I am studying " Abstract Algebra " written by dummit/foote. In page 23, This book defines the concept of dihedral group as follows:

For each n = 3, 4, 5, etc... , The set of symmetries of a regular n-gon, where a symmetry is any rigid motion of the n-gon which can be effected by taking a copy back on the original n-gon so it exaclty covers it.

I can understand what it means but I have some curious about this definition because this definition is not rigorous for me. What is the definition of rigid motion? what is copy back? Can we admit this kind of vague terminology in math?

I tried to find other book which describing the concept symmetry. In the book " A first course in Abstract Algebra, this book defines a symmetry of geometrical figure as a rearrangement of the figure preserving the arrangement of its sides and vertices as well as its distance and angles. I thought that this definition is also depending on our intuition... and not rigourous if we compare this with set theory, real analysis, or other mathematical definition...

So, In summary... Q1. I understood that symmetry is a transformation which preserves shape, angle, distances... in intuitive meaning Did I understand well? Q2. If I understood well, what is a precise and rigorous definition of symmetry? as we did in set theory, real analysis, etc... Q3. Can we deal geometry by using a precise and rigorous method only using axiom, set, logic ??

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    $\begingroup$ As far as dihedral groups are concerned, a precise definition of $D_{2n}$ is the group generated by two elements $r$ and $s$ subject to the relations $r^n=s^2=(sr)^2=1$ $\endgroup$ – leibnewtz Apr 1 '18 at 6:11
  • $\begingroup$ @leibnewtz I don't think that is a satisfactory way to define the dihedral groups. They are one of the first class of examples encountered in elementary group theory courses, whereas the theory of group presentations is comparatively advanced and generally not treated formally in elementary courses. $\endgroup$ – Derek Holt Apr 1 '18 at 10:17
  • $\begingroup$ @DerekHolt I think that indicates a problem with elementary group theory courses rather than a problem with the definition. Besides, the OP says they're learning this material from DF, which gets into presentations quickly $\endgroup$ – leibnewtz Apr 1 '18 at 16:24
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    $\begingroup$ Congratulations for asking for/about rigor. But when learning, understanding precedes rigor. This definition seems to me just fine for page 23 of a beginning abstract algebra book. You can see just what the symmetries are (rotations and reflections) and begin to grasp the shape of the dihedral group. @fredgoodman 's answer shows how this informal geometric definition can be made rigorous. $\endgroup$ – Ethan Bolker Apr 1 '18 at 16:57
  • $\begingroup$ @EthanBolker For me at least, when I'm learning a new subject it's often easier to first learn rigorous and formal definitions, and then to build up an intuition for the subject at hand. When it comes to research however, the story is not so simple... $\endgroup$ – leibnewtz Apr 2 '18 at 5:12

I agree with your assessment- much of the language used when introducing symmetry is hand-wavy and imprecise. When you see the word "symmetry," you should think "automorphism." That is, a bijection from an object to itself that preserves the underlying relations.

The Dihedral group is the automorphism group of the cycle graph. Formally, the cycle graph $C_{n}$ (for $n \geq 3$) has vertex set $[n] = \{1, \ldots, n\}$, and edge set: $$\{ \{i, i+1\} : i \in [n-1] \} \cup \{ \{ 1, n\} \}$$

That is, we label the vertices of $C_{n}$ using the labels $1, 2, \ldots, n$. A vertex labeled $i$ is adjacent to the vertices labeled $i-1$ and $i+1$, where the indices are taken modulo $n$.

Now a graph automorphism of $G(V, E)$ is a bijection $\varphi : V(G) \to V(G)$ such that if the edge $ij \in E(G)$, then $\varphi(i) \varphi(j) \in E(G)$. That is, a graph automorphism is a bijection on the graph's vertex set and also a graph homomorphism. (Notice that a graph isomorphism and group isomorphism are defined similarly.)

The Dihedral group is formally defined as: $$ D_{2n} = \langle r, s : r^{n} = s^{2} = 1, rs = sr^{-1} \rangle $$

Here, $r$ is the rotation map $r : V(C_{n}) \to V(C_{n})$ given by: $r(i) = i+1$, where the indices of the cycle are again taken modulo $n$. Now $s : V(G) \to V(G)$ is the reflection map, which fixes vertex $1$ and swaps vertices $2$ and $n-1$. The reflection map also swaps the other vertices of $C_{n}$ appropriately (e.g., $3 \leftrightarrow n-2, 4 \leftrightarrow n-3$, etc.), so that $s$ is a graph homomorphism.

When working with the Dihedral group abstractly, the presentation above is often enough. To get a better feel for how $D_{2n}$ acts on the vertices of the cycle graph $C_{n}$, it is helpful to use the permutation representation of $D_{2n}$. We can write: $$r = (1, 2, 3, \ldots, n) $$

And: $$s = \prod_{i=2}^{\lfloor n/2 \rfloor} (i, n-i+1)$$

The permutation representation is more precise and concise than my description of the two maps.

A final point- I did not formally prove that $D_{2n} \cong \text{Aut}(C_{n})$, and nor does Dummit and Foote in Chapter 1. To do so formally, you need the Orbit-Stabilizer theorem. Though you can likely convince yourself that $D_{2n} \cong \text{Aut}(C_{n})$ intuitively.

I hope this helps!

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    $\begingroup$ This is correct, but doesn't address the spirit of the OP's question, which asks if the geometric definition s/he's been given is rigorous or rigorizable. $\endgroup$ – Ethan Bolker Apr 1 '18 at 16:51
  • $\begingroup$ I think this answer shows that $D_{2n} \cong \langle s, r \rangle \leq S_n$, right? That is, the dihedral group of order $2n$ is a subgroup of $S_n$. $\endgroup$ – EpsilonDelta Jul 3 at 13:57

To begin with, you haven't accurately quoted the passage in D & F, page 23. They write "a symmetry is any rigid motion of the $n$-gon which can be effected by taking a copy of the $n$-gon, moving this copy in any fashion in 3-space and placing the copy back on the original $n$-gon so it exactly covers [the original copy]."

This is intuitive to you because you can imagine actually doing it physically with two paper models of the $n$-gon, one taped to your table and one free to move in space.

It also has an absolutely precise mathematical meaning. You are to consider "rigid motions" of all of 3-space which carry the $n$-gon to itself, i.e. rigid motions $T : \mathbb R^3 \to \mathbb R^3$ such that $T(P) = P$, where $P$ denotes the $n$-gon.

And what is a rigid motion of $\mathbb R^3$? It is an orientation preserving isometry, equivalently a map $x \mapsto A x + b$, where $A$ is a rotation of 3-space. Moreover, a rotation of 3-space is a linear isometry with determinant equal to 1.

Now one can show that if $T$ is a rigid motion of $\mathbb R^3$ taking $P$ to $P$, then $T$ must take vertices of $P$ to vertices and must take the center of mass of $P$ to itself. Without loss of generality, put the center of mass at the origin. Then $T$ must fix the origin and therefore $T$ is a rotation of 3-space.

Thus one ends up with the following:

A symmetry of the $n$-gon $P$ is a rotation of 3-space that takes $P$ onto $P$.

Alternatively, one could define symmetries of $P$ just to be isometries of $P$, regarded as a metric subspace of Euclidean 3-space. But one can show that each such symmetry extends to a rigid motion of 3-space, so this does not result in a more general notion of symmetry.

Reference: this book, sections 1.4, 2.3, 11.1.


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