A'A}$
  2. the rotation about $A$ that maps the translated image of $B'$ to $B$
  3. the reflection at $AB$ if necesary for the third point

Performing these movements after the given movement leaves $A,B,C$ fixed in the end and the only movement that leaves a (nondegenerate) triangle fixed is the identity.

In the case of a regular $n$-gon being mapped to itself, the center must remain fixed(!) Together with two vertices we thus know th eeffect of the movement on three noncollinera points as required.

Alternatively, if you feel uncomfortable about the center being ficed: If we know where a vertex $A$ and a neighbour(!) vertex $B$ go, we also know that "the other" neighbour $C$ (assuming the polygon has more than two vertices) of $A$ must be mapped to "the other" neighbour of the image of $A$

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  • $\begingroup$ Thanks! I understood both explanations. Seems that I have forgotten what I know about isometries of $\mathbb{R}^2$, thanks for the refresh $\endgroup$ – Belgi Oct 23 '13 at 18:59

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