Dihedral groups: Relationship between symmetries and rigid motions

Question

In Dummit & Foote, $D_{2n},\ n \geqslant 3$ is the set of symmetries of a regular $n$-gon, where a symmetry is a 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 then placing the copy back on the original $n$-gon so it exactly covers it.

Question 1: What does it mean by moving this copy in any fashion? How can I find the inverse of every rigid motion?

Using this definition, the book calculates the order of $D_{2n}$, it reasons as follows:

Given any vertex $i$, there is a symmetry which sends vertex 1 into position $i$. Since vertex 2 is adjacent to vertex $1$, vertex 2 must end up in position $i+1$ or $i-1$ by some symmetry. Thus there are $2n$ positions the ordered pair of vertices 1,2 may be sent to upon applying symmetries.

Question 2: It is clear that in this case I can achieve those symmetries by rotating the $n$-gon or first reflecting then rotating the $n$-gon. But when I consider the group of the rigid motion of a tetrahedron using the same reasoning, how can I justify that every symmetry can actually be achieved by some sort of rigid motion?

Answer 1

Given an object $X$ that lives in some Euclidean space $\Bbb R^n$ (that is, $X \subseteq \Bbb R^n$) for some $n$, the symmetries of $X$ are defined as isometries (or "rigid motions") of $\Bbb R^n$ that send $X$ to itself.

An isometry of $\Bbb R^n$ is a distance-preserving function from $\Bbb R^n$ to itself. That's why the motion is "rigid," no distances are changed in the process of transforming (unlike transforming a piece of paper into a crumpled up ball, or a circle to an ellipse).

In other words, a function $T\colon \Bbb R^n \to \Bbb R^n$ is an isometry if $d(x, y) = d(T(x), T(y))$ for any points $x, y \in \Bbb R^n$, where $d(x, y) = \sqrt{(x_1 - y_1)^2 + \ldots + (x_n - y_n)^2}$ is the usual distance function of $\Bbb R^n$.

Given a set $X \subseteq \Bbb R^n$, we say an isometry $T$ is a symmetry of $X$ if $T(X) = X$ (which doesn't necessarily mean that $T$ fixes each point of $X$, only that the overall "footprint" of $X$ isn't changed).

In the case of $\Bbb R^2$, the (non identity) isometries come in four fun flavors:

• Rotations, which have exactly one fixed point and preserve orientation (i.e., send clockwise loops to clockwise loops),

• Reflections, which have a line of fixed points and reverse orientation (clocks in mirrors tick counterclockwise),

• Translations, which have no fixed points and preserve orientation, and

• Glide reflections, which have no fixed points and reverse orientations. A glide reflection is a composition of a reflection and a translation in the same direction of the axis of reflection. Footprints are an example of glide reflections in action:
  

In general, the inverse of an isometry is an isometry of the same geometric type: To undo a rotation, rotate backwards. Reflections are self-inverse. To undo a translation, translate backwards. I'll let you think about how to undo glide reflection like "flip across the $x$-axis and shift right $1$ unit".

I'll note that in $\Bbb R^2$, objects with only finitely many symmetries like regular $n$-gons (and unlike an infinite checkerboard) have only rotations and reflections as symmetries. This also happens to be true of the tetrahedron (but not of the cube. If the cube is $[-1, 1]^3$, then the antipode map $x \mapsto -x$ is a symmetry of the cube, but it's neither a rotation nor reflection, but instead inversion in a point).

The idea of picking up an $n$-gon and putting it down in the same place is a good mental picture, but it's not something that makes it into the actual definitions.

Answer 2

One way is to think of $D_{2n}$ is permutations of the vertices of the (regular) $n$-gon (that is elements of $S_n$) that preserve adjacency.

For example, with $n = 5$, suppose we send $1 \mapsto 3$. Since the second vertex is adjacent to the first, we have to send $2$ to either $2$ or $4$. Let's examine both possibilities:

If $2 \mapsto 2$, then $5$ which started out adjacent to $1$, has to map to $4$ (since the possible adjacent vertex positions to $3$ are $2$ and $4$, and $2$ is already taken). In a similar fashion the other vertex which started out adjacent to $2$ (that is, $3$) must map to $1$ (since $3$ has already been mapped onto by $1$).

So we have:

$1\mapsto 3\\2 \mapsto 2\\3 \mapsto 1\\4 \mapsto 5\\5 \mapsto 4$

which is the element $(1\ 3)(4\ 5) \in S_5$ (we have explicitly found out where four vertices go, and the last has to go the the one "left over" since permutations are bijective maps).

On the other hand, if $2 \mapsto 4$, then preserving adjacency means we must have $3 \mapsto 5$, $4 \mapsto 1$ and $5 \mapsto 2$. This is the permutation:

$(1\ 3\ 5\ 2\ 4) = (1\ 2\ 3\ 4\ 5)^2$.

A similar discussion holds for $1 \mapsto k$, for each $k \in\{1,2,3,4,5\}$ We find that we either get the identity, a five-cycle, or a double swap. The five-cycles are the rotations (which preserve adjacency and also "before and after"), and the double swaps are the reflections (which preserve adjacency, but reverse "before and after").

For the general $n$-gon there are a few wrinkles: $D_{2n}$ behaves slightly differently for even and odd $n$. Although the entire group is still generated by an $n$-cycle and a product of disjoint double-swaps (for $n > 3$), powers of an $n$-cycle may not be $n$-cycles, for they may break down into "sub-cycles" for powers that are a divisor of $n$. If $n$ is odd, the "swaps" (reflections) fix a single vertex, if $n$ is even they fix a pair of opposite vertices.

One can also concretely visualize these transformations in $\Bbb R^2$ as $2 \times 2$ rotation matrices about the origin (through angles of $2k\pi/n$ for $k = 0,1,\dots,n-1$), together with each rotation times the $x$-axis reflection:

$R_x = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$.