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A dihedral group is the group of symmetries of a regular polygon, including both rotations and reflections. Here comes my question:

How can I show that rotations and reflections are the only symmetries of a regular polygon?

I have found no proof for such a statement when the dihedral groups are introduced in lots of textbooks. It might be a starting point to show first that an isometry (symmetry) on a regular polygon maps the vertices to vertices and then the center must be a fixed point of the isometry. But I get stuck with showing that reflections are the only possible isometries besides the rotations.

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A symmetry must map vertex 1 to some vertex $i$. Vertex 2 must be mapped to an ajacent vertex, $i+1$ or $i-1$. Once you know the images of 1 and 2, the other vertices are fixed. If 2 is mapped to $i+1$ you have the rotations. If 2 is mapped to $i-1$, you have the reflections (This might require some more arguments, at the elementary level at which this question is posed.)

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  • $\begingroup$ How can you formally show that a symmetry maps a vertex to a vertex? $\endgroup$
    – Paolo
    Jul 26, 2020 at 15:23
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    $\begingroup$ @Paolo, as you ask for formal proof, you have to first formally state what you mean by symmetry, and by vertex. Informally: if you see a polygon you can see where the vertices are. A symmetry maps the polygon to itself, so a vertex cannot move to a place where there was no vertex. $\endgroup$
    – Günter Rote
    Jul 28, 2020 at 5:52
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An isometry must fix the center of a polygon (thought of as a solid polygon) and so the only isometries of the plane which could induce a symmetry of the polygon are reflections, and rotations. (This assumes that we know the isometries of the plane are generated by translations, and reflections across lines through the origin)

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  • $\begingroup$ So one might also need to show that the isometries of a regular polygon must be induced from those of the plane? $\endgroup$
    – user9464
    Sep 3, 2013 at 1:25
  • $\begingroup$ I suppose that needs to be proved, but I imagine it follows rather easily from the fact that isometries of the circle extend to isometries of the disk, extend to isometries of the plane. $\endgroup$
    – Dan Rust
    Sep 3, 2013 at 1:27

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