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The question is

Describe the elements of the group of proper symmetries of a regular tetrahedron as a subgroup of $S_4$. Show that -- besides the identity -- there are two sorts of rotations -- one fixing only one vertex and the other fixing no vertex.

I don't understand what the elements are referring to.

I've seen in different places that $S_4$ is supposed to be the symmetric group on $n$ elements, but my textbook says "The symmetry groups of the Platonic solids are quite interesting finite groups called the tetrahedral group ($A_4$) of proper symmetries of a tetrahedron, the octahedral group ($S_4$) of proper symmetries of an octahedron or cube"? Honestly, I might just be completely lost on this. The first question "Describe the elements of the group of proper symmetries of a regular tetrahedron as a subgroup of $S_4$" makes no sense to me.

Are the symmetries on the tetrahedron the elements of the group of proper symmetries? Is that how it works?

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    $\begingroup$ Welcome to Mathematics Stack Exchange. Think of permuting the four vertices of the tetrahedron $\endgroup$ Oct 26, 2020 at 0:36

2 Answers 2

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  • There are different ways of rotating a tetrahedron in such a way that it looks the same as it did before.
  • If you label the vertices of the tetrahedron, you can label the rotations by describing how you move each vertex from one position to another. For example, if the four vertices are A, B,C,D, you might pivot one face of the triangle, sending $B\rightarrow C\rightarrow D \rightarrow B$, and keeping vertex $A$ fixed in space. This is one sort of rotation.
  • Any permutation of the four vertices thereby corresponds to a potential rotation of the tetrahedron. Some of the symmetries are proper rotations, meaning they do not involve a mirror reflections. (such as the mirror reflection $B\leftrightarrow C$)
  • So the group of proper rotations $(A_4)$ of the tetrahedron can be described as the subgroup of all vertex permutations $(S_4)$ that don't induce a mirror reflection.
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  • $\begingroup$ So when they refer to the "group of proper symmetries of a regular tetrahedron as a subgroup of $S_4$" it's $A_4$? And are the elements of $A_4$ the two rotations that are being discussed in the question? $\endgroup$
    – daikons
    Oct 26, 2020 at 1:33
  • $\begingroup$ You can confirm by imagining rotating a tetrahedron that there are lots of proper rotations, like [ABCD -> ACDB], [ABCD->CABD], and [ABCD->DBDA]. (I think there are twelve total). An arbitrary vertex permutation can fix 0, 1, 2, 3, elts (or 4 elements if the identity). Your job is to show that proper rotations come from all the permutations that fix 0 elts or 1 elt; the rest of the permutations are all improper rotations because they induce a reflection; thus all the proper rotations fall into two categories. Yes, the proper rotations collectively form the alternating group $A_4$. $\endgroup$
    – user326210
    Oct 26, 2020 at 1:48
  • $\begingroup$ What exactly are elts? $\endgroup$
    – daikons
    Oct 26, 2020 at 2:01
  • $\begingroup$ elts/elements = The four vertices of the tetrahedron. $\endgroup$
    – user326210
    Oct 26, 2020 at 2:02
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I think what they refer to as "proper symmetry" is having a physical tetrahedron in real life and rotating it. Using this interpretation, we notice that any configuration can be specified by only $2$ of it's vertex, the other $2$ then only have one choice, giving us $4\cdot 3=12$ elements.

We know that this group is a subgroup of $S_4$, and the only subgroup of order $12$ is $A_4$.

To show that there are only $2$ types of rotation. For a non-trivial element, if it fixes $2$ vertices, then it's a odd permutation, not in $A_4$. If it fixes $3$, then it's fixes the last one, so it's the identity.

Obviously those both rotation of fixing one or no points are both achievable, so those are the only type.

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  • $\begingroup$ Wait I'm not following how its 12 elements? How does it become $4 \cdot 3$? $\endgroup$
    – daikons
    Oct 26, 2020 at 4:24

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