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From Paolo Aluffi's "Algebra: Chapter 0", question II.2.8:

Calculate the order of the symmetry groups for platonic solids.

I can easily look this up, and some tutorials give the actual groups outright, but I still don't understand how they get the answer. Specifically, I do not know how to avoid double-counting and also how to make sure I'm covering ALL the symmetries. I'm stuck, and this is what I've tried.

As an example, we could look at the tetrahedron. I know ahead of time that the answer is $24$, and before looking at the answer I can already give that as an upper-bound since it has four vertices and its potential vertex permutations are a subgroup of $S_4$.

"Okay, $3$ rotations on $4$ vertices is $12$ rotations total, times $2$ for reflection is $24$... but that doesn't seem right, some of those rotations are identity. Do I multiply the vertex rotations together? That gives me $3^4 = 81$, which is obviously wrong! Can vertex rotations provide the same symmetries as reflection? They can't do that in the Dihedral groups..." And so on. Plus, I don't get as nice an upper-bound for the rest of the platonic solids.

This is for self-study. My goal isn't the answer, but the method. I'm specifically looking for the order of the group, not the group itself. Please make sure the explanations use math that an undergraduate student would be comfortable with, because I think that's my level wrt Algebra.

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  • $\begingroup$ Do you know group actions? $\endgroup$
    – Clément Guérin
    Commented Apr 3, 2018 at 15:09
  • $\begingroup$ We haven't gotten to group actions in the book yet, but I kinda know them from the last time I tried learning group theory. It's the isomorphism of a group's elements and some automorphisms, right? $\endgroup$
    – Larry B.
    Commented Apr 3, 2018 at 15:32
  • $\begingroup$ Not exactly, one way to see this (this may be what you are saying but I wouldn't say it like this) a group $G$ acts on $X$ if you simply give a group morphism $\rho:G\to X$. Specifically one can say that $g\in G$ sends $x\in X$ to $g\cdot x$. If I bring that up, it is because one way to count the number of group elements of these isometry groups is to show that the group of isometry of a regular polyhedron acts freely transitively on the set of flags of the regular polyhedron. $\endgroup$
    – Clément Guérin
    Commented Apr 3, 2018 at 15:37
  • $\begingroup$ Anyway I will add an answer without this stuff (but for self-study it is a nice goal to read enough about group action to make sense of my last sentence). $\endgroup$
    – Clément Guérin
    Commented Apr 3, 2018 at 15:38
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    $\begingroup$ @ClémentGuérin That's not the right definition of a group action. Unless you meant $\rho: G \times X \to X$, or that $\rho$ is a group homomorpshim $G \to Aut(X)$ $\endgroup$
    – Exit path
    Commented Apr 6, 2018 at 15:47

2 Answers 2

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Let $\mathcal{P}$ be a polyhedron, $G$ be its isometry group and $G^+$ be the subgroup of direct isometries (determinant $1$).

  • First point the determinant map $\det: G\to \mathbb{C}^*$ has image either $\{1\}$ or $\{\pm 1\}$ and its kernel is $G^+$ by definition. As a result, the index of $G^+$ inside $G$ is either $1$ or $2$. Remark that it suffices to exhibit one element in $G$ not in $G^+$ to justify that the index $G^+$ inside $G$ is $2$.

  • In all the cases you are interested in, there is always a reflexion plane so that $G\neq G^+$ and therefore the index $G^+$ inside $G$ is $2$. In particular $|G|=2|G^+|$. So we have now reduced the problem of counting the isometries of $\mathcal{P}$ to counting the direct isometries.

  • What is nice about (non-trivial) direct isometries is that they are completely and uniquely determined by an axis of rotation and an angle of rotation around this axis.

  • Therefore to count the number of direct isometries it suffices to count the number of axes and what are the admissible angles around each of them.

  • if $g$ is a non-trivial element of $G^+$, I claim that the axis of $g$ must either go through the middle of a face, the middle of an edge or a vertex of the polyhedron.

  • Now we have everything we need to make the counting: you consider all possible spots to have an axe going through and count the number of non-trivial rotations around this axis.

For the tetrahedron, an axis going through a vertex and the center of gravity of the tetrahedron must go to the opposite face, you have $4$ such axes (basically because you have four vertices). Around each axe you have three edges so you can just rotate of angle $2\pi/3$ or $-2\pi/3$. All in it gives you $8$ elements. Then you consider the edges, you have $6$ of them and any axe going through the middle of an edge and the center of gravity of the tetrahedron must go through the opposite edge, so that you get $3$ axes and around it you can only turn it by an angle of $\pi$, so you get $3$ rotations from there. $3+8+1$ (don't forget the trivial one) gives you $12$ elements in $G^+$ and thus $24$ elements in $G$.

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    $\begingroup$ Let me see if I understand: ignoring $\det = -1$ (orientation-reversing transforms), all the symmetries are rotations, either about the vertices or edge midpoints or faces. Count the non-identity rotations, and then add one for identity. Multiply total rotations by two for orientation reversal. Is that right? $\endgroup$
    – Larry B.
    Commented Apr 3, 2018 at 17:00
  • $\begingroup$ @LarryB. Yes exactly. $\endgroup$
    – Clément Guérin
    Commented Apr 3, 2018 at 17:07
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Let $F$ be the count of faces. Those all are $N$-gonal. Then you have for the group order $G=2NF$. (Here "$2$" is the number of vertices per edge.)

Dually you could considered $V$ to be the vertex count, all of which have $S$ edges incident. Then you have likewise $G=2SV$. (Here "$2$" is the number of faces per edge.)

More general, consider any $D$-dimensional Platonic solid. Let $F_{D-1}$ be the facet count. Denote by $N_{k-1,k}$ the amount $k-1$-dimensional sub-elements at each $k$-dimensional one. Then you will have

$G=F_{D-1}\prod\limits_{1 \le k \lt D}N_{k-1,k}$

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  • $\begingroup$ That's a succinct formula, but why does it work? $\endgroup$
    – Larry B.
    Commented Apr 6, 2018 at 15:12
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    $\begingroup$ Platonic solids are the regular polyhedra (or more general: polytopes). Thus you have symmetry of the flags, i.e. incidence tuples of some face with some of its edges with one of its vertices (and similarily for the higher dimensional analogues). Therefore you ought map any face onto any other - yielding factor $F$. Then you can turn any face around its own symmetry center - yielding the further factor $N=N_{1,2}$. And finally you can reflect each edge (turn the edge within 1-dimensional space) - yielding the final factor $2=N_{0,1}$. $\endgroup$
    – Dr. Richard Klitzing
    Commented Apr 6, 2018 at 15:43

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