Geometrically determining the symmetry of a tetrahedron (using solid geometry, not group theory)

Question

It is well established that a regular tetrahedron has 12 orientation preserving symmetries, the group $A_4$.

To better understand these symmetries, I set out to identify them geometrically: Prove that the rigid (orientation preserving) motions which send a regular tetrahedron to itself are precisely

1. The identity
2. Rotations of $\pm 120^\circ$ around a line through the tetrahedron's center and a vertex
3. Rotations of $180^\circ$ around a line through the tetrahedron's center and the midpoint of an edge.

My proof is below, to which I request verification and feedback, in particular the portion between the "[*]" symbols, which I'd like verification of or suggestions to make it (or its exposition) simpler or more clear.

Note: This question does not ask for lists of these symmetries, visualizations, or linear algebra or coordinate based derivations, but rather a synthetic, geometric proof of exactly which motions send the tetrahedron to itself.

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Proof: Let us call a such a motion an admissible motion. Since a regular tetrahedron has a unique center $C$, any admissible motion must keep $C$ as a fixed point. In $\mathbb R^3$, the motions with a fixed point are precisely the rotations around a line $\ell$ including this point; hence, an admissible motion must be a rotation around a line $\ell$ passing through $C$. Furthermore, since $\ell$ penetrates the tetrahedron, it must pass through exactly two fixed points $S$ and $S'$on the surface of the tetrahedron. Such a rotation will be admissible if and only if it sends faces to faces.

Recall that a rotation around $\ell$ sends a plane [added: not containing $\ell$] to itself if and only if the plane is normal to $\ell$. Similarly, if the rotation sends a line [added: other than $\ell$] to itself, this line must be normal to $\ell$.

Suppose $S$ is a vertex. Then exactly one face $F$ is normal to $\ell$. Since $F$ will remain in its plane, an admissible rotation must send $F$ to itself. This happens if and only if the angle of rotation $\theta$ is $\pm 120^\circ$. Such a rotation sends the remaining faces to each other cyclically and is therefore admissible.

Now, suppose instead that fixed point $S$ is on an edge $E$ but is not a vertex. Then $S$ is on only one edge, so an admissible rotation must send this edge to itself. This happens precisely when $\theta = 180^\circ$, $\ell$ is normal to $E$, and $S$ is the midpoint of $E$. By symmetry, $S'$ will be the midpoint of the unique edge $E'$ non-adjacent to $E$.

We now show that any such rotation is admissible. [*] Given any two non-adjacent edges, the four faces of the tetrahedron are precisely the triangles with a side as one edge and their remaining (opposite) vertex on an endpoint of the other edge. Since rotation about $\ell$ sends $E$ and $E'$ to themselves, it swaps pairs of these faces [*], which shows that it sends faces to faces and is admissible.

Finally, fixed point $S$ might be on a face $F$ but not an edge. Since $S$ is on only one face, an admissible rotation must send $F$ to itself, requiring $\ell$ to be normal to $F$, which can only happen if $\ell$ goes through a vertex, which reduces to the first case above. This completes the proof.

Answer 1

Nobody was touching this question, so i will say some words on the problem, there was a bounty, so there is a special interest in having a parallel opinion and feed-back. (This too long for a comment, became an answer...)

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One may use the same plan, but structured to be "more convincing". Since i have to argue geometrically, we observe first that geometrically we have:

• four vertices (or $0$-simplices),
• six edges (or $1$-simplices),
• four faces (or $2$-simplices), (and one tetrahedral part, the one $3$-simplex,) and each (admissible, orientatin-preserving) symmetry $\sigma$ has to preserve the geometry in each dimension, the incidence relations, and the duality vertex-(opposite )face, respectively edge-(opposite )edge.

OP considers the cases by starting with the observation, that $\sigma$ must be a rotation, take its fixed line, intersect it with the tetrahedron boundary (only the faces), get two points, and now consider cases depending on the position of these points. Yes, one can do so.

But i would go an other way, in order to write less. We separate the cases by asking if $\sigma$ acting on the vertices - and by duality on the faces - has a fixed vertex $S=S_\sigma$ - and a fixed face $F_\sigma$, seen as a set. If this is the case, then $\sigma$ induces a symmetry of the equilateral triangle $F_\sigma$, so it is a $k\cdot 120^\circ$-rotation of it, $k=0,1,2$, and the case is finished.

In the else-case, no vertex is fixed. (The point not is to eliminate by a simple argument the permutation of the vertices which is a $4$-cycle, and does not respect the orientation. I would take the OP-points $S,S'$, each not a vertex, but why are they each on an edge with a quick argument?)

OP concludes (with no direct argument) that one of the two points $S,S'$ is on an edge, and not a vertex. Say $S$ is this point, and it is on the edge $AB$. Let $A',B'$ be the other two vertices. The rotation axis $SS'$ must pass through the center $C$, and $SC$ intersects the edge $A'B'$ in its mid point, so $S'$ is this mid point of $A'B'$. Now i would argue as follows. Let $M_{XY}$ be the mid point of an edge $XY$. So $S=M_{AB}$, $S'=M_{A'B'}$.

So we must have a rotation around $(l)=SS'$ that preserves the four mid points in the plane through $C$ perpendicular on $(l)$. Only $90^\circ$-multiples are thus allowed. But a $\pm 90^\circ$-rotation moves $A$ not in a vertex, so we have to take the $180^\circ$-rotation, which brings $A\leftrightarrow B$, and $A'\leftrightarrow B'$. With this argument (replacing the paragraph between the stars) there is nothing to object. We can also look how faces may move, but this is an argument depending on the visualization, to have it rigorous, one has to consider vertices or edges...

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We have finished, but maybe one more word should be said about the "else-case", e.g. why is $S$ an edge mid point? Well, with the above argumentation we have found a total of $1+4\cdot 2+3$ admissible symmetries / rotations.

• The $1$ stays for the identity,
• the $4\times 2$ stay for picking one vertex among four, then using the one or the opposite $120^\circ$-rotation that does the same with the opposite face, an equilateral triangle,
• the three $180^\circ$-rotations around each of the lines through the center and two opposite edge mid points.

There are totally $24$ possible permutations of the tetrahedron, (and we can further geometrically list them all by composing with the symmetry that switches two fixed vertices, letting the other two where they are), so we have covered all chances.