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I can see that why the order of the group of symmetries of a regular tetrahedron is $12$ : Roughly speaking, each time one of ${\{1,2,3,4}\}$ is on 'top' and we do to the other $3$ as we did in a regular triangle. But I need to know the elements of the group and I don't have not only a physical regular tetrahedron but also my imagination is not strong. Videos like $1$ or $2$ didn't help, because I can't understand for example what happens to the other $2$ points when the top point and one of the surface points is exchanged.

I would appreciate any simple clear explanation.

Added - For next permutations (with order $3$) now simply we consider the ‘surface’ under the other three-points, i.e. each time $4$ on top is replaced by $1$ or $2$ or $3$. And, we perform rotations as we were doing when $4$ was on top. So we will get:

‘$4$’ on top : ${\{(123),(132)}\}$

‘$1$’ on top : ${\{(234),(243)}\}$

‘$2$’ on top : ${\{(134),(143)}\}$

‘$3$’ on top : ${\{(124),(142)}\}$.

But what about $(12)(34)$, $(13)(24)$ and $(14)(23)$? How to get them?

Jyrki Lahtonen
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2 Answers2

12

One of them

enter image description here

Use different axes (connecting the midpoints of two opposite edges of the tetrahedron) to get the other products of two disjoint 2-cycles as 180 degree rotations.

Here is the same animation with the cube surrounding the tetrahedron shown as a wireframe. The axis of rotation joins the centers of two opposite faces of the cube

enter image description here

Jyrki Lahtonen
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  • Sorry I didn't get neither the picture nor the explanation. For example consider (12)(34). In order to change change 1 and 2 with each other, there must be a 180-degree rotation in some axis that pass through the midpoint of 1 and 2 as well. Why this axis of rotation is the axis of rotation which interchange 3 and 4? –  Mar 19 '16 at 19:18
  • The axis of rotation connects the midpoints of 12 and 34. – Jyrki Lahtonen Mar 19 '16 at 19:22
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    Another way of thinking about this is to view it as a rotation of the cube surrounding the tetrahedron (the 4 vertices of the tetrahedron are also vertices of the cube) – Jyrki Lahtonen Mar 19 '16 at 19:25
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    It's brilliant! Thank you. Only thing that I struggle with is that "being a rigid shape is sufficient that after exchange of 1 and 2, and, 3 and 4 by one rigid axis the final configuration coincides with the original one"? –  Mar 19 '16 at 19:31
  • That's a good question. It should become clear that the said rotation is a symmetry, when I add the surrounding cube to the picture. Gimme 5-10 minutes. – Jyrki Lahtonen Mar 19 '16 at 19:33
  • in order to interchange 1 and 2, or, 3 and 4, the mentioned axis must be perpendicular to each of the lines joining 1 and 2, or, 3 and 4. I think that's enough. But I can't prove the perpendicularity. –  Mar 19 '16 at 19:47
  • @Liebe Those two lines are diagonals of two opposite faces of the surrounding cube. The axis is perpendicular to both of them, because the axis is perpendicular to both those faces. – Jyrki Lahtonen Mar 19 '16 at 19:49
  • Simply explained and perfect answer. Thank you very much indeed. :) –  Mar 19 '16 at 19:51
  • @ Jyrki Lahtonen nice explanations and animation. Which software do you use ? – Jean Marie Mar 19 '16 at 21:21
  • @JeanMarie Mathematica. I only know a tiny fraction of its features, but I learned enough to do these animations when I was teaching courses on introductory abstract algebra and group theory. – Jyrki Lahtonen Mar 19 '16 at 21:44
  • @Jyrki Lahtonen nice explanation sir, thank you, sir, one question , is rigid motion of 3d diagram are equivalent to only symmetries of that diagram? – A learner Mar 01 '21 at 14:30
  • That mean rigid motion of $2$d diagram n-gon is clearly dihedral group? – A learner Mar 01 '21 at 14:33
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    That depends on the definitions @Alearner. To reflect a regular $n$-gon you either need to use a third dimension to bring the back to the front, or break it into "atoms" (more like invididual points). A reflection (with respect to a codimension one hyperplane) reverses the handedness, be it 2D or 3D. I don't know if reflections are counted among the set of rigid motions? – Jyrki Lahtonen Mar 01 '21 at 15:37
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    A quick sampling of search results suggests that many people (targeting high schoolers and such) include reflections among rigid motions @Alearner. In that case you do get $D_n$ as the rigid motions preserving a regular $n$-gon. In this thread we only consider handedness preserving rigid motions. That is, rotations. – Jyrki Lahtonen Mar 02 '21 at 09:55
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    However, I am inclined to trust Wikipedia that specifies rigid motions as handedness and distance preserving affine transformations. Caveat: I don't know how pre-college teachers use these concepts. Better to check. – Jyrki Lahtonen Mar 02 '21 at 10:08
  • @Jyrki Lahtonen thank you,sir, for the reply. – A learner Mar 03 '21 at 03:39
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Consider the set of half edges of the tetrahedron. It is easy to convince yourself that the symmetry group of the solid allows you to move from one such half edge to any other in exactly one way. Since there are six edges, there are twelve half edges, and this implies that the group of symmetries has exactly 12 elements.

Now the group obviously permutes the four vertices. Since there is exactly one subgroup of order 12 in $S_4$, namely the alternating group, it must be our group.

Mariano Suárez-Álvarez
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    The argument to count the elements of the group works for all regular solids. For example, as a dodecahedron has 30 edges, it has 60 half edges and, therefore, its group of symmetries has order 60. – Mariano Suárez-Álvarez Mar 19 '16 at 19:30