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Fraleigh(7th) Theorem 12.5: Every finite group $G$ of isometries of the plane is isomorphic to either $Z_n$ or to a dihedral group $D_n$ for some positive integer $n$. (Note: An isometry of $\mathbb{R}^2$ is a permutation of $\mathbb{R}^2$ that preserves distances.)

Proof: First we show that there is a fixed point by all of $G$. Let $G=\{f_1,\cdots,f_m\}$ and let $(x_i,y_i):=f_i(0,0)$. Then the point $P=(\bar{x},\bar{y})=\left( \dfrac{x_1+\cdots+x_m}{m},\dfrac{y_1+\cdots+y_m}{m}\right)$ is the centroid of the set $S=\{(x_i,y_i)\ |\ i=1,\cdots,m\}$. The isometries in $G$ permute the points in $S$ among themselves, since if $f_i f_j=f_k$ then $f_i(x_j,y_j)=f_if_j(0,0)=f_k(0,0)=(x_k,y_k)$. It can be shown that the centroid of a set of points is uniquely determined by its distances from the points, and since each isometry in $G$ just permutes the set $S$, it must leave the centroid $(\bar{x},\bar{y})$ fixed. Thus $G$ consists of the identity, rotations about $P$, and reflections across a line through $P$. ......

First, I can't understand the bold text. How can I change the phrase "the centroid of a set of points is uniquely determined by its distances from the points" into mathmatical words? And thus why the centroid is left fixed by all of $G$?

Second, What is the name of the above theorem? I saw some sites calling it Leonardo's theorem, but there is no search result of this name in wiki. Is there any other name?

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I think the statement about distances means the following. Denote $u_1, \ldots, u_n$ the points of $S$, $\overline{u}$ its centroid, and $d_i = |\!|\overline{u}- u_i|\!|$ the distance from $\overline{u}$ to $u_i$. Then the centroid of $S$ is the only point $v$ such that $|\!|v-u_i|\!| = d_i$ for all $1 \le i \le n$. To prove that, you need to see that the position of a point is already unique if you know its distance to two distinct points such that the three points form a line. –  Joel Cohen Jul 14 '11 at 6:21
    
@Joel: I thought that way first, too, but that's not sufficient to draw the claimed conclusion. Here you are allowed to permute the distances $d_i$, because the group action permutes the points $u_i$. The equation $||v-u_i||=d_{\pi(i)}$ may hold for some point $v\neq\overline{u}$, if $\pi$ is not the identity permutation. In fact, if $n=2$ it is easy to find counterexamples, but those are not relevant here. –  Jyrki Lahtonen Jul 14 '11 at 7:01
    
Rats. I meant that the solution $v$ to the system $||v-u_i||=d_{\pi(i)}$ is not unique for an arbitrary set of distances. –  Jyrki Lahtonen Jul 14 '11 at 8:30
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I think the statement about distances is basically what Joel Cohen said except strengthened by ignoring the indices. I.e. Fraleigh is claiming that the centroid is the only point in the plane whose set of distances from the points $(x_i,y_i)$ is equal as a set to the set of $d_i$. –  Ben Blum-Smith Jul 14 '11 at 21:15
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@Ben: Your argument depends on elements of $G$ being affine (or linear), so it is essentially my answer below. IOW it is perfectly possible for a bijective mapping of the plane to itself to permute a set of points without keeping its centroid fixed. It is not possible for an isometry to do that, but you have to prove that first. user6312 does prove this fact in his answer. May be Fraleigh is leading up to this somehow - I cannot tell without access to his book. –  Jyrki Lahtonen Jul 15 '11 at 8:49
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2 Answers

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Let $u_1$ be a point in the plane. Think of $u_1$ as a vector, so we won't have to use two coordinates all the time. In the quoted passage, $u_1=(0,0)$. Then the passage looked at the orbit of $u_1$ under $G$, and considered the centroid of this orbit. But we can work with any starting point, and the idea works in the same way for $3$-dimensional space.

So let the orbit of $u_1$ consist of the points $u_1$, $u_2$, $\dots$, $u_n$. (The orbit of $u_1$ is finite, since the group is finite.) An isometry in $G$ permutes the $u_i$.

The centroid $\overline{u}$ of the orbit is defined by $$\overline{u}=\frac{u_1+u_2+\cdots +u_n}{n}.$$

We want to show that $\overline{u}$ is a fixed point of every isometry in $G$.

Here is a proof, probably a lot longer than necessary. First we need a lemma.

Lemma Let $\phi$ be an isometry, and let $x$ and $y$ be points. Let $0 \le t \le 1$, and suppose that $z=tx+(1-t)y$. Then $\phi(z)=t\phi(x)+(1-t)\phi(y)$.

Proof: The isometry $\phi$ preserves distances. Let $d$ be the distance between $x$ and $y$. Then $z$ is at distance $(1-t)d$ from $x$ and distance $td$ from $y$. It follows that $\phi(z)$ is at distance $(1-t)d$ from $\phi(x)$ and distance $td$ from $\phi(y)$. However, since the distance between $\phi(x)$ and $\phi(y)$ is $d$, this means that $\phi(z)=t\phi(x)+(1-t)\phi(y)$, since there is exactly one point that is simultaneously at distance $(1-t)d$ from $\phi(x)$ and at distance $td$ from $\phi(y)$.

Theorem: Let $\phi$ be an isometry, let $u_1$, $u_2$, $\dots$, $u_n$ be points, and let $\overline{u}$ be their centroid. Then $$\phi(\overline{u})=\frac{\phi(u_1)+\phi(u_2)+\cdots \phi(u_n)}{n}$$ (or, more informally, $\phi$ preserves centroids).

Proof: By induction on $n$. The case $n=1$ gives a nice easy start, but the lemma also takes care of $n=2$. We show that if the result holds at $n$, it holds at $n+1$.

Let the centroid of the first $n$ points be $\overline{c}_n$, and let the centroid of the $n+1$ points be $\overline{c}_{n+1}$.

Note that $$\overline{c}_{n+1} =\frac{n}{n+1}\overline{c}_n +\frac{1}{n+1}u_{n+1}.$$ (This is easily verified by substituting the definition of centroid, but it is also physically obvious.)

Now we can apply the lemma, since $\overline{c}_{n+1}$ has been expressed in the form $t\overline{c}_n +(1-t)u_{n+1}$.

We find that $$\phi(\overline{c}_{n+1})=\frac{n}{n+1}\phi(\overline{c}_n) +\frac{1}{n+1}\phi(u_{n+1}).$$ Now by the induction hypothesis $$\phi(\overline{c}_n)=\frac{\phi(u_1)+\phi(u_2)+\cdots \phi(u_n)}{n}.$$ Simplify the right-hand side. We obtain the centroid of the $\phi(u_i)$.

Comment: Took longer than I thought it would! Probably wouldn't have done it if I had known. But it is done in gruesome detail, and I have a wordiness problem. There is undoubtedly a way to short-circuit all this. The above argument used only the definition of isometry.

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In isolation it does sound like the bold text needs a justification. But do you know at this point that any isometry of the plane is affine? In other words that the elements of the group are of the form $f_j:\vec{x}\mapsto L_j(\vec{x})+\vec{y}_j$ for some linear mapping $L_j$ (that actually also needs to be an isometry) and a constant vector $\vec{y}_j$? If you do, then the claim does follow, because the centroid is an affine linear combination of the points in the orbit with coefficients all equal to $1/m$, and that particular combination is stable. Write $\vec{x}_i=f_i(0,0)$. Then for all $j$ $$ f_j(P)=L_j(P)+\vec{y}_j=L_j\left(\frac{\sum_{i=1}^m\vec{x}_i}m\right)+\vec{y}_j =\frac1m\sum_{i=1}^m (L(\vec{x}_i)+\vec{y_j})=\frac1m\sum_{i=1}^mf_j(\vec{x}_i)=P, $$ because we can split the vector $\vec{y}_j$ into $m$ equal parts, and in the last step the mapping $f_j$ just permutes the vectors $\vec{x}_i$.

I'm not happy with this argument, because it sounds like there is something more elegant out there.

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I have posted something enormously less elegant, but that starts from roughly nothing. –  André Nicolas Jul 14 '11 at 6:01
    
@user6312: At least you start out with a Lemma proving that an isometry is affine on line segments, which is, of course, enough to reach the conclusion! –  Jyrki Lahtonen Jul 14 '11 at 6:27
    
Yes, I went at it kind of mechanically, starting with midpoint, and then grinding. –  André Nicolas Jul 14 '11 at 6:41
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