How to find the order of the group of rigid motions of platonic solids in $\mathbb{R}^3$? The following appear as exercises in Dummit and Foote's Algebra (Section $1.2$ - Dihedral Groups):


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*Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a tetrahedron. Show that $|G| = 12$

*Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a cube. Show that $|G| = 24$

*Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of an octahedron. Show that $|G| = 24$

*Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a dodecahedron. Show that $|G| = 60$

*Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of an icosahedron. Show that $|G| = 60$

From this answer I figured that rigid motions are orientation-preserving isometries, i.e. reflections are not allowed.
So, for a tetrahedron, I thought of symmetry axes passing through a vertex and the centroid of the opposite face. There are four such axes (let's call them $A,B,C,D$). Along every axis, we can define $1_i, r_i, r_i^2$ as three rotations with $r_i^3= 1$, the identity element ($i=A,B,C,D$). Since there are four such axes, $|G| = 3\times 4 = 12$. Is this fine, or am I missing something? I'm slightly concerned about the fact that $1_A,1_B,1_C,1_D$ may all possibly be the same (since they are identity transformations), and that I am over-counting?
Minor question (detour): Are identity transformations corresponding to different axes different, or the same?
For the cube, I did the following:

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*For every pair of opposite faces, we have an axis of symmetry. There are $3$ such pairs, hence $3$ such axes (say $A,B,C,D$). About each axis we define $1,r_i,r_i^2,r_i^3$ with $r_i^4 = 1$ where $i=A,B,C,D$.

*There are four body diagonals (say $E,F,G,H$), and about each diagonal (symmetry axis) we define $1,r_j,r_j^2$ with $r_j^3= 1$ where $j=E,F,G,H$.

In view of the above calculations, we have $|G| = 3\times 4 + 4\times 3 = 24$.
Using this method becomes difficult hereafter, for larger solids. It is not easy to identify all the axes of symmetry by hand. Moreover, the only group I have learned about in some detail at this point is $D_{2n}$, so please do not give solutions such as "the required group $G$ is isomorphic to a known and well-studied group $X$, and we know $|X| = ?$ so $|G| = ?$"
I think it boils down to having a good way to count all the distinct rigid motions. Could someone help me with this?
I came across James Ha's solutions here, but I don't understand how the solutions presented in the PDF are equivalent to mine even for the tetrahedron and cube cases. It'd be nice if someone can help me see the equivalence, and also tell me how to proceed with the other platonic solids! Thanks a lot!
 A: There is no substitute for cutting four equal equilateral triangles out of cardboard and taping them together to make a tetrahedron. Once you've done this, put a fingertip in the center of an edge and another fingertip at the center of the opposite edge. Then spin the tetrahedron about the axis joining your fingertips. You should find that a $180^\circ$ rotation brings the tetrahedron back to itself. In my experience, this is hard to visualize until you've done it physically.
There are three such pairs of opposite edges and hence three such $180^\circ$ rotations. These, together with the identity and eight rotations of $\pm120^\circ$ about various axes joining the centroid of a face to the opposite vertex account for all the rotational symmetries of the tetrahedron.
The other Platonic solids have similar $180^\circ$ rotations. But if you just want a count, you can do something simpler. Start with one face of the solid facing you with fixed orientation (say one edge horizontal). If it's an $m$-sided face, there are $m$ edges that can be the horizontal one, and these $m$ orientations can all be obtained from one another by rotating about the center of the face. Now if the solid has $f$ faces, any of the $f$ can be brought to the "facing you" position by a rotation. So there should be $mf$ rotational symmetries. This accounts for everything.
The answer of orangeskid is similar, but even simpler, than this one. Start with an edge facing you, oriented horizontally. Let the horizontal plane containing this edge be such that it bisects the dihedral angle between the two faces that meet along that edge. (In other words, from your perspective, these two faces, which slant away from you, will appear equal.) Now you can do the $180^\circ$ rotation discussed above, but you can also bring any other edge of the solid to the "facing you" position by a rotation. So there are $2e$ symmetries.
A: To add some elaboration to existing answers, and additional comments:
As orangeskid mentions, you can infer the size of the symmetry group from the number of transformations between two edges. Here is a way to see this more clearly:
Consider directed edges on the polyhedron, which consist of a vertex and an edge emanating from that vertex (or equivalently, an edge with one of its endpoints distinguished). If we have $e$ edges, then we have $2e$ of these directed edges. Because we're using Platonic solids, every one of these can be taken to any other (this follows pretty easily from most definitions of the Platonic solids, but should be pretty intuitive).
But once we know that one directed edge $(v_1,e_1)$ goes to another directed edge $(v_2,e_2)$, we have completely specified the rotation: once we move $v_1$ to $v_2$, we've constrained the possible rotations to a single axis around which things can turn (since we have a point that is now immobile), and only one of those ways to rotate it will move $e_1$ to $e_2$.
In particular, this means that a rotation is uniquely specified by where it takes a single directed edge; since each of the $2e$ possibilities gives a unique rotation, there must be $2e$ possible rotations total.
(If we permit orientation-reversing transformations, there are twice as many; for every way to take a directed edge to another, we get a second transformation fixing that directed edge by reflecting about it.)
As for the identity transformations fixing an axis, these are all the same identity transformation; they leave the shape unchanged.

To more clearly spell out the types of (orientation-preserving) rotations possible for each possible platonic solid:
For every platonic solid, the possible rotations are either a nontrivial rotation about a vertex, a $180^\circ$ rotation about an edge, a nontrivial rotation about a face, or the identity transformation.
For the tetrahedron, faces are opposite vertices, so there are $4\cdot (3-1)$ nontrivial vertex/face rotations, $1$ identity, and $3$ edge-flips ($6$ edges, but two used per flip), for a total of $12$.
For the cube, there are $8\cdot (3-1)/2$ vertex rotations, $6\cdot(4-1)/2$ face rotations, $12/2$ edge flips, and $1$ identity, for a total of $24$.
For the octahedron, there are $6\cdot(4-1)/2$ vertex rotations, $8\cdot (3-1)/2$ face rotations, $12/2$ edge flips, and $1$ identity, for a total of $24$.
For the dodecahedron, there are $20\cdot(3-1)/2$ vertex rotations, $12\cdot(5-1)/2$ face rotations, $30/2$ edge flips, and $1$ identity, for a total of $60$.
For the icosahedron, there are $12\cdot(5-1)/2$ vertex rotations, $20\cdot(3-1)/2$ face rotations, $30/2$ edge flips, and $1$ identity, for a total of $60$.
A: For polyhedra in $3$ space you can show that an edge $a$ can be taken to  anoother edge $b$ by  $2$   orientation-preserving transformation of the solid (get one, and then can also rotate around $b$) . If you consider all transformations, then there are $4$ such transformations.transformations.
Therefore, $|G_{+}(S)| = 2 e$, $|G(S)|= 4 e$, where $e$ is the number of edges of $S$.
