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Last edited by hbghlyj 2023-3-12 01:15College Geometry: Using the Geometer’s Sketchpad
THEOREM 10.1 (Leonardo's Theorem) A finite symmetry group for a figure in the plane must be either the cyclic group $C_{n}$ or the dihedral group $D_{n}$.
This theorem says that the size of the symmetry group can vary, but that there are only two options for its type: symmetry based solely on rotations or symmetry with both rotations and reflections.
LEMMA 10.1 A finite symmetry group has a point that is fixed for every one of its symmetries.
Proof. The proof will use coordinates. Suppose the finite symmetry group is $\{f_1,⋯,f_n\}$, and for convenience let $f_1$ be the identity. Pick any point $P_1$ and calculate
\begin{array}{c}f_{1}\left(P_{1}\right)=P_{1} \\ f_{2}\left(P_{1}\right)=P_{2} \\ f_{3}\left(P_{1}\right)=P_{3} \\ \vdots \\ f_{n}\left(P_{1}\right)=P_{n}\end{array}
Form the new point
$$
P=\frac{P_{1}+P_{2}+\ldots+P_{n}}{n} .
$$
$P$ is called the center of gravity for this collection of points.
Now, $f_1(P) = P$ because $f_1$ is the identity isometry. What happens for $f_2(P)$? First think about what $f_2$ does to the list of points.\begin{array}{c}f_{2}\left(P_{1}\right)=f_{2}\left(f_{1}\left(P_{1}\right)\right)=\left(f_{2} \circ f_{1}\right)\left(P_{1}\right) \\ f_{2}\left(P_{2}\right)=f_{2}\left(f_{2}\left(P_{1}\right)\right)=\left(f_{2} \circ f_{2}\right)\left(P_{1}\right) \\ f_{2}\left(P_{3}\right)=f_{2}\left(f_{3}\left(P_{1}\right)\right)=\left(f_{2} \circ f_{3}\right)\left(P_{1}\right) \\ \vdots \\ f_{2}\left(P_{n}\right)=f_{2}\left(f_{n}\left(P_{1}\right)\right)=\left(f_{2} \circ f_{n}\right)\left(P_{1}\right)\end{array}We proved the compositions $\left(f_{2} \circ f_{1}\right),\left(f_{2} \circ f_{1}\right), \ldots,\left(f_{2} \circ f_{n}\right)$ are different from each other. This list of compositions fills the $f_2$ row of the group table with $n$ different results. Therefore the list $f_2(P_1),f_2(P_2),\ldots,f_2(P_n)$ contains all of the original $n$ points, though in a different order. Calculating the center of gravity for the points $f_2(P_1),f_2(P_2),\ldots,f_2(P_n)$ is thus the same as calculating the original center of gravity $P$. So the center of gravity remains fixed for $f_2$, i.e. $f_2(P) = P$.
The same reasoning works for $f_3,\ldots,f_n$, showing that this particular point $P$ is fixed for every symmetry in this group. □
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Since the members of a finite symmetry group have at least one fixed point, they must be either rotations or reflections. Even the identity falls into these categories, for the identity isometry can be interpreted as a rotation of 0°. Later we will examine symmetry groups that include translations and glide reflections.
Because of this lemma, we know that these will be infinite groups.
Proof of Theorem 10.1 Let us start small and work upwards.
Case 1 Suppose the finite symmetry group has a single rotation — which must be the identity — and has no reflections. This is the cyclic group $C_{1}$.
Case 2 Suppose the finite symmetry group has a single rotation and has one reflection. This is the dihedral group $D_{1}$.
Case 3 Suppose the finite symmetry group has a single rotation and has more than one reflection. Let us focus our attention on two of the reflections. Since these two reflections share a fixed point, their mirror lines intersect. Recall that the composition of two reflections in intersecting lines is a rotation. This cannot be the identity rotation (Why not?) so this case is impossible.
Case 4 Suppose the finite symmetry group has more than one rotation and has no reflections. Let $R_{\alpha}$ be the rotation with the smallest angle, modulo $360^{\circ}$. It turns out that every other rotation in this group can be generated by a power of $R_{\alpha}$. To prove this claim, assume that $R_{\theta}$ is a rotation in the symmetry group but that $R_{\theta}$ is not a power of $R_{\alpha}$. This means that $\theta$ is not a multiple of $\alpha$. Consequently $\theta$ lies between two consecutive multiples of $\alpha$. Expressed algebraically, there is a nonnegative integer $k$ so that $k \alpha<\theta<(k+1) \alpha$. Then $0<(\theta-k \alpha)<\alpha$. However, having $R_{\theta}$ in the symmetry group means that the rotation $R_{(\theta-k \alpha)}$ is in the group, since $R_{(\theta-k \alpha)}=R_{\theta} \circ\left(R_{\alpha}^{-1}\right)^{k}$. This contradicts the choice of $\alpha$ as the smallest positive angle for a rotation, and shows that $\theta$ must be a multiple of $\alpha$.
Let $n$ be the smallest power of $R_{\alpha}$ that equals the identity. This symmetry group is the cyclic group $C_{n}$.
Case 5 Suppose the finite symmetry group has more than one rotation and has at least one reflection. Call this reflection $F_{1}$ and once again let $R_{\alpha}$ be the rotation with the smallest angle, with $R_{\alpha}^{n}={\rm identity}$. Notice that $R_{\alpha}^{j} \circ F_{1}$ is an opposite isometry, so it must be a reflection. Every choice of $j$ produces a reflection, and every choice produces a different reflection. (Why is this true? Think about a column in a group table.)
Are these all of the reflections? Assume $F_{2}$ is a reflection not equal to $F_{1}$. The reflection $F_{2}$ has the same fixed point as $F_{1}$, and their mirror lines intersect there, so the composition $F_{2} \circ F_{1}$ is a rotation, say $R_{k \alpha}$. Then $F_{2}=R_{k \alpha} \circ F_{1}$, which is already accounted for.
Thus there are $n$ reflections in addition to the $n$ rotations. This is the dihedral group $D_{n}$. |
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