# Platonic Solids

It´s a theorem that there exist only five platonic solids ( up to similarity). I was searching some proofs of this, but I could not. I want to see some proof of this, specially one that uses principally group theory.

Here´s the definition of Platonic solid Wikipedia Platonic solids

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What definition of platonic solid do you have in mind? – Mariano Suárez-Alvarez Apr 6 '12 at 2:47
(There is no proof that only uses group theory... Any proof depends on geometric input) – Mariano Suárez-Alvarez Apr 6 '12 at 2:48
Ok I edited, you are right – Pilot Apr 6 '12 at 2:50
@MarianoSuárez-Alvarez What is your reasoning for claiming there is no purely group-theoretic proof? It seems to me that the geometry could get in easily enough if you observe that the symmetry group of a polyhedron must be a subgroup of some symmetry group of $\mathbb R^3$, I guess $SL^\pm(3,{\mathbb R})$ or something. There might be an argument to be made that there is no purely group-theoretic proof; if so I would be interested to see it. – MJD Apr 6 '12 at 4:18
Platonic solids are not "group-theoretical objects", whatever that may be, so at some point or another some geometry will have to come in. – Mariano Suárez-Alvarez Apr 6 '12 at 4:20

The proof I know doesn't use group theory. At each vertex, there must be the same number $\ge 3$ of the same regular polygon meeting, and the angles must add to less than $180^{\circ}$. The only possibilities are three, four, or five triangles, three squares, or three pentagons. Then observe that each of these generates only one solid by Euler's formula for the number of faces and regularity forcing where each face goes.

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You do need to prove somehow that each possibility provides only one solid, too. – Mariano Suárez-Alvarez Apr 6 '12 at 2:49
@MarianoSuárez-Alvarez: That was the intended point of the last sentence, but it wasn't explicit. I have added to it. – Ross Millikan Apr 6 '12 at 2:53
This is a good proof. It works even for non-Euclidean geometries. – Grumpy Parsnip Apr 6 '12 at 3:12

From Regular Polytopes by Coxeter, let a spherical graph have $N_0$ vertices, $N_1$ edges, and $N_2$ faces. Euler's formula reads $$N_0 - N_1 + N_2 = 2. \; \; \; \; \; (1.61)$$ Now, suppose our graph is regular, each face has $p$ sides, each vertex has $q$ surrounding faces. Then both $p,q \geq 3.$ Next, $$q N_0 = 2 N_1 = p N_2 . \; \; \; \; \; (1.71)$$ Put them together, $$\frac{1}{N_1} = \frac{1}{p} + \frac{1}{q} - \frac{1}{2}. \; \; \; \; \; (1.72)$$ As $N_1$ is positive, and $p,q \geq 3,$ the possible solutions to $$\frac{1}{p} + \frac{1}{q} > \frac{1}{2}$$ are $$\{p,q\} =\{3,3\}, \; \; \{3,4\}, \; \; \{4,3\}, \; \; \{3,5\}, \; \; \{5,3\}.$$
Note that the spherical dual graph to $\{p,q\}$ is $\{q,p\},$ while the tetrahedron $\{3,3\}$ is self dual.

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Here's a group theoretic proof that one can use plus some euclidean geometry.

Let $X$ be a regular platonic solid. Now I believe you can let a finite subgroup of $SO_3$ act on $X$ by acting on the faces, edges or vertices. Now to each face you can draw a line perpendicular to that face, and let $SO_3$ act on that line instead. We will call such a line a pole. Similarly one can let $SO_3$ act on the poles above an edge and a vertex. So now we have that $SO_3$ is acting on the set of poles associated to a face,edge or vertex. If $p$ is a pole above a vertex, let

$$|G_p| = r_p = \text{number of faces that meet at a vertex}.$$

If $p'$ is a pole above an edge, let

$$|G_{p'}| = r_{p'} = \text{number of faces that meet at an edge} = 2.$$

Finally if $p''$ is a pole above a face, let

$$|G_{p''}| = r_{p''} = \text{number of sides a face has} = n.$$

Now it is not hard to show (I can provide a proof of this) that

$$\sum_{\text{over all poles p,p' or p''}}(r_p - 1) = 2|G| - 2$$

where $G$ is the group of rotational symmetries of $X$. In fact the proof of the formula above is group theoretic: One looks at the order of group elements and orders of stabilisers.

Now at the same time we know that the number of poles is equal to $V + F + E$, where $V$ is the number of vertices, $F$ the number of faces and $E$ the number of edges. If you use this information and plug it into our formula above, we have that

$$kV + nF + 2E - (V + F + E) = 2|G| - 2.$$

At the same time, the Orbit - Stabiliser Theorem gives us that $kV = nF = 2E = |G|$. Hence

$$\begin{eqnarray} 3nF - V - F - E &=& 2|G| - 2 \\ \\ \implies \frac{3nF}{|G|} - \frac{1}{k} - \frac{1}{n} - \frac{1}{2} &=& 2 - \frac{2}{|G|}. \end{eqnarray}$$

However $3nF/|G| = 3$, so that upon simplifying we have

$$\begin{eqnarray} \frac{1}{2} + \frac{2}{|G|} &=& \frac{1}{k} + \frac{1}{n} \\ \implies \frac{1}{k} + \frac{1}{n} &>& 2. \end{eqnarray}$$

The task now is reduced to finding integers that satisfy that inequality above. Since $k,n \geq 3$, the only possible integer solutions are $k=3, n=3$ or $k=3,n=4$ or $k=3, n=5$ or $k=4,n=3$ or $k=5,n=3.$

In the first case for example, we have a regular polyhedron made out of an equilateral triangle, with 3 faces (made out of equilateral triangles) meeting at a vertex. The tetrahedron does satisfy these requirements, but as Mariano has noted above it remains to check that the tetrahedron is the only one that satisfies this. Similarly one has to check that the other 4 cases only give the octahedron, cube, dodecahedron and icosahedron. I leave this to you to check!

$\textbf{Edit:}$ Mariano has told me that the proof that there is a finite subgroup of $SO_3$ acting on $X$ is not trivial.

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The claim that there is a finite subgroup of $SO(3)$ acting is critical, and needs proof :) – Mariano Suárez-Alvarez Apr 6 '12 at 5:45
The classification of finite subgroups of $\textrm{SO}(3)$ is a good exercise, though. – Zhen Lin Jul 9 '12 at 17:20