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This is a question from Artin's algebra textbook.

The tetrahedral group of rotations has 1 element of order 1, 8 elements of order 3 (rotations of $120^°$ around a vertex), and 3 elements of order 2 (rotations of $180^o$ around the axis through the midpoints of opposite edges).

Is there a way to find the class equation without using brute force on $A_4$?

J. W. Tanner
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user35687
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3 Answers3

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It has 12 elements. Any force you use won't be brutish.

The conjugacy classes in $S_n$ are easy to describe. And you can prove that each conjugacy class in $S_n$ which consists of even permutations (i.e., which is contained in $A_n$) will either stay a conjugacy class in $A_n$ or split in half (see here). For $A_4$, there's only one conjugacy class (of size 8) that even has the potential to split in half. You can just check if that happens.

user555203
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    Your wording is confusing, because $S_n$ has even and odd permutations – J. W. Tanner Feb 11 '19 at 21:25
  • Thanks -- I guess the parentheses don't do what I wanted them to do. I mean the ones which consist of even permutations, i.e., the only ones which are actually in $A_n$. I'll make an edit. – user555203 Feb 11 '19 at 22:23
  • Yes, your phrase "which consists of even permutations" is restrictive, not parenthetical. The wording is better now. Thank you – J. W. Tanner Feb 11 '19 at 23:03
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Artin gives a nice explanation regarding the icosahedral group I, and one can take a similar approach to the tetrahedral group T. The explanation is geometrical, so it is helpful to actually hold in hand a tetrahedron with numbered faces.

The rotations about a pole passing through a vertex and the opposite face have 3 possible positions, so they form a subgroup of order 3. There are 4 vertices, 4 separate such poles, so we have 4 subgroups of order 3. Each subgroup has the identity and 2 other elements, so there are 8 elements of order 3.

In addition, there are 3 subgroups of order 2, each a rotation about a pole passing through two opposite edges. Each of these subgroups has the identity and one other element, so there are 3 elements of order 2.

So the "order" equation is 12 = 1+3+8. As for the case of I, we need now to ask if these numbers split further for the conjugacy class equation.

The subgroups of order 2 are all in one conjugacy class for the same reason as in the explanation Artin gives for those of I. That is, each subgroup of order 2 is the stabilizer of an edge (actually 2 edges). The group T operates transitively on the edges. Geometrically, conjugation by some element g that moves an edge acts as follows. It moves an edge to another, then an element of the subgroup that stabilizes that other edge acts on it, and then g-1 moves the edge back. So conjugation takes a subgroup that stabilizes two edges to any of the other subgroups that stabilize two edges.

If the subgroups of order 2 are conjugates, and they have only 1 non-identity element, then the nonidentity elements are all conjugate elements. So, the 3 elements of order 2 are in one conjugacy class.

The 4 subgroups of order 3 are all conjugate also, for the same reason as those of order 2. But here, the elements of order 3 cannot all be conjugate, because the order of a conjugacy class must divide the order of the group, and 8 does not divide 12. On the other hand, if we take an element in one subgroup, and conjugation brings it to all the subgroups of order 3, then the conjugacy class of that element must be at least 4. So the 8 elements of order 3 split into 2 conjugacy classes of size 4, and the class equation must be 12=1+3+4+4.

mgspan
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The class equation of $A_4$ is well known to be $12=1+4+4+3$, since the cycle structure is preserved by conjugation. In this case, the class of $3$-cycles of size $8$ splits.

The subgroup of the tetrahedral group you refer to is also well known to be $A_4$.